c++ - C ++：16進数から10進数への変換

Question

hex^{（16進数）}をdec^{（10進数）}に簡単に変換する方法を探しています。私はこれを行う簡単な方法を次のように見つけました：

int k = 0x265;
cout << k << endl;

しかし、それでは入力できません265。とにかくそれがそのように機能するためにありますか？

入力： 265

出力： 613

とにかくそれをすることはありますか？

注：私は試しました：

int k = 0x, b;
cin >> b;
cout << k + b << endl;

そしてそれは動作しません。

Answer 1

#include <iostream>
#include <iomanip>
#include <sstream>

int main()
{
    int x, y;
    std::stringstream stream;

    std::cin >> x;
    stream << x;
    stream >> std::hex >> y;
    std::cout << y;

    return 0;
}

Answer 2

std::hexマニピュレーターを使用する：

#include <iostream>
#include <iomanip>

int main()
{
    int x;
    std::cin >> std::hex >> x;
    std::cout << x << std::endl;

    return 0;
}

Answer 3

さて、Cの方法は次のようなものかもしれません...

#include <stdlib.h>
#include <stdio.h>

int main()
{
        int n;
        scanf("%d", &n);
        printf("%X", n);

        exit(0);
}

Answer 4

文字列を使用し、ASCIIテーブルで10進数に変換するソリューションは次のとおりです。

#include <iostream>
#include <string>
#include "math.h"
using namespace std;
unsigned long hex2dec(string hex)
{
    unsigned long result = 0;
    for (int i=0; i<hex.length(); i++) {
        if (hex[i]>=48 && hex[i]<=57)
        {
            result += (hex[i]-48)*pow(16,hex.length()-i-1);
        } else if (hex[i]>=65 && hex[i]<=70) {
            result += (hex[i]-55)*pow(16,hex.length( )-i-1);
        } else if (hex[i]>=97 && hex[i]<=102) {
            result += (hex[i]-87)*pow(16,hex.length()-i-1);
        }
    }
    return result;
}

int main(int argc, const char * argv[]) {
    string hex_str;
    cin >> hex_str;
    cout << hex2dec(hex_str) << endl;
    return 0;
}

Answer 5

私はこれを使用します：

template <typename T>
bool fromHex(const std::string& hexValue, T& result)
{
    std::stringstream ss;
    ss << std::hex << hexValue;
    ss >> result;

    return !ss.fail();
}

Answer 6

    std::cout << "Enter decimal number: " ;
    std::cin >> input ;

    std::cout << "0x" << std::hex << input << '\n' ;

boolean、float、またはintの入力を追加すると、intメイン関数呼び出しで返されます...

With function templates, based on argument types, C generates separate functions to handle each type of call appropriately. All function template definitions begin with the keyword template followed by arguments enclosed in angle brackets < and >. A single formal parameter T is used for the type of data to be tested.

Consider the following program where the user is asked to enter an integer and then a float, each uses the square function to determine the square. With function templates, based on argument types, C generates separate functions to handle each type of call appropriately. All function template definitions begin with the keyword template followed by arguments enclosed in angle brackets < and >. A single formal parameter T is used for the type of data to be tested.

Consider the following program where the user is asked to enter an integer and then a float, each uses the square function to determine the square.

#include <iostream>
 using namespace std;
template <class T>      // function template
T square(T);    /* returns a value of type T and accepts                  type T     (int or float or whatever) */
  void main()
{
int x, y;
float w, z;
cout << "Enter a integer:  ";
cin >> x;
y = square(x);
cout << "The square of that number is:  " << y << endl;
cout << "Enter a float:  ";
cin >> w;
z = square(w);
cout << "The square of that number is:  " << z << endl;
}

template <class T>      // function template
T square(T u) //accepts a parameter u of type T (int or float)
{
return u * u;
}

Here is the output:

Enter a integer:  5
The square of that number is:  25
Enter a float:  5.3
The square of that number is:  28.09

Answer 7

This should work as well.

#include <ctype.h>
#include <string.h>

template<typename T = unsigned int>
T Hex2Int(const char* const Hexstr, bool* Overflow)
{
    if (!Hexstr)
        return false;
    if (Overflow)
        *Overflow = false;

    auto between = [](char val, char c1, char c2) { return val >= c1 && val <= c2; };
    size_t len = strlen(Hexstr);
    T result = 0;

    for (size_t i = 0, offset = sizeof(T) << 3; i < len && (int)offset > 0; i++)
    {
        if (between(Hexstr[i], '0', '9'))
            result = result << 4 ^ Hexstr[i] - '0';
        else if (between(tolower(Hexstr[i]), 'a', 'f'))
            result = result << 4 ^ tolower(Hexstr[i]) - ('a' - 10); // Remove the decimal part;
        offset -= 4;
    }
    if (((len + ((len % 2) != 0)) << 2) > (sizeof(T) << 3) && Overflow)
        *Overflow = true;
    return result;
}

The 'Overflow' parameter is optional, so you can leave it NULL.

Example:

auto result = Hex2Int("C0ffee", NULL);
auto result2 = Hex2Int<long>("DeadC0ffe", NULL);

Answer 8

only use:

cout << dec << 0x;