你可以创建一个带有部分特化的函数traits：

template <auto func, std::size_t I>
struct param_type;

template <typename Ret, typename... Args, Ret (*func)(Args...), std::size_t I>
struct param_type<func, I>
{
    using type = std::tuple_element_t<I, std::tuple<Args...>>;
};

// C-ellipsis version aka printf-like functions
template <typename Ret, typename... Args, Ret (*func)(Args..., ...), std::size_t I>
struct param_type<func, I>
{
    using type = std::tuple_element_t<I, std::tuple<Args...>>;
};

Demo

另一种std::tuple_element方式通过函数声明（仅声明，无需定义）

template <std::size_t Ind, typename R, typename ... Args>
std::tuple_element_t<Ind-1u, std::tuple<std::remove_reference_t<Args>...>>
   pt_helper (R(Args...));

所以param_type简单地变成

template <auto X, std::size_t Ind>
struct param_type
 { using type = decltype( pt_helper<Ind>(X) ); };

观察pt_helper声明中的Ind-1u：你要求param_type返回带有参数3的函数的第三个参数类型：通常，在C/C++世界中，索引从零开始计数。我建议你接受param_type<foo, 2>::type是char;在这种情况下，您必须删除-1。
也观察std::remove_reference_t。因为您希望param_type<foo, 3>::type是char（不带引用）而不是char &（带引用）。
无论如何，下面是一个完整的编译示例

#include <tuple>
#include <string>
#include <type_traits>

void foo (std::string const& a, int b, char &c)
 { }

template <std::size_t Ind, typename R, typename ... Args>
std::tuple_element_t<Ind-1u, std::tuple<std::remove_reference_t<Args>...>>
   pt_helper (R(Args...));

template <auto X, std::size_t Ind>
struct param_type
 { using type = decltype( pt_helper<Ind>(X) ); };

int main ()
 {
   using T = typename param_type<foo, 3u>::type;

   static_assert( std::is_same_v<T, char>, "!" );
 }

在本例中，您可以检查2种类型是否与std::is_same<T,U>::value相同

if (std::is_same<param_type<foo, 3>::type,char>::value) {
    /*stuff*/
}else {
    /*stuff2*/
}

查看更多关于is_same

有点晚了，但看到我自己的完整（免费）函数traits库here（生产级，完整文档-比Boost版本小得多，也更完整）。将其应用于您的代码（演示here）：

#include <string>
#include <iostream>

// See https://github.com/HexadigmSystems/FunctionTraits
#include "TypeTraits.h"

void foo(std::string const& a, int b, char &c)
{
    // ...
}

int main()
{
    ////////////////////////////////////////////////
    // Everything in "TypeTraits.h" above declared
    // in this namespace
    ////////////////////////////////////////////////
    using namespace StdExt;
    
    using F = decltype(foo);
    
    ////////////////////////////////////////////////////////////
    // 3rd type in "foo()" (passing 2 as the zero-based index)
    // so yields "char &" (the arg's exact type). Just apply
    // "std::remove_reference_t" as well if you want to remove
    // the "&" (resulting in "char").
    ////////////////////////////////////////////////////////////
    using Arg3InFoo_t = ArgType_t<F, 2>;

    ////////////////////////////////////////////////////////////////////
    // Display above type (its user-friendly name). Alternatively you
    // can use "ArgTypeName_v" instead of "TypeName_v" (former just
    // defers to the latter).
    //
    // Note: Following call to "tcout" (also declared in "TypeTraits.h"
    // above) always resolves to "std::cout" on non-Microsoft platforms
    // and usually "std::wcout" on Microsoft platforms (when compiling
    // for UTF-16 in that environment which is usually the case). You
    // can directly call "std::cout" or "std::wcout" instead if you
    // prefer, assuming you know which platform you're always running
    // on.
    ////////////////////////////////////////////////////////////////////
    tcout << TypeName_v<Arg3InFoo_t>;
    
    return 0;
}