4

クラスの関数を生成するためのプリプロセッサ ディレクティブをいくつか作成しました。Boost 1.55 では、すべて正常に動作します。Boost 1.57 に変更しようとすると、奇妙なコンパイル エラーが発生します。

プログラムは次のとおりです。

#include <iostream>

#include <boost/preprocessor.hpp>

#ifndef BOOST_PP_VARIADICS
  #define BOOST_PP_VARIADICS
#endif

#define ADD_CONST_REF(op, data, elem) elem const &

#define FUNCTION_DECL(r, data, tup) \
  double BOOST_PP_TUPLE_ELEM(0, tup) \
    BOOST_PP_IF(BOOST_PP_LIST_IS_NIL(BOOST_PP_TUPLE_TO_LIST(BOOST_PP_TUPLE_ELEM(1, tup))), \
                (), \
                BOOST_PP_SEQ_TO_TUPLE( \
                  BOOST_PP_SEQ_TRANSFORM(ADD_CONST_REF, \
                                         _, \
                                         BOOST_PP_TUPLE_TO_SEQ(BOOST_PP_LIST_TO_TUPLE(BOOST_PP_TUPLE_TO_LIST(BOOST_PP_TUPLE_ELEM(1, tup)))) \
                  ) \
                ) \
    ) \
  const;

#define MY_FUNCTIONS(seq) \
  BOOST_PP_SEQ_FOR_EACH(FUNCTION_DECL, _, seq)

class Foo
{
public:
  MY_FUNCTIONS( \
    ((a, (int))) \
    ((b)       ) \
  )
};

#define ADD_ARG(op, data, elem) elem const & BOOST_PP_CAT(a_, elem)

#define MY_FUNCTION_BEGIN(tup) \
  double Foo::BOOST_PP_TUPLE_ELEM(0, tup) \
    BOOST_PP_IF(BOOST_PP_LIST_IS_NIL(BOOST_PP_TUPLE_TO_LIST(BOOST_PP_TUPLE_ELEM(1, tup))), \
                (), \
                BOOST_PP_SEQ_TO_TUPLE( \
                  BOOST_PP_SEQ_TRANSFORM( \
                    ADD_ARG, \
                    _, \
                    BOOST_PP_TUPLE_TO_SEQ(BOOST_PP_LIST_TO_TUPLE(BOOST_PP_TUPLE_TO_LIST(BOOST_PP_TUPLE_ELEM(1, tup)))) \
                  ) \
                ) \
    ) const \
  {

#define MY_FUNCTION_END() }

MY_FUNCTION_BEGIN((a, (int)))
{
  std::cout << "Foo::a(" << a_int << ")\n";
}
MY_FUNCTION_END()

MY_FUNCTION_BEGIN((b))
{
  std::cout << "Foo::b()\n";
}
MY_FUNCTION_END()

int main()
{
  Foo f;
  f.a(42);
  f.b();

  return 0;
}

gcc 4.9.1 では、次の出力が得られます。

test.cpp:33:1: error: macro "BOOST_PP_SEQ_ELEM_III" requires 2 arguments, but only 1 given
   )
 ^
test.cpp:33:1: error: macro "BOOST_PP_VARIADIC_ELEM_2" requires 4 arguments, but only 2 given
test.cpp:60:1: error: macro "BOOST_PP_SEQ_ELEM_III" requires 2 arguments, but only 1 given
 MY_FUNCTION_BEGIN((b))
 ^
test.cpp:60:1: error: macro "BOOST_PP_VARIADIC_ELEM_2" requires 4 arguments, but only 2 given

clang 3.5.0 は私に言います:

test.cpp:30:3: error: too few arguments provided to function-like macro invocation
  MY_FUNCTIONS( \
  ^
test.cpp:25:25: note: expanded from macro 'MY_FUNCTIONS'
  BOOST_PP_SEQ_FOR_EACH(FUNCTION_DECL, _, seq)
                        ^
..../boost.1_57_0/boost/preprocessor/seq/for_each.hpp:26:67: note: expanded from macro 'BOOST_PP_SEQ_FOR_EACH'
#    define BOOST_PP_SEQ_FOR_EACH(macro, data, seq) BOOST_PP_FOR((macro, data, seq (nil)), BOOST_PP_SEQ_FOR_EACH_P, BOOST_PP_SEQ_FOR_EACH_O, BOOST_PP_SEQ_FOR_EACH_M)
                                                                  ^
..../boost.1_57_0/boost/preprocessor/repetition/detail/for.hpp:22:78: note: expanded from macro 'BOOST_PP_FOR_1'
# define BOOST_PP_FOR_1(s, p, o, m) BOOST_PP_FOR_1_C(BOOST_PP_BOOL(p(2, s)), s, p, o, m)
                                                                             ^
note: (skipping 18 expansions in backtrace; use -fmacro-backtrace-limit=0 to see all)
packages/boost/boost_i4/boost/preprocessor/seq/elem.hpp:22:39: note: expanded from macro 'BOOST_PP_SEQ_ELEM'
#    define BOOST_PP_SEQ_ELEM(i, seq) BOOST_PP_SEQ_ELEM_I(i, seq)
                                      ^
packages/boost/boost_i4/boost/preprocessor/seq/elem.hpp:41:45: note: expanded from macro 'BOOST_PP_SEQ_ELEM_I'
#        define BOOST_PP_SEQ_ELEM_I(i, seq) BOOST_PP_SEQ_ELEM_II(BOOST_PP_SEQ_ELEM_ ## i seq)
                                            ^
packages/boost/boost_i4/boost/preprocessor/seq/elem.hpp:43:62: note: expanded from macro 'BOOST_PP_SEQ_ELEM_II'
#    define BOOST_PP_SEQ_ELEM_II(im) BOOST_PP_SEQ_ELEM_III(im)
                                                             ^
packages/boost/boost_i4/boost/preprocessor/seq/elem.hpp:44:13: note: macro 'BOOST_PP_SEQ_ELEM_III' defined here
#    define BOOST_PP_SEQ_ELEM_III(x, _) x

Boost 1.57 に変更した後の問題を知っている人はいますか?

4

2 に答える 2

0

C++11 可変個引数マクロを使用した実用的なソリューションを見つけました。ここにあります：

#include <iostream>
#include <string>

#ifndef BOOST_PP_VARIADICS
  #define BOOST_PP_VARIADICS
#endif

#include <boost/preprocessor.hpp>

#define FUNCTION_DECL_IMPL_1(name) \
  double name() const;

#define FUNCTION_ARGS_IMPL_1(arg1) arg1 const &
#define FUNCTION_ARGS_IMPL_2(arg1, arg2) arg1 const &, arg2 const &
#define FUNCTION_ARGS_IMPL_3(arg1, arg2, arg3) arg1 const &, arg2 const &, arg3 const &

#define FUNCTION_ARGS(...) \
  BOOST_PP_OVERLOAD(FUNCTION_ARGS_IMPL_, __VA_ARGS__)(__VA_ARGS__)

#define FUNCTION_DECL_IMPL_2(name, args) \
  double name( \
    FUNCTION_ARGS(BOOST_PP_TUPLE_REM()args) \
    ) const;

#define FUNCTION_DECL_VAR(...) \
  BOOST_PP_OVERLOAD(FUNCTION_DECL_IMPL_, __VA_ARGS__)(__VA_ARGS__)

#define FUNCTION_DECL(r, data, tup) \
  FUNCTION_DECL_VAR(BOOST_PP_TUPLE_REM()tup)

#define MY_FUNCTIONS(seq) \
  BOOST_PP_SEQ_FOR_EACH(FUNCTION_DECL, _, seq)

class Foo
{
public:
  MY_FUNCTIONS( \
    ((a, (int))) \
    ((b)) \
    ((c, (double, std::string))) \
  )
};

#define MY_FUNCTION_BEGIN_IMPL_1(name) \
  double Foo::name() const {

#define FUNCTION_ARGS2_IMPL_1(arg1) arg1 const & a_arg1
#define FUNCTION_ARGS2_IMPL_2(arg1, arg2) arg1 const & a_arg1, arg2 const & a_arg2
#define FUNCTION_ARGS2_IMPL_3(arg1, arg2, arg3) arg1 const & a_arg1, arg2 const & a_arg2, arg3 const & a_arg3

#define FUNCTION_ARGS2(...) \
  BOOST_PP_OVERLOAD(FUNCTION_ARGS2_IMPL_, __VA_ARGS__)(__VA_ARGS__)

#define MY_FUNCTION_BEGIN_IMPL_2(name, args) \
  double Foo::name( \
    FUNCTION_ARGS2(BOOST_PP_TUPLE_REM()args) \
    ) const {

#define MY_FUNCTION_BEGIN(...) \
  BOOST_PP_OVERLOAD(MY_FUNCTION_BEGIN_IMPL_, __VA_ARGS__)(__VA_ARGS__)

#define MY_FUNCTION_END() }

MY_FUNCTION_BEGIN(a, (int))
{
  std::cout << "Foo::a(" << a_arg1 << ")\n";
}
MY_FUNCTION_END()

MY_FUNCTION_BEGIN(b)
{
  std::cout << "Foo::b()\n";
}
MY_FUNCTION_END()

MY_FUNCTION_BEGIN(c, (double, std::string))
{
  std::cout << "Foo::c(" << a_arg1 << "," << a_arg2 << ")\n";
}
MY_FUNCTION_END()

int main()
{
  Foo f;
  f.a(42);
  f.b();
  f.c(3.1415, "wohaa");

  return 0;
}
于 2015-02-24T13:34:38.343 に答える