预先定义一个可变函数指针参数

问题描述

我有一个函数 foo ，它接受一个可变函数指针作为其参数。

I have a function foo that takes a variadic function pointer as its argument.

使用using在函数声明之前定义参数的类型。

I would like to use "using" to define the argument's type prior to the function declaration.

template <typename ... vARGS>
using TFuncType = void(*)(vARGS ... V_args);

template <typename ... vARGS>
void foo(TFuncType<vARGS ...> funcptr) {}

void bar(int i) {}

int main() {
  foo(&bar); // This line fails to compile.
}

这不编译。错误（通过clang使用c ++ 1z）是：

This doesn't compile. The error (via clang using c++1z) is:

/make/proj/test/variadic-funcparam-deduce2.cpp:39:5: error: no matching function for call to 'foo'
foo(&bar);
^~~
/make/proj/test/variadic-funcparam-deduce2.cpp:33:36: note: candidate template ignored: substitution failure [with vARGS = int]
template <typename ... vARGS> void foo(TFuncType<vARGS ...> funcptr) {}

替换失败？

如果我在foo（）中明确写入类型，我可以成功编译：

I can successfully compile if I explicitly write the type inside foo():

template <typename ... vARGS>
void foo(void(*funcptr)(vARGS ... V_args)) {}

但我不能得到初始（使用）版本工作，即使明确指定模板参数，并使用预先转换的 TFuncType< int> 即：

But I cannot get the initial ("using") version to work even when explicitly specifying the template parameters, and using a pre-casted TFuncType<int> for the argument, i.e.:

int main() {
  TF_call<int> fptr = &bar; // This line is OK.
  foo<int>(fptr);
}

有人知道这里有什么吗？

Does anyone know what's up here?

使用typedef'd（using）可变参数和/或我缺少的函数指针有什么奇怪吗？

Is there something strange about using typedef'd ("using") variadics and/or function pointers that I'm missing?

推荐答案

我相信这可能与我从复制的以下文本相关来自14.5.7 [temp.alias]段落2中的C ++标准：

I believe this may be related to the following text which I copied from this answer that itself takes from the C++ standard in 14.5.7 [temp.alias] paragraph 2:

如果我要解释这一点，那意味着GCC接受代码实际上是不合格的。

If I'm interpreting that right, it means that GCC accepting the code is actually non-conforming.

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