c++ - 未定義の参照/未解決の外部シンボルエラーとは何ですか？それを修正するにはどうすればよいですか？

Question

未定義の参照/未解決の外部シンボルエラーとは何ですか？一般的な原因とその修正/防止方法は何ですか？

Answer 1

C ++プログラムのコンパイルは、2.2で指定されているように、いくつかのステップで行われます （参照用のKeith Thompsonのクレジット）：

翻訳の構文規則の優先順位は、次のフェーズで指定されます[脚注を参照]。

1. 物理ソースファイルの文字は、必要に応じて、実装で定義された方法で、基本的なソース文字セット（行末インジケーターに改行文字を導入）にマップされます。[をちょきちょきと切る]
2. バックスラッシュ文字（\）の直後に改行文字が続く各インスタンスが削除され、物理ソース行を接続して論理ソース行を形成します。[をちょきちょきと切る]
3. ソースファイルは、前処理トークン（2.5）と空白文字のシーケンス（コメントを含む）に分解されます。[をちょきちょきと切る]
4. 前処理ディレクティブが実行され、マクロ呼び出しが展開され、_Pragma単項演算子式が実行されます。[をちょきちょきと切る]
5. 文字リテラルまたは文字列リテラルの各ソース文字セットメンバー、および文字リテラルまたは非生文字列リテラルの各エスケープシーケンスとユニバーサル文字名は、実行文字セットの対応するメンバーに変換されます。[をちょきちょきと切る]
6. 隣接する文字列リテラルトークンは連結されます。
7. トークンを区切る空白文字は重要ではなくなりました。各前処理トークンはトークンに変換されます。（2.7）。結果のトークンは、構文的および意味的に分析され、変換単位として変換されます。[をちょきちょきと切る]
8. 翻訳された翻訳単位とインスタンス化単位は、次のように組み合わされます。[SNIP]
9. すべての外部エンティティ参照が解決されます。ライブラリコンポーネントは、現在の翻訳で定義されていないエンティティへの外部参照を満たすためにリンクされています。このようなトランスレータの出力はすべて、実行環境での実行に必要な情報を含むプログラムイメージに収集されます。（強調鉱山）

[脚注]実装は、実際には異なるフェーズが一緒に折りたたまれる可能性がありますが、これらの別々のフェーズが発生するかのように動作する必要があります。

指定されたエラーは、コンパイルのこの最後の段階で発生します。これは、最も一般的にはリンクと呼ばれます。これは基本的に、一連の実装ファイルをオブジェクトファイルまたはライブラリにコンパイルし、それらを連携させたいことを意味します。

でシンボルを定義aしたとしa.cppます。さて、そのシンボルをb.cpp 宣言して使用しました。リンクする前に、そのシンボルがどこかで定義されていると単純に想定しますが、それはまだどこであるかを気にしません。リンクフェーズでは、シンボルを見つけて正しくリンクしb.cppます（実際には、シンボルを使用するオブジェクトまたはライブラリにリンクします）。

Microsoft Visual Studioを使用している場合は、プロジェクトがファイルを生成することがわかり.libます。これらには、エクスポートされたシンボルのテーブルとインポートされたシンボルのテーブルが含まれています。インポートされたシンボルは、リンク先のライブラリに対して解決され、エクスポートされたシンボルは、それを使用するライブラリ.lib（存在する場合）に提供されます。

他のコンパイラ/プラットフォームにも同様のメカニズムが存在します。

一般的なエラーメッセージはerror LNK2001、Microsoft Visual Studioのerror LNK1120場合error LNK2019は、、 GCCの場合はsymbolNameです。undefined reference to

コード：

struct X
{
   virtual void foo();
};
struct Y : X
{
   void foo() {}
};
struct A
{
   virtual ~A() = 0;
};
struct B: A
{
   virtual ~B(){}
};
extern int x;
void foo();
int main()
{
   x = 0;
   foo();
   Y y;
   B b;
}

GCCで次のエラーが生成されます：

/home/AbiSfw/ccvvuHoX.o: In function `main':
prog.cpp:(.text+0x10): undefined reference to `x'
prog.cpp:(.text+0x19): undefined reference to `foo()'
prog.cpp:(.text+0x2d): undefined reference to `A::~A()'
/home/AbiSfw/ccvvuHoX.o: In function `B::~B()':
prog.cpp:(.text._ZN1BD1Ev[B::~B()]+0xb): undefined reference to `A::~A()'
/home/AbiSfw/ccvvuHoX.o: In function `B::~B()':
prog.cpp:(.text._ZN1BD0Ev[B::~B()]+0x12): undefined reference to `A::~A()'
/home/AbiSfw/ccvvuHoX.o:(.rodata._ZTI1Y[typeinfo for Y]+0x8): undefined reference to `typeinfo for X'
/home/AbiSfw/ccvvuHoX.o:(.rodata._ZTI1B[typeinfo for B]+0x8): undefined reference to `typeinfo for A'
collect2: ld returned 1 exit status

およびMicrosoftVisualStudioでの同様のエラー：

1>test2.obj : error LNK2001: unresolved external symbol "void __cdecl foo(void)" (?foo@@YAXXZ)
1>test2.obj : error LNK2001: unresolved external symbol "int x" (?x@@3HA)
1>test2.obj : error LNK2001: unresolved external symbol "public: virtual __thiscall A::~A(void)" (??1A@@UAE@XZ)
1>test2.obj : error LNK2001: unresolved external symbol "public: virtual void __thiscall X::foo(void)" (?foo@X@@UAEXXZ)
1>...\test2.exe : fatal error LNK1120: 4 unresolved externals

一般的な原因は次のとおりです。

• 適切なライブラリ/オブジェクトファイルへのリンクまたは実装ファイルのコンパイルの失敗
• 宣言された未定義の変数または関数。
• クラスタイプのメンバーに関する一般的な問題
• テンプレートの実装は表示されません。
• シンボルはCプログラムで定義され、C++コードで使用されました。
• モジュール/dll間でメソッド/クラスを誤ってインポート/エクスポートします。（MSVS固有）
• 循環ライブラリの依存関係
• `WinMain @16'への未定義の参照
• 相互依存するライブラリの順序
• 同じ名前の複数のソースファイル
• #pragma（Microsoft Visual Studio）を使用するときに.lib拡張子を入力ミスまたは含めない
• テンプレートの友達の問題
• 一貫性のないUNICODE定義
• const変数の宣言/定義に「extern」がありません（C ++のみ）

Answer 2

Class members:

A pure virtual destructor needs an implementation.

Declaring a destructor pure still requires you to define it (unlike a regular function):

struct X
{
    virtual ~X() = 0;
};
struct Y : X
{
    ~Y() {}
};
int main()
{
    Y y;
}
//X::~X(){} //uncomment this line for successful definition

This happens because base class destructors are called when the object is destroyed implicitly, so a definition is required.

virtual methods must either be implemented or defined as pure.

This is similar to non-virtual methods with no definition, with the added reasoning that the pure declaration generates a dummy vtable and you might get the linker error without using the function:

struct X
{
    virtual void foo();
};
struct Y : X
{
   void foo() {}
};
int main()
{
   Y y; //linker error although there was no call to X::foo
}

For this to work, declare X::foo() as pure:

struct X
{
    virtual void foo() = 0;
};

Non-virtual class members

Some members need to be defined even if not used explicitly:

struct A
{ 
    ~A();
};

The following would yield the error:

A a;      //destructor undefined

The implementation can be inline, in the class definition itself:

struct A
{ 
    ~A() {}
};

or outside:

A::~A() {}

If the implementation is outside the class definition, but in a header, the methods have to be marked as inline to prevent a multiple definition.

All used member methods need to be defined if used.

A common mistake is forgetting to qualify the name:

struct A
{
   void foo();
};

void foo() {}

int main()
{
   A a;
   a.foo();
}

The definition should be

void A::foo() {}

static data members must be defined outside the class in a single translation unit:

struct X
{
    static int x;
};
int main()
{
    int x = X::x;
}
//int X::x; //uncomment this line to define X::x

An initializer can be provided for a static const data member of integral or enumeration type within the class definition; however, odr-use of this member will still require a namespace scope definition as described above. C++11 allows initialization inside the class for all static const data members.

Answer 3

Answer 4

Declared but did not define a variable or function.

A typical variable declaration is

extern int x;

As this is only a declaration, a single definition is needed. A corresponding definition would be:

int x;

For example, the following would generate an error:

extern int x;
int main()
{
    x = 0;
}
//int x; // uncomment this line for successful definition

Similar remarks apply to functions. Declaring a function without defining it leads to the error:

void foo(); // declaration only
int main()
{
   foo();
}
//void foo() {} //uncomment this line for successful definition

Be careful that the function you implement exactly matches the one you declared. For example, you may have mismatched cv-qualifiers:

void foo(int& x);
int main()
{
   int x;
   foo(x);
}
void foo(const int& x) {} //different function, doesn't provide a definition
                          //for void foo(int& x)
                          

Other examples of mismatches include

• Function/variable declared in one namespace, defined in another.
• Function/variable declared as class member, defined as global (or vice versa).
• Function return type, parameter number and types, and calling convention do not all exactly agree.

The error message from the compiler will often give you the full declaration of the variable or function that was declared but never defined. Compare it closely to the definition you provided. Make sure every detail matches.

Answer 5

The order in which interdependent linked libraries are specified is wrong.

The order in which libraries are linked DOES matter if the libraries depend on each other. In general, if library A depends on library B, then libA MUST appear before libB in the linker flags.

For example:

// B.h
#ifndef B_H
#define B_H

struct B {
    B(int);
    int x;
};

#endif

// B.cpp
#include "B.h"
B::B(int xx) : x(xx) {}

// A.h
#include "B.h"

struct A {
    A(int x);
    B b;
};

// A.cpp
#include "A.h"

A::A(int x) : b(x) {}

// main.cpp
#include "A.h"

int main() {
    A a(5);
    return 0;
};

Create the libraries:

$ g++ -c A.cpp
$ g++ -c B.cpp
$ ar rvs libA.a A.o 
ar: creating libA.a
a - A.o
$ ar rvs libB.a B.o 
ar: creating libB.a
a - B.o

Compile:

$ g++ main.cpp -L. -lB -lA
./libA.a(A.o): In function `A::A(int)':
A.cpp:(.text+0x1c): undefined reference to `B::B(int)'
collect2: error: ld returned 1 exit status
$ g++ main.cpp -L. -lA -lB
$ ./a.out

So to repeat again, the order DOES matter!

Answer 6

what is an "undefined reference/unresolved external symbol"

I'll try to explain what is an "undefined reference/unresolved external symbol".

note: i use g++ and Linux and all examples is for it

For example we have some code

// src1.cpp
void print();

static int local_var_name; // 'static' makes variable not visible for other modules
int global_var_name = 123;

int main()
{
    print();
    return 0;
}

and

// src2.cpp
extern "C" int printf (const char*, ...);

extern int global_var_name;
//extern int local_var_name;

void print ()
{
    // printf("%d%d\n", global_var_name, local_var_name);
    printf("%d\n", global_var_name);
}

Make object files

$ g++ -c src1.cpp -o src1.o
$ g++ -c src2.cpp -o src2.o

After the assembler phase we have an object file, which contains any symbols to export. Look at the symbols

$ readelf --symbols src1.o
  Num:    Value          Size Type    Bind   Vis      Ndx Name
     5: 0000000000000000     4 OBJECT  LOCAL  DEFAULT    4 _ZL14local_var_name # [1]
     9: 0000000000000000     4 OBJECT  GLOBAL DEFAULT    3 global_var_name     # [2]

I've rejected some lines from output, because they do not matter

So, we see follow symbols to export.

[1] - this is our static (local) variable (important - Bind has a type "LOCAL")
[2] - this is our global variable

src2.cpp exports nothing and we have seen no its symbols

Link our object files

$ g++ src1.o src2.o -o prog

and run it

$ ./prog
123

Linker sees exported symbols and links it. Now we try to uncomment lines in src2.cpp like here

// src2.cpp
extern "C" int printf (const char*, ...);

extern int global_var_name;
extern int local_var_name;

void print ()
{
    printf("%d%d\n", global_var_name, local_var_name);
}

and rebuild an object file

$ g++ -c src2.cpp -o src2.o

OK (no errors), because we only build object file, linking is not done yet. Try to link

$ g++ src1.o src2.o -o prog
src2.o: In function `print()':
src2.cpp:(.text+0x6): undefined reference to `local_var_name'
collect2: error: ld returned 1 exit status

It has happened because our local_var_name is static, i.e. it is not visible for other modules. Now more deeply. Get the translation phase output

$ g++ -S src1.cpp -o src1.s

// src1.s
look src1.s

    .file   "src1.cpp"
    .local  _ZL14local_var_name
    .comm   _ZL14local_var_name,4,4
    .globl  global_var_name
    .data
    .align 4
    .type   global_var_name, @object
    .size   global_var_name, 4
global_var_name:
    .long   123
    .text
    .globl  main
    .type   main, @function
main:
; assembler code, not interesting for us
.LFE0:
    .size   main, .-main
    .ident  "GCC: (Ubuntu 4.8.2-19ubuntu1) 4.8.2"
    .section    .note.GNU-stack,"",@progbits

So, we've seen there is no label for local_var_name, that's why linker hasn't found it. But we are hackers :) and we can fix it. Open src1.s in your text editor and change

.local  _ZL14local_var_name
.comm   _ZL14local_var_name,4,4

to

    .globl  local_var_name
    .data
    .align 4
    .type   local_var_name, @object
    .size   local_var_name, 4
local_var_name:
    .long   456789

i.e. you should have like below

    .file   "src1.cpp"
    .globl  local_var_name
    .data
    .align 4
    .type   local_var_name, @object
    .size   local_var_name, 4
local_var_name:
    .long   456789
    .globl  global_var_name
    .align 4
    .type   global_var_name, @object
    .size   global_var_name, 4
global_var_name:
    .long   123
    .text
    .globl  main
    .type   main, @function
main:
; ...

we have changed the visibility of local_var_name and set its value to 456789. Try to build an object file from it

$ g++ -c src1.s -o src2.o

ok, see readelf output (symbols)

$ readelf --symbols src1.o
8: 0000000000000000     4 OBJECT  GLOBAL DEFAULT    3 local_var_name

now local_var_name has Bind GLOBAL (was LOCAL)

link

$ g++ src1.o src2.o -o prog

and run it

$ ./prog 
123456789

ok, we hack it :)

So, as a result - an "undefined reference/unresolved external symbol error" happens when the linker cannot find global symbols in the object files.

Answer 7

Symbols were defined in a C program and used in C++ code.

The function (or variable) void foo() was defined in a C program and you attempt to use it in a C++ program:

void foo();
int main()
{
    foo();
}

The C++ linker expects names to be mangled, so you have to declare the function as:

extern "C" void foo();
int main()
{
    foo();
}

Equivalently, instead of being defined in a C program, the function (or variable) void foo() was defined in C++ but with C linkage:

extern "C" void foo();

and you attempt to use it in a C++ program with C++ linkage.

If an entire library is included in a header file (and was compiled as C code); the include will need to be as follows;

extern "C" {
    #include "cheader.h"
}

Answer 8

If all else fails, recompile.

I was recently able to get rid of an unresolved external error in Visual Studio 2012 just by recompiling the offending file. When I re-built, the error went away.

This usually happens when two (or more) libraries have a cyclic dependency. Library A attempts to use symbols in B.lib and library B attempts to use symbols from A.lib. Neither exist to start off with. When you attempt to compile A, the link step will fail because it can't find B.lib. A.lib will be generated, but no dll. You then compile B, which will succeed and generate B.lib. Re-compiling A will now work because B.lib is now found.

Answer 9

Template implementations not visible.

Unspecialized templates must have their definitions visible to all translation units that use them. That means you can't separate the definition of a template to an implementation file. If you must separate the implementation, the usual workaround is to have an impl file which you include at the end of the header that declares the template. A common situation is:

template<class T>
struct X
{
    void foo();
};

int main()
{
    X<int> x;
    x.foo();
}

//differentImplementationFile.cpp
template<class T>
void X<T>::foo()
{
}

To fix this, you must move the definition of X::foo to the header file or some place visible to the translation unit that uses it.

Specialized templates can be implemented in an implementation file and the implementation doesn't have to be visible, but the specialization must be previously declared.

For further explanation and another possible solution (explicit instantiation) see this question and answer.

Answer 10

This is one of most confusing error messages that every VC++ programmers have seen time and time again. Let’s make things clarity first.

A. What is symbol? In short, a symbol is a name. It can be a variable name, a function name, a class name, a typedef name, or anything except those names and signs that belong to C++ language. It is user defined or introduced by a dependency library (another user-defined).

B. What is external? In VC++, every source file (.cpp,.c,etc.) is considered as a translation unit, the compiler compiles one unit at a time, and generate one object file(.obj) for the current translation unit. (Note that every header file that this source file included will be preprocessed and will be considered as part of this translation unit)Everything within a translation unit is considered as internal, everything else is considered as external. In C++, you may reference an external symbol by using keywords like extern, __declspec (dllimport) and so on.

C. What is “resolve”? Resolve is a linking-time term. In linking-time, linker attempts to find the external definition for every symbol in object files that cannot find its definition internally. The scope of this searching process including:

• All object files that generated in compiling time
• All libraries (.lib) that are either explicitly or implicitly specified as additional dependencies of this building application.

This searching process is called resolve.

D. Finally, why Unresolved External Symbol? If the linker cannot find the external definition for a symbol that has no definition internally, it reports an Unresolved External Symbol error.

E. Possible causes of LNK2019: Unresolved External Symbol error. We already know that this error is due to the linker failed to find the definition of external symbols, the possible causes can be sorted as:

1. Definition exists

For example, if we have a function called foo defined in a.cpp:

int foo()
{
    return 0;
}

In b.cpp we want to call function foo, so we add

void foo();

to declare function foo(), and call it in another function body, say bar():

void bar()
{
    foo();
}

Now when you build this code you will get a LNK2019 error complaining that foo is an unresolved symbol. In this case, we know that foo() has its definition in a.cpp, but different from the one we are calling(different return value). This is the case that definition exists.

2. Definition does not exist

If we want to call some functions in a library, but the import library is not added into the additional dependency list (set from: Project | Properties | Configuration Properties | Linker | Input | Additional Dependency) of your project setting. Now the linker will report a LNK2019 since the definition does not exist in current searching scope.

Answer 11

Incorrectly importing/exporting methods/classes across modules/dll (compiler specific).

MSVS requires you to specify which symbols to export and import using __declspec(dllexport) and __declspec(dllimport).

This dual functionality is usually obtained through the use of a macro:

#ifdef THIS_MODULE
#define DLLIMPEXP __declspec(dllexport)
#else
#define DLLIMPEXP __declspec(dllimport)
#endif

The macro THIS_MODULE would only be defined in the module that exports the function. That way, the declaration:

DLLIMPEXP void foo();

expands to

__declspec(dllexport) void foo();

and tells the compiler to export the function, as the current module contains its definition. When including the declaration in a different module, it would expand to

__declspec(dllimport) void foo();

and tells the compiler that the definition is in one of the libraries you linked against (also see 1)).

You can similary import/export classes:

class DLLIMPEXP X
{
};

Answer 12

undefined reference to WinMain@16 or similar 'unusual' main() entry point reference (especially for visual-studio).

You may have missed to choose the right project type with your actual IDE. The IDE may want to bind e.g. Windows Application projects to such entry point function (as specified in the missing reference above), instead of the commonly used int main(int argc, char** argv); signature.

If your IDE supports Plain Console Projects you might want to choose this project type, instead of a windows application project.

---

Here are case1 and case2 handled in more detail from a real world problem.

Answer 13

Also if you're using 3rd party libraries make sure you have the correct 32/64 bit binaries

Answer 14

Microsoft offers a #pragma to reference the correct library at link time;

#pragma comment(lib, "libname.lib")

In addition to the library path including the directory of the library, this should be the full name of the library.

Answer 15

Visual Studio NuGet package needs to be updated for new toolset version

I just had this problem trying to link libpng with Visual Studio 2013. The problem is that the package file only had libraries for Visual Studio 2010 and 2012.

The correct solution is to hope the developer releases an updated package and then upgrade, but it worked for me by hacking in an extra setting for VS2013, pointing at the VS2012 library files.

I edited the package (in the packages folder inside the solution's directory) by finding packagename\build\native\packagename.targets and inside that file, copying all the v110 sections. I changed the v110 to v120 in the condition fields only being very careful to leave the filename paths all as v110. This simply allowed Visual Studio 2013 to link to the libraries for 2012, and in this case, it worked.

Answer 16

Suppose you have a big project written in c++ which has a thousand of .cpp files and a thousand of .h files.And let's says the project also depends on ten static libraries. Let's says we are on Windows and we build our project in Visual Studio 20xx. When you press Ctrl + F7 Visual Studio to start compiling the whole solution ( suppose we have just one project in the solution )

What's the meaning of compilation ?

• Visual Studio search into file .vcxproj and start compiling each file which has the extension .cpp. Order of compilation is undefined.So you must not assume that the file main.cpp is compiled first
• If .cpp files depends on additional .h files in order to find symbols that may or may not be defined in the file .cpp
• If exists one .cpp file in which the compiler could not find one symbol, a compiler time error raises the message Symbol x could not be found
• For each file with extension .cpp is generated an object file .o and also Visual Studio writes the output in a file named ProjectName.Cpp.Clean.txt which contains all object files that must be processed by the linker.

The Second step of compilation is done by Linker.Linker should merge all the object file and build finally the output ( which may be an executable or a library)

Steps In Linking a project

• Parse all the object files and find the definition which was only declared in headers ( eg: The code of one method of a class as is mentioned in previous answers, or event the initialization of a static variable which is member inside a class)
• If one symbol could not be found in object files he also is searched in Additional Libraries.For adding a new library to a project Configuration properties -> VC++ Directories -> Library Directories and here you specified additional folder for searching libraries and Configuration properties -> Linker -> Input for specifying the name of the library. -If the Linker could not find the symbol which you write in one .cpp he raises a linker time error which may sound like error LNK2001: unresolved external symbol "void __cdecl foo(void)" (?foo@@YAXXZ)

Observation

1. Once the Linker find one symbol he doesn't search in other libraries for it
2. The order of linking libraries does matter.
3. If Linker finds an external symbol in one static library he includes the symbol in the output of the project.However, if the library is shared( dynamic ) he doesn't include the code ( symbols ) in output, but Run-Time crashes may occur

How To Solve this kind of error

Compiler Time Error :

• Make sure you write your c++ project syntactical correct.

Linker Time Error

• Define all your symbol which you declare in your header files
• Use #pragma once for allowing compiler not to include one header if it was already included in the current .cpp which are compiled
• Make sure that your external library doesn't contain symbols that may enter into conflict with other symbols you defined in your header files
• When you use the template to make sure you include the definition of each template function in the header file for allowing the compiler to generate appropriate code for any instantiations.

Answer 17

A bug in the compiler/IDE

I recently had this problem, and it turned out it was a bug in Visual Studio Express 2013. I had to remove a source file from the project and re-add it to overcome the bug.

Steps to try if you believe it could be a bug in compiler/IDE:

• Clean the project (some IDEs have an option to do this, you can also manually do it by deleting the object files)
• Try start a new project, copying all source code from the original one.

Answer 18

Use the linker to help diagnose the error

Most modern linkers include a verbose option that prints out to varying degrees;

• Link invocation (command line),
• Data on what libraries are included in the link stage,
• The location of the libraries,
• Search paths used.

For gcc and clang; you would typically add -v -Wl,--verbose or -v -Wl,-v to the command line. More details can be found here;

• Linux ld man page.
• LLVM linker page.
• "An introduction to GCC" chapter 9.

For MSVC, /VERBOSE (in particular /VERBOSE:LIB) is added to the link command line.

• The MSDN page on the /VERBOSE linker option.

Answer 19

Linked .lib file is associated to a .dll

I had the same issue. Say i have projects MyProject and TestProject. I had effectively linked the lib file for MyProject to the TestProject. However, this lib file was produced as the DLL for the MyProject was built. Also, I did not contain source code for all methods in the MyProject, but only access to the DLL's entry points.

To solve the issue, i built the MyProject as a LIB, and linked TestProject to this .lib file (i copy paste the generated .lib file into the TestProject folder). I can then build again MyProject as a DLL. It is compiling since the lib to which TestProject is linked does contain code for all methods in classes in MyProject.

Answer 20

Since people seem to be directed to this question when it comes to linker errors I am going to add this here.

One possible reason for linker errors with GCC 5.2.0 is that a new libstdc++ library ABI is now chosen by default.

If you get linker errors about undefined references to symbols that involve types in the std::__cxx11 namespace or the tag [abi:cxx11] then it probably indicates that you are trying to link together object files that were compiled with different values for the _GLIBCXX_USE_CXX11_ABI macro. This commonly happens when linking to a third-party library that was compiled with an older version of GCC. If the third-party library cannot be rebuilt with the new ABI then you will need to recompile your code with the old ABI.

So if you suddenly get linker errors when switching to a GCC after 5.1.0 this would be a thing to check out.

Answer 21

Your linkage consumes libraries before the object files that refer to them

• You are trying to compile and link your program with the GCC toolchain.
• Your linkage specifies all of the necessary libraries and library search paths
• If libfoo depends on libbar, then your linkage correctly puts libfoo before libbar.
• Your linkage fails with undefined reference to something errors.
• But all the undefined somethings are declared in the header files you have #included and are in fact defined in the libraries that you are linking.

Examples are in C. They could equally well be C++

A minimal example involving a static library you built yourself

my_lib.c

#include "my_lib.h"
#include <stdio.h>

void hw(void)
{
    puts("Hello World");
}

my_lib.h

#ifndef MY_LIB_H
#define MT_LIB_H

extern void hw(void);

#endif

eg1.c

#include <my_lib.h>

int main()
{
    hw();
    return 0;
}

You build your static library:

$ gcc -c -o my_lib.o my_lib.c
$ ar rcs libmy_lib.a my_lib.o

You compile your program:

$ gcc -I. -c -o eg1.o eg1.c

You try to link it with libmy_lib.a and fail:

$ gcc -o eg1 -L. -lmy_lib eg1.o 
eg1.o: In function `main':
eg1.c:(.text+0x5): undefined reference to `hw'
collect2: error: ld returned 1 exit status

The same result if you compile and link in one step, like:

$ gcc -o eg1 -I. -L. -lmy_lib eg1.c
/tmp/ccQk1tvs.o: In function `main':
eg1.c:(.text+0x5): undefined reference to `hw'
collect2: error: ld returned 1 exit status

A minimal example involving a shared system library, the compression library libz

eg2.c

#include <zlib.h>
#include <stdio.h>

int main()
{
    printf("%s\n",zlibVersion());
    return 0;
}

Compile your program:

$ gcc -c -o eg2.o eg2.c

Try to link your program with libz and fail:

$ gcc -o eg2 -lz eg2.o 
eg2.o: In function `main':
eg2.c:(.text+0x5): undefined reference to `zlibVersion'
collect2: error: ld returned 1 exit status

Same if you compile and link in one go:

$ gcc -o eg2 -I. -lz eg2.c
/tmp/ccxCiGn7.o: In function `main':
eg2.c:(.text+0x5): undefined reference to `zlibVersion'
collect2: error: ld returned 1 exit status

And a variation on example 2 involving pkg-config:

$ gcc -o eg2 $(pkg-config --libs zlib) eg2.o 
eg2.o: In function `main':
eg2.c:(.text+0x5): undefined reference to `zlibVersion'

What are you doing wrong?

In the sequence of object files and libraries you want to link to make your program, you are placing the libraries before the object files that refer to them. You need to place the libraries after the object files that refer to them.

Link example 1 correctly:

$ gcc -o eg1 eg1.o -L. -lmy_lib

Success:

$ ./eg1 
Hello World

Link example 2 correctly:

$ gcc -o eg2 eg2.o -lz

Success:

$ ./eg2 
1.2.8

Link the example 2 pkg-config variation correctly:

$ gcc -o eg2 eg2.o $(pkg-config --libs zlib) 
$ ./eg2
1.2.8

The explanation

Reading is optional from here on.

By default, a linkage command generated by GCC, on your distro, consumes the files in the linkage from left to right in commandline sequence. When it finds that a file refers to something and does not contain a definition for it, to will search for a definition in files further to the right. If it eventually finds a definition, the reference is resolved. If any references remain unresolved at the end, the linkage fails: the linker does not search backwards.

First, example 1, with static library my_lib.a

A static library is an indexed archive of object files. When the linker finds -lmy_lib in the linkage sequence and figures out that this refers to the static library ./libmy_lib.a, it wants to know whether your program needs any of the object files in libmy_lib.a.

There is only object file in libmy_lib.a, namely my_lib.o, and there's only one thing defined in my_lib.o, namely the function hw.

The linker will decide that your program needs my_lib.o if and only if it already knows that your program refers to hw, in one or more of the object files it has already added to the program, and that none of the object files it has already added contains a definition for hw.

If that is true, then the linker will extract a copy of my_lib.o from the library and add it to your program. Then, your program contains a definition for hw, so its references to hw are resolved.

When you try to link the program like:

$ gcc -o eg1 -L. -lmy_lib eg1.o

the linker has not added eg1.o to the program when it sees -lmy_lib. Because at that point, it has not seen eg1.o. Your program does not yet make any references to hw: it does not yet make any references at all, because all the references it makes are in eg1.o.

So the linker does not add my_lib.o to the program and has no further use for libmy_lib.a.

Next, it finds eg1.o, and adds it to be program. An object file in the linkage sequence is always added to the program. Now, the program makes a reference to hw, and does not contain a definition of hw; but there is nothing left in the linkage sequence that could provide the missing definition. The reference to hw ends up unresolved, and the linkage fails.

Second, example 2, with shared library libz

A shared library isn't an archive of object files or anything like it. It's much more like a program that doesn't have a main function and instead exposes multiple other symbols that it defines, so that other programs can use them at runtime.

Many Linux distros today configure their GCC toolchain so that its language drivers (gcc,g++,gfortran etc) instruct the system linker (ld) to link shared libraries on an as-needed basis. You have got one of those distros.

This means that when the linker finds -lz in the linkage sequence, and figures out that this refers to the shared library (say) /usr/lib/x86_64-linux-gnu/libz.so, it wants to know whether any references that it has added to your program that aren't yet defined have definitions that are exported by libz

If that is true, then the linker will not copy any chunks out of libz and add them to your program; instead, it will just doctor the code of your program so that:-

• At runtime, the system program loader will load a copy of libz into the same process as your program whenever it loads a copy of your program, to run it.

• At runtime, whenever your program refers to something that is defined in libz, that reference uses the definition exported by the copy of libz in the same process.

Your program wants to refer to just one thing that has a definition exported by libz, namely the function zlibVersion, which is referred to just once, in eg2.c. If the linker adds that reference to your program, and then finds the definition exported by libz, the reference is resolved

But when you try to link the program like:

gcc -o eg2 -lz eg2.o

the order of events is wrong in just the same way as with example 1. At the point when the linker finds -lz, there are no references to anything in the program: they are all in eg2.o, which has not yet been seen. So the linker decides it has no use for libz. When it reaches eg2.o, adds it to the program, and then has undefined reference to zlibVersion, the linkage sequence is finished; that reference is unresolved, and the linkage fails.

Lastly, the pkg-config variation of example 2 has a now obvious explanation. After shell-expansion:

gcc -o eg2 $(pkg-config --libs zlib) eg2.o

becomes:

gcc -o eg2 -lz eg2.o

which is just example 2 again.

I can reproduce the problem in example 1, but not in example 2

The linkage:

gcc -o eg2 -lz eg2.o

works just fine for you!

(Or: That linkage worked fine for you on, say, Fedora 23, but fails on Ubuntu 16.04)

That's because the distro on which the linkage works is one of the ones that does not configure its GCC toolchain to link shared libraries as-needed.

Back in the day, it was normal for unix-like systems to link static and shared libraries by different rules. Static libraries in a linkage sequence were linked on the as-needed basis explained in example 1, but shared libraries were linked unconditionally.

This behaviour is economical at linktime because the linker doesn't have to ponder whether a shared library is needed by the program: if it's a shared library, link it. And most libraries in most linkages are shared libraries. But there are disadvantages too:-

• It is uneconomical at runtime, because it can cause shared libraries to be loaded along with a program even if doesn't need them.

• The different linkage rules for static and shared libraries can be confusing to inexpert programmers, who may not know whether -lfoo in their linkage is going to resolve to /some/where/libfoo.a or to /some/where/libfoo.so, and might not understand the difference between shared and static libraries anyway.

This trade-off has led to the schismatic situation today. Some distros have changed their GCC linkage rules for shared libraries so that the as-needed principle applies for all libraries. Some distros have stuck with the old way.

Why do I still get this problem even if I compile-and-link at the same time?

If I just do:

$ gcc -o eg1 -I. -L. -lmy_lib eg1.c

surely gcc has to compile eg1.c first, and then link the resulting object file with libmy_lib.a. So how can it not know that object file is needed when it's doing the linking?

Because compiling and linking with a single command does not change the order of the linkage sequence.

When you run the command above, gcc figures out that you want compilation + linkage. So behind the scenes, it generates a compilation command, and runs it, then generates a linkage command, and runs it, as if you had run the two commands:

$ gcc -I. -c -o eg1.o eg1.c
$ gcc -o eg1 -L. -lmy_lib eg1.o

So the linkage fails just as it does if you do run those two commands. The only difference you notice in the failure is that gcc has generated a temporary object file in the compile + link case, because you're not telling it to use eg1.o. We see:

/tmp/ccQk1tvs.o: In function `main'

instead of:

eg1.o: In function `main':

See also

The order in which interdependent linked libraries are specified is wrong

Putting interdependent libraries in the wrong order is just one way in which you can get files that need definitions of things coming later in the linkage than the files that provide the definitions. Putting libraries before the object files that refer to them is another way of making the same mistake.

Answer 22

A wrapper around GNU ld that doesn't support linker scripts

Some .so files are actually GNU ld linker scripts, e.g. libtbb.so file is an ASCII text file with this contents:

INPUT (libtbb.so.2)

Some more complex builds may not support this. For example, if you include -v in the compiler options, you can see that the mainwin gcc wrapper mwdip discards linker script command files in the verbose output list of libraries to link in. A simple work around is to replace the linker script input command file with a copy of the file instead (or a symlink), e.g.

cp libtbb.so.2 libtbb.so

Or you could replace the -l argument with the full path of the .so, e.g. instead of -ltbb do /home/foo/tbb-4.3/linux/lib/intel64/gcc4.4/libtbb.so.2

Answer 23

Befriending templates...

Given the code snippet of a template type with a friend operator (or function);

template <typename T>
class Foo {
    friend std::ostream& operator<< (std::ostream& os, const Foo<T>& a);
};

The operator<< is being declared as a non-template function. For every type T used with Foo, there needs to be a non-templated operator<<. For example, if there is a type Foo<int> declared, then there must be an operator implementation as follows;

std::ostream& operator<< (std::ostream& os, const Foo<int>& a) {/*...*/}

Since it is not implemented, the linker fails to find it and results in the error.

To correct this, you can declare a template operator before the Foo type and then declare as a friend, the appropriate instantiation. The syntax is a little awkward, but is looks as follows;

// forward declare the Foo
template <typename>
class Foo;

// forward declare the operator <<
template <typename T>
std::ostream& operator<<(std::ostream&, const Foo<T>&);

template <typename T>
class Foo {
    friend std::ostream& operator<< <>(std::ostream& os, const Foo<T>& a);
    // note the required <>        ^^^^
    // ...
};

template <typename T>
std::ostream& operator<<(std::ostream&, const Foo<T>&)
{
  // ... implement the operator
}

The above code limits the friendship of the operator to the corresponding instantiation of Foo, i.e. the operator<< <int> instantiation is limited to access the private members of the instantiation of Foo<int>.

Alternatives include;

• Allowing the friendship to extend to all instantiations of the templates, as follows;

  template <typename T>
  class Foo {
      template <typename T1>
      friend std::ostream& operator<<(std::ostream& os, const Foo<T1>& a);
      // ...
  };
  

• Or, the implementation for the operator<< can be done inline inside the class definition;

  template <typename T>
  class Foo {
      friend std::ostream& operator<<(std::ostream& os, const Foo& a)
      { /*...*/ }
      // ...
  };
  

Note, when the declaration of the operator (or function) only appears in the class, the name is not available for "normal" lookup, only for argument dependent lookup, from cppreference;

A name first declared in a friend declaration within class or class template X becomes a member of the innermost enclosing namespace of X, but is not accessible for lookup (except argument-dependent lookup that considers X) unless a matching declaration at the namespace scope is provided...

There is further reading on template friends at cppreference and the C++ FAQ.

Code listing showing the techniques above.

---

As a side note to the failing code sample; g++ warns about this as follows

warning: friend declaration 'std::ostream& operator<<(...)' declares a non-template function [-Wnon-template-friend]

note: (if this is not what you intended, make sure the function template has already been declared and add <> after the function name here)

Answer 24

When your include paths are different

Linker errors can happen when a header file and its associated shared library (.lib file) go out of sync. Let me explain.

How do linkers work? The linker matches a function declaration (declared in the header) with its definition (in the shared library) by comparing their signatures. You can get a linker error if the linker doesn't find a function definition that matches perfectly.

Is it possible to still get a linker error even though the declaration and the definition seem to match? Yes! They might look the same in source code, but it really depends on what the compiler sees. Essentially you could end up with a situation like this:

// header1.h
typedef int Number;
void foo(Number);

// header2.h
typedef float Number;
void foo(Number); // this only looks the same lexically

Note how even though both the function declarations look identical in source code, but they are really different according to the compiler.

You might ask how one ends up in a situation like that? Include paths of course! If when compiling the shared library, the include path leads to header1.h and you end up using header2.h in your own program, you'll be left scratching your header wondering what happened (pun intended).

An example of how this can happen in the real world is explained below.

Further elaboration with an example

I have two projects: graphics.lib and main.exe. Both projects depend on common_math.h. Suppose the library exports the following function:

// graphics.lib    
#include "common_math.h" 
   
void draw(vec3 p) { ... } // vec3 comes from common_math.h

And then you go ahead and include the library in your own project.

// main.exe
#include "other/common_math.h"
#include "graphics.h"

int main() {
    draw(...);
}

Boom! You get a linker error and you have no idea why it's failing. The reason is that the common library uses different versions of the same include common_math.h (I have made it obvious here in the example by including a different path, but it might not always be so obvious. Maybe the include path is different in the compiler settings).

Note in this example, the linker would tell you it couldn't find draw(), when in reality you know it obviously is being exported by the library. You could spend hours scratching your head wondering what went wrong. The thing is, the linker sees a different signature because the parameter types are slightly different. In the example, vec3 is a different type in both projects as far as the compiler is concerned. This could happen because they come from two slightly different include files (maybe the include files come from two different versions of the library).

Debugging the linker

DUMPBIN is your friend, if you are using Visual Studio. I'm sure other compilers have other similar tools.

The process goes like this:

1. Note the weird mangled name given in the linker error. (eg. draw@graphics@XYZ).
2. Dump the exported symbols from the library into a text file.
3. Search for the exported symbol of interest, and notice that the mangled name is different.
4. Pay attention to why the mangled names ended up different. You would be able to see that the parameter types are different, even though they look the same in the source code.
5. Reason why they are different. In the example given above, they are different because of different include files.

[1] By project I mean a set of source files that are linked together to produce either a library or an executable.

EDIT 1: Rewrote first section to be easier to understand. Please comment below to let me know if something else needs to be fixed. Thanks!

Answer 25

Inconsistent UNICODE definitions

A Windows UNICODE build is built with TCHAR etc. being defined as wchar_t etc. When not building with UNICODE defined as build with TCHAR defined as char etc. These UNICODE and _UNICODE defines affect all the "T" string types; LPTSTR, LPCTSTR and their elk.

Building one library with UNICODE defined and attempting to link it in a project where UNICODE is not defined will result in linker errors since there will be a mismatch in the definition of TCHAR; char vs. wchar_t.

The error usually includes a function a value with a char or wchar_t derived type, these could include std::basic_string<> etc. as well. When browsing through the affected function in the code, there will often be a reference to TCHAR or std::basic_string<TCHAR> etc. This is a tell-tale sign that the code was originally intended for both a UNICODE and a Multi-Byte Character (or "narrow") build.

To correct this, build all the required libraries and projects with a consistent definition of UNICODE (and _UNICODE).

1. This can be done with either;

   #define UNICODE
   #define _UNICODE
   

2. Or in the project settings;

   Project Properties > General > Project Defaults > Character Set

3. Or on the command line;

   /DUNICODE /D_UNICODE
   

The alternative is applicable as well, if UNICODE is not intended to be used, make sure the defines are not set, and/or the multi-character setting is used in the projects and consistently applied.

Do not forget to be consistent between the "Release" and "Debug" builds as well.

Answer 26

Clean and rebuild

A "clean" of the build can remove the "dead wood" that may be left lying around from previous builds, failed builds, incomplete builds and other build system related build issues.

In general the IDE or build will include some form of "clean" function, but this may not be correctly configured (e.g. in a manual makefile) or may fail (e.g. the intermediate or resultant binaries are read-only).

Once the "clean" has completed, verify that the "clean" has succeeded and all the generated intermediate file (e.g. an automated makefile) have been successfully removed.

This process can be seen as a final resort, but is often a good first step; especially if the code related to the error has recently been added (either locally or from the source repository).

Answer 27

Missing "extern" in const variable declarations/definitions (C++ only)

For people coming from C it might be a surprise that in C++ global constvariables have internal (or static) linkage. In C this was not the case, as all global variables are implicitly extern (i.e. when the static keyword is missing).

Example:

// file1.cpp
const int test = 5;    // in C++ same as "static const int test = 5"
int test2 = 5;

// file2.cpp
extern const int test;
extern int test2;

void foo()
{
 int x = test;   // linker error in C++ , no error in C
 int y = test2;  // no problem
}

correct would be to use a header file and include it in file2.cpp and file1.cpp

extern const int test;
extern int test2;

Alternatively one could declare the const variable in file1.cpp with explicit extern

Answer 28

Even though this is a pretty old questions with multiple accepted answers, I'd like to share how to resolve an obscure "undefined reference to" error.

Different versions of libraries

I was using an alias to refer to std::filesystem::path: filesystem is in the standard library since C++17 but my program needed to also compile in C++14 so I decided to use a variable alias:

#if (defined _GLIBCXX_EXPERIMENTAL_FILESYSTEM) //is the included filesystem library experimental? (C++14 and newer: <experimental/filesystem>)
using path_t = std::experimental::filesystem::path;
#elif (defined _GLIBCXX_FILESYSTEM) //not experimental (C++17 and newer: <filesystem>)
using path_t = std::filesystem::path;
#endif

Let's say I have three files: main.cpp, file.h, file.cpp:

• file.h #include's <experimental::filesystem> and contains the code above
• file.cpp, the implementation of file.h, #include's "file.h"
• main.cpp #include's <filesystem> and "file.h"

Note the different libraries used in main.cpp and file.h. Since main.cpp #include'd "file.h" after <filesystem>, the version of filesystem used there was the C++17 one. I used to compile the program with the following commands:

$ g++ -g -std=c++17 -c main.cpp -> compiles main.cpp to main.o
$ g++ -g -std=c++17 -c file.cpp -> compiles file.cpp and file.h to file.o
$ g++ -g -std=c++17 -o executable main.o file.o -lstdc++fs -> links main.o and file.o

This way any function contained in file.o and used in main.o that required path_t gave "undefined reference" errors because main.o referred to std::filesystem::path but file.o to std::experimental::filesystem::path.

Resolution

To fix this I just needed to change <experimental::filesystem> in file.h to <filesystem>.

Answer 29

When linking against shared libraries, make sure that the used symbols are not hidden.

The default behavior of gcc is that all symbols are visible. However, when the translation units are built with option -fvisibility=hidden, only functions/symbols marked with __attribute__ ((visibility ("default"))) are external in the resulting shared object.

You can check whether the symbols your are looking for are external by invoking:

# -D shows (global) dynamic symbols that can be used from the outside of XXX.so
nm -D XXX.so | grep MY_SYMBOL 

the hidden/local symbols are shown by nm with lowercase symbol type, for example t instead of `T for code-section:

nm XXX.so
00000000000005a7 t HIDDEN_SYMBOL
00000000000005f8 T VISIBLE_SYMBOL

You can also use nm with the option -C to demangle the names (if C++ was used).

Similar to Windows-dlls, one would mark public functions with a define, for example DLL_PUBLIC defined as:

#define DLL_PUBLIC __attribute__ ((visibility ("default")))

DLL_PUBLIC int my_public_function(){
  ...
}

Which roughly corresponds to Windows'/MSVC-version:

#ifdef BUILDING_DLL
    #define DLL_PUBLIC __declspec(dllexport) 
#else
    #define DLL_PUBLIC __declspec(dllimport) 
#endif

More information about visibility can be found on the gcc wiki.

---

When a translation unit is compiled with -fvisibility=hidden the resulting symbols have still external linkage (shown with upper case symbol type by nm) and can be used for external linkage without problem if the object files become part of a static libraries. The linkage becomes local only when the object files are linked into a shared library.

To find which symbols in an object file are hidden run:

>>> objdump -t XXXX.o | grep hidden
0000000000000000 g     F .text  000000000000000b .hidden HIDDEN_SYMBOL1
000000000000000b g     F .text  000000000000000b .hidden HIDDEN_SYMBOL2

Answer 30

Functions or class-methods are defined in source files with the inline specifier.

An example:-

main.cpp

#include "gum.h"
#include "foo.h"

int main()
{
    gum();
    foo f;
    f.bar();
    return 0;
}

foo.h (1)

#pragma once

struct foo {
    void bar() const;
};

gum.h (1)

#pragma once

extern void gum();

foo.cpp (1)

#include "foo.h"
#include <iostream>

inline /* <- wrong! */ void foo::bar() const {
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

gum.cpp (1)

#include "gum.h"
#include <iostream>

inline /* <- wrong! */ void gum()
{
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

If you specify that gum (similarly, foo::bar) is inline at its definition then the compiler will inline gum (if it chooses to), by:-

• not emitting any unique definition of gum, and therefore
• not emitting any symbol by which the linker can refer to the definition of gum, and instead
• replacing all calls to gum with inline copies of the compiled body of gum.

As a result, if you define gum inline in a source file gum.cpp, it is compiled to an object file gum.o in which all calls to gum are inlined and no symbol is defined by which the linker can refer to gum. When you link gum.o into a program together with another object file, e.g. main.o that make references to an external symbol gum, the linker cannot resolve those references. So the linkage fails:

Compile:

g++ -c  main.cpp foo.cpp gum.cpp

Link:

$ g++ -o prog main.o foo.o gum.o
main.o: In function `main':
main.cpp:(.text+0x18): undefined reference to `gum()'
main.cpp:(.text+0x24): undefined reference to `foo::bar() const'
collect2: error: ld returned 1 exit status

You can only define gum as inline if the compiler can see its definition in every source file in which gum may be called. That means its inline definition needs to exist in a header file that you include in every source file you compile in which gum may be called. Do one of two things:

Either don't inline the definitions

Remove the inline specifier from the source file definition:

foo.cpp (2)

#include "foo.h"
#include <iostream>

void foo::bar() const {
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

gum.cpp (2)

#include "gum.h"
#include <iostream>

void gum()
{
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

Rebuild with that:

$ g++ -c  main.cpp foo.cpp gum.cpp
imk@imk-Inspiron-7559:~/develop/so/scrap1$ g++ -o prog main.o foo.o gum.o
imk@imk-Inspiron-7559:~/develop/so/scrap1$ ./prog
void gum()
void foo::bar() const

Success.

Or inline correctly

Inline definitions in header files:

foo.h (2)

#pragma once
#include <iostream>

struct foo {
    void bar() const  { // In-class definition is implicitly inline
        std::cout << __PRETTY_FUNCTION__ << std::endl;
    }
};
// Alternatively...
#if 0
struct foo {
    void bar() const;
};
inline void foo::bar() const  {
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}
#endif

gum.h (2)

#pragma once
#include <iostream>

inline void gum() {
    std::cout << __PRETTY_FUNCTION__ << std::endl;
}

Now we don't need foo.cpp or gum.cpp:

$ g++ -c main.cpp
$ g++ -o prog main.o
$ ./prog
void gum()
void foo::bar() const

Answer 31

An “<strong>Undefined Reference” error occurs when we have a reference to object name (class, function, variable, etc.) in our program and the linker cannot find its definition when it tries to search for it in all the linked object files and libraries.

Thus when the linker cannot find the definition of a linked object, it issues an “undefined reference” error. As clear from definition, this error occurs in the later stages of the linking process. There are various reasons that cause an “undefined reference” error.

Some possible reason(more frequent):

#1) No Definition Provided For Object

This is the simplest reason for causing an “undefined reference” error. The programmer has simply forgotten to define the object.

Consider the following C++ program. Here we have only specified the prototype of function and then used it in the main function.

#include <iostream>
int func1();
int main()
{
     
    func1();
}

Output:

main.cpp:(.text+0x5): undefined reference to 'func1()'
collect2: error ld returned 1 exit status

So when we compile this program, the linker error that says “undefined reference to ‘func1()’” is issued.

In order to get rid of this error, we correct the program as follows by providing the definition of the function func1. Now the program gives the appropriate output.

#include <iostream>
using namespace std;
int func1();
 
int main()
{
     
    func1();
}
int func1(){
    cout<<"hello, world!!";
}

Output:

hello, world!!

#2) Wrong Definition (signatures don’t match) Of Objects Used

Yet another cause for “undefined reference” error is when we specify wrong definitions. We use any object in our program and its definition is something different.

Consider the following C++ program. Here we have made a call to func1 (). Its prototype is int func1 (). But its definition does not match with its prototype. As we see, the definition of the function contains a parameter to the function.

Thus when the program is compiled, the compilation is successful because of the prototype and function call match. But when the linker is trying to link the function call with its definition, it finds the problem and issues the error as “undefined reference”.

#include <iostream>
using namespace std;
int func1();
int main()
{
     
    func1();
}
int func1(int n){
    cout<<"hello, world!!";
}

Output:

main.cpp:(.text+0x5): undefined reference to 'func1()'
collect2: error ld returned 1 exit status

Thus to prevent such errors, we simply cross-check if the definitions and usage of all the objects are matching in our program.

#3) Object Files Not Linked Properly

This issue can also give rise to the “undefined reference” error. Here, we may have more than one source files and we might compile them independently. When this is done, the objects are not linked properly and it results in “undefined reference”.

Consider the following two C++ programs. In the first file, we make use of the “print ()” function which is defined in the second file. When we compile these files separately, the first file gives “undefined reference” for the print function, while the second file gives “undefined reference” for the main function.

int print();
int main()
{
    print();
}

Output:

main.cpp:(.text+0x5): undefined reference to 'print()'
collect2: error ld returned 1 exit status

int print() {
    return 42;
}

Output:

(.text+0x20): undefined reference to 'main'
collect2: error ld returned 1 exit status

The way to resolve this error is to compile both the files simultaneously (For example, by using g++).

Apart from the causes already discussed, “undefined reference” may also occur because of the following reasons.

#4) Wrong Project Type

When we specify wrong project types in C++ IDEs like the visual studio and try to do things that the project does not expect, then, we get “undefined reference”.

#5) No Library

If a programmer has not specified the library path properly or completely forgotten to specify it, then we get an “undefined reference” for all the references the program uses from the library.

#6) Dependent Files Are Not Compiled

A programmer has to ensure that we compile all the dependencies of the project beforehand so that when we compile the project, the compiler finds all the dependencies and compiles successfully. If any of the dependencies are missing then the compiler gives “undefined reference”.

Apart from the causes discussed above, the “undefined reference” error can occur in many other situations. But the bottom line is that the programmer has got the things wrong and in order to prevent this error they should be corrected.

Answer 32

Some typo errors to consider: (happened to me as a beginner a lot)

• If you are using a class: check if you have not forgetton "classname::" before the function name inside your cpp file where you define the the function.
• If you use forward declaration: be sure to declare the right type. E.g.: if you want to forward declare a "struct" use "struct" and not "class".

Answer 33

Different architectures

You may see a message like:

library machine type 'x64' conflicts with target machine type 'X86'

In that case, it means that the available symbols are for a different architecture than the one you are compiling for.

On Visual Studio, this is due to the wrong "Platform", and you need to either select the proper one or install the proper version of the library.

On Linux, it may be due to the wrong library folder (using lib instead of lib64 for instance).

On MacOS, there is the option of shipping both architectures in the same file. It may be that the link expects both versions to be there, but only one is. It can also be an issue with the wrong lib/lib64 folder where the library is picked up.

Answer 34

When you are using a wrong compiler to build your program

If you are using gcc or clang compiler suites, you should use the right compiler driver according to the language you are using. Compile and link C++ program using g++ or clang++. Using gcc or clang instead will cause references to the C++ standard library symbols to be undefined. Example:

$ gcc -o test test.cpp    
/usr/lib/gcc/x86_64-pc-linux-gnu/10.2.0/../../../../x86_64-pc-linux-gnu/bin/ld: /tmp/ccPv7MvI.o: warning: relocation against `_ZSt4cout' in read-only section `.text' 
/usr/lib/gcc/x86_64-pc-linux-gnu/10.2.0/../../../../x86_64-pc-linux-gnu/bin/ld: /tmp/ccPv7MvI.o: in function `main': test.cpp:(.text+0xe): undefined reference to `std::cout' 
/usr/lib/gcc/x86_64-pc-linux-gnu/10.2.0/../../../../x86_64-pc-linux-gnu/bin/ld: test.cpp:(.text+0x13): undefined reference to `std::basic_ostream<char, std::char_traits<char> >& std::operator<< <std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&, char const*)' 
/usr/lib/gcc/x86_64-pc-linux-gnu/10.2.0/../../../../x86_64-pc-linux-gnu/bin/ld: /tmp/ccPv7MvI.o: in function `__static_initialization_and_destruction_0(int, int)': 
test.cpp:(.text+0x43): undefined reference to `std::ios_base::Init::Init()' 
/usr/lib/gcc/x86_64-pc-linux-gnu/10.2.0/../../../../x86_64-pc-linux-gnu/bin/ld: test.cpp:(.text+0x58): undefined reference to `std::ios_base::Init::~Init()' 
/usr/lib/gcc/x86_64-pc-linux-gnu/10.2.0/../../../../x86_64-pc-linux-gnu/bin/ld: warning: creating DT_TEXTREL in a PIE 
collect2: error: ld returned 1 exit status

Answer 35

This issue occurred for me by declaring the prototype of a function in a header file:

int createBackground(VertexArray rVA, IntRect arena);

But then defining the function elsewhere using a reference with the first parameter:

int createBackground(VertexArray& rVA, IntRect arena) {}

The fact that the prototype was not using a reference in the first parameter, while the definition was, was causing this issue. When I changed both to properly match either containing a reference or not containing a reference, the issue was fixed.

Cheers.

Answer 36

My example:

header file

const string GMCHARACTER("character");
class GameCharacter : public GamePart
{
    private:
        string name;
        static vector<GameCharacter*> characterList;
    public:
        GameCharacter(cstring nm, cstring id) :
            GamePart(GMCHARACTER, id, TRUE, TRUE, TRUE),
            name(nm)
            { }
...
}

.cpp file:

vector<GameCharacter*> characterList;
...

This produced an "undefined" loader error because "characterList" was declared as a static member variable, but was defined as a global variable.

I added this because -- while someone else listed this case in a long list of things to look out for -- that listing did not give examples. This is an example of something more to look for, especially in C++.

Answer 37

In my case, The syntax was correct, yet I had that error when a class calls a second class within the same DLL. the CPP file of the second class had the wrong Properties->Item Type inside visual studio, in my case it was set to C/C++ Header instead of the correct one C/C++ compiler so the compiler didn't compile the CPP file while building it and cause the Error LNK2019

here's an example, Assuming the syntax is correct you should get the error by changing the item type in properties

//class A header file 
class ClassB; // FORWARD DECLERATION
class ClassA
{
public:
ClassB* bObj;
ClassA(HINSTANCE hDLL) ;
//  member functions
}
--------------------------------
//class A cpp file   
ClassA::ClassA(HINSTANCE hDLL) 
{
    bObj = new ClassB();// LNK2019 occures here 
    bObj ->somefunction();// LNK2019 occures here
}
/*************************/

//classB Header file
struct mystruct{}
class ClassB{
public:
ClassB();
mystruct somefunction();
}

------------------------------
//classB cpp file  
/* This is the file with the wrong property item type in visual studio --C/C++ Header-*/
ClassB::somefunction(){}
ClassB::ClassB(){}