c++提取函数指针的模板参数

发布于 2025-01-10 07:09:49 字数 612 浏览 0 评论 0原文

有没有办法提取/内省模板函数指针以获取其模板参数类型和大小？

更一般地说，C++ 中什么样的 Callables/Invocables 能够通过内省来获取模板参数的类型信息以及技术是什么？

编辑：我想做的是将可调用对象作为模板参数传递给其他类，并携带该可调用对象所具有的有关其采用的模板参数的所有信息。然后我以一种更编译时和功能友好的方式调用这个可调用函数。

template<typename T>
T SomeFunc() {return T{};}

constexpr auto some_func_fp = &SomeFunc<int>;
// can we do some introspection to get the types 
// and size of the template arguments of some templated function pointers?

原文

Is there a way to extract/introspect on templated function pointer to get its template argument types and size?

More generally speaking, what kind of Callables/Invocables in C++ are able to be introspected to get the type info of template args and what are the techiniques?

Edit:
What I'm trying to do is to pass a callable object to some other class as a template parameter and carrying all the info this callable has regarding what template parameters it takes. Then I invoke this callable in a more compile-time and functional friendly way.

https://godbolt.org/z/q3f98177Y

template<typename T>
T SomeFunc() {return T{};}

constexpr auto some_func_fp = &SomeFunc<int>;
// can we do some introspection to get the types 
// and size of the template arguments of some templated function pointers?

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心病无药医 2025-01-17 07:09:49

一般来说，函数的类型（和指针类型）不能与模板参数无关。（在你的情况下，你可以检测返回类型）

#include <type_traits>

template<typename T>
int SomeFunc(){return 0;}

auto fi = &SomeFunc<int>;
auto ff = &SomeFunc<float>;

// fi has same type as ff, int(*)()
static_assert(std::is_same_v<decltype(fi),decltype(ff)>);

You can't, in general, the function's type (and the pointer type) has nothing to do with the template parameter. (in your case you can detect the return type though)

#include <type_traits>

template<typename T>
int SomeFunc(){return 0;}

auto fi = &SomeFunc<int>;
auto ff = &SomeFunc<float>;

// fi has same type as ff, int(*)()
static_assert(std::is_same_v<decltype(fi),decltype(ff)>);

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