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参数列表中间的可变模板参数

template-meta-programming variadic-templates templates c++11 c++

Felk · 技术社区 · 6 年前

我正试图指定一个将泛型函数作为参数的函数。函数的定义如下:

template <typename TRet, typename... TsArgs>
using Fun = TRet (*)(TsArgs...);

如何指定将该泛型函数作为参数的泛型函数?我试过这个:

template<typename TRet, typename... TsArgs, Fun<TRet, TsArgs...> F>
TRet wrap(TsArgs... args) {
  return F(args...);
}

要包装此函数:

bool foo(int x, double y) {
    return x < y;
}

这样地:

Fun<bool, int, double> func = wrap<bool, int, double, foo>;

然而,不幸的是,这并不能编译。GCC 8.1有以下错误信息:

<source>:16:35: error: no matches converting function 'wrap' to type 'Fun<bool, int, double>' {aka 'bool (*)(int, double)'}
     Fun<bool, int, double> func = wrap<bool, int, double, foo>;
                                   ^~~~~~~~~~~~~~~~~~~~~~~~~~~~

而clang 6有以下错误:

<source>:16:35: error: address of overloaded function 'wrap' does not match required type 'bool (int, double)'
    Fun<bool, int, double> func = wrap<bool, int, double, foo>;
                                  ^~~~~~~~~~~~~~~~~~~~~~~~~~~~

但是,如果我替换 TsArgs 具有 int, double 就像签名一样 foo() 它编译得很好,让我相信参数列表中间的变量模板参数并不像我期望的那样工作。否则我怎么能实现我的目标呢?

这是mcve:

template <typename TRet, typename... TsArgs>
using Fun = TRet (*)(TsArgs...);

template<typename TRet, typename... TsArgs, Fun<TRet, TsArgs...> F>
TRet wrap(TsArgs... args) {
    return F(args...);
}

bool foo(int x, double y) {
    return x < y;
}

int main() {
    Fun<bool, int, double> func = wrap<bool, int, double, foo>;
    return 0;
}

2 回复 | 直到 6 年前

max66 6 年前

auto

wrapHelper

template <typename T, T>
struct wrapHelper;

template <typename TRet, typename... TsArgs, Fun<TRet, TsArgs...> F>
struct wrapHelper<Fun<TRet, TsArgs...>, F>
 {
   static TRet func (TsArgs ... args)
    { return F(args...); }
 };

wrapper

template <auto X>
struct wrap : public wrapHelper<decltype(X), X>
 { };

TRet TsArgs... foo

Fun<bool, int, double> func = wrap<bool, int, double>::func<foo>;

Fun<bool, int, double> func = wrap<foo>::func;

auto func = wrap<foo>::func;

#include <iostream>
#include <type_traits>

template <typename TRet, typename... TsArgs>
using Fun = TRet (*)(TsArgs...);

bool foo(int x, double y)
 { return x < y; }

template <typename T, T>
struct wrapHelper;

template <typename TRet, typename... TsArgs, Fun<TRet, TsArgs...> F>
struct wrapHelper<Fun<TRet, TsArgs...>, F>
 {
   static TRet func (TsArgs ... args)
    { return F(args...); }
 };

template <auto X>
struct wrap : public wrapHelper<decltype(X), X>
 { };

int main()
 {   
   auto func { wrap<foo>::func };

   static_assert( std::is_same<decltype(func), Fun<bool, int, double>>{} );

   std::cout << func(1, 2.0) << std::endl;
 }

max66 6 年前

wrap TRet TArgs... func() Fun<TRet, TsArgs...>

template <typename TRet, typename... TsArgs>
struct wrap
 {
   template <Fun<TRet, TsArgs...> F>
   static TRet func (TsArgs ... args)
    { return F(args...); }
 };

Fun<bool, int, double> func = wrap<bool, int, double>::func<foo>;

#include <iostream>

template <typename TRet, typename... TsArgs>
using Fun = TRet (*)(TsArgs...);

template <typename TRet, typename... TsArgs>
struct wrap
 {
   template <Fun<TRet, TsArgs...> F>
   static TRet func (TsArgs ... args)
    { return F(args...); }
 };

bool foo(int x, double y)
 { return x < y; }

int main()
 {
   Fun<bool, int, double> func = wrap<bool, int, double>::func<foo>;

   std::cout << func(1, 2.0) << std::endl;
 }