C++ extract argument from function signature

Question

Given a template function of the signature

template <typename F>
void fn(F f);

I would like to extract the parameter pack of the given function signature F, if a callable function was given to fn().

Unlike in foo0() I am not able to do this in foo1() and foo2() in the following example (simplified to a single parameter instead of a parameter pack):

#include <type_traits>

template <typename Sig> struct argument_of;
template <typename R, typename Arg> struct argument_of<R(Arg)> { typedef Arg type; };

void bar(int);

void foo0() {
    static_assert(std::is_same<int, typename argument_of<decltype(bar)>::type>::value, "!");
}

template <typename F>
void foo1(F) {
    static_assert(std::is_same<int, typename argument_of<F>::type>::value, "!");
}

template <typename F>
void foo2(F f) {
    static_assert(std::is_same<int, typename argument_of<decltype(f)>::type>::value, "!");
}

int main() {
    foo0();
    foo1(bar);
    foo2(bar);
}

See live example.

What would be the proper way if the calling style of fn() remains as given for foo1() and foo2() in the example above?

Yes, great. Would you mind making an answer out of this which also explains why this specialization is necessary? — user2525536, Mar 02 '19 at 06:28

kmdreko · Accepted Answer · 2019-03-02T07:17:51.677

You just need a specialization to handle a function pointer type as well:

template <typename Sig> struct argument_of;
template <typename R, typename Arg> struct argument_of<R(Arg)> { typedef Arg type; };
template <typename R, typename Arg> struct argument_of<R(*)(Arg)> { typedef Arg type; };

Its a small distinction, and I hope that I get my terminology right, but the symbol bar is a non-overloaded function of type void(int) whereas the variable f (or unnamed F-parameter), is a function pointer of type void(*)(int). Variables and functions are fundamentally different in C++ and the type-system has this distinction.

Its also worth noting that another variant that you may wish to specialize for is a function reference which would look like void(&)(int). A function reference type can be deduced in situations like decltype(*f) or decltype((bar)) (note the extra parenthesis).

There's also the accepted answer [here](https://stackoverflow.com/questions/34437557/difference-between-voidint-void-int) that briefly talks about the differences in terms of value categories and is worth a read. — kmdreko, Mar 02 '19 at 06:55
Good hint. So I could have done it [this](http://cpp.sh/9kruno) way as well, using `std::remove_pointer` on `F` or `decltype(f)` before resolving it with `argument_of`. — user2525536, Mar 02 '19 at 07:39

C++ extract argument from function signature

1 Answers1