未作为最后一个参数传递时的模板参数包扣除 [英] Template parameter pack deduction when not passed as last parameter

问题描述

考虑以下代码:

#include <iostream>
#include <functional>

template<typename... Args>
void testfunc(const std::function<void (float, Args..., char)>& func)
{

}

int main(int argc, char* argv[])
{
    auto func = [](float, int, char) {};
    auto sfunc = static_cast<std::function<void (float, int, char)>>(func);
    testfunc<int>(sfunc);

    return 0;
}

我明确指定类型是因为( https://stackoverflow.com/a/40476083 ):

I specify the type explicitly because (https://stackoverflow.com/a/40476083):

当参数包没有出现在参数的最后声明，它是一个非推论上下文.非推论上下文意味着必须显式地指定模板参数.

When a parameter pack doesn't appear last in the parameter declaration, it is a non-deduced context. A non-deduced context means that the template arguments have to be given explicitly.

MSVC成功编译了它，而gcc和clang都拒绝了该代码:

MSVC successfully compiles it, while both gcc and clang reject the code:

source_file.cpp: In function ‘int main(int, char**)’:
source_file.cpp:14:24: error: no matching function for call to ‘testfunc(std::function<void(float, int, char)>&)’
     testfunc<int>(sfunc);
                        ^
source_file.cpp:5:6: note: candidate: template<class ... Args> void testfunc(const std::function<void(float, Args ..., char)>&)
 void testfunc(const std::function<void (float, Args..., char)>& func)
      ^
source_file.cpp:5:6: note:   template argument deduction/substitution failed:
source_file.cpp:14:24: note:   mismatched types ‘char’ and ‘int’
     testfunc<int>(sfunc);
                        ^
source_file.cpp:14:24: note:   ‘std::function<void(float, int, char)>’ is not derived from ‘const std::function<void(float, Args ..., char)>’

现在让我们稍作更改-让我们从本地 func 中删除 int 参数，从而使模板参数包变为空:

Let's now make a slight change - let's remove the int argument from our local func, thereby causing the template argument pack to become empty:

#include <iostream>
#include <functional>

template<typename... Args>
void testfunc(const std::function<void (float, Args..., char)>& func)
{

}

int main(int argc, char* argv[])
{
    auto func = [](float, char) {};
    auto sfunc = static_cast<std::function<void (float, char)>>(func);
    testfunc<>(sfunc);

    return 0;
}

这一次，所有三个编译器都将代码视为错误.已通过 http://rextester.com/l/cpp_online_compiler_gcc 和本地Visual Studio安装进行了测试.

This time, all three compilers reject the code as incorrect. Tested with http://rextester.com/l/cpp_online_compiler_gcc and a local Visual Studio installation.

问题:

1. 在第一种情况下谁是正确的?
2. 如何达到理想的效果-即如何明确指定(可能为空)参数包?

推荐答案

我们可以阻止推论:

template<typename... Args>
void testfunc(const block_deduction<std::function<void (float, Args..., char)>>& func)

使用

template<class T>
struct tag_t{using type=T;};

template<class T>
using block_deduction=typename tag_t<T>::type;

现在是 Args ... 处于非推论上下文中.

and now Args... is in a non-deduced context.

您可以使用SFINAE进行更奇特的操作，并省略 char ，然后测试 char 在 Args ... 的末尾，但是似乎太过分了.

You can do fancier things with SFINAE and omitting char then testing that char is at the end of the Args..., but that seems overkill.

当gcc和clang与MSVC意见不一致时，我会向甜甜圈下注，这是不对的.但是我没有标准地去证实这一点.

I will bet dollars to donuts that when gcc and clang disagree with MSVC, MSVC isn't right. But I have not standard delved to confirm that.

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