CODEWITHSUNDEEP
c++templatesc++14
Tomashu
Tomashu

Reputation: 543

Get function return type in template

How can I get return type for any function passed to template?
I don't know how to convert between template<typename T> and template<typename Result, typename Args...>:

template<typename T>
void print_name(T f)
{
    static_assert(internal::is_function_pointer<T>::value
            || std::is_member_function_pointer<T>::value,
            "T must be function or member function pointer.");
    typename decltype(f(...)) Result; // ???
    typename std::result_of<T>()::type Result; // ???
    printf("%s\n", typeid(Result).name());
}

void f_void() {}
int f_int(int x) { return 0; }
float f_float(int x, int y) { return 0.f; }
struct X { int f(int x, float y) { return 0; } };

int main()
{
    print_name(f_void);
    print_name(f_int);
    print_name(f_float);
    print_name(&X::f);
    return 0;
}

How can i get type Result inside function print_name?

Upvotes: 13

Views: 14217

Answers (2)

noseems
noseems

Reputation: 151

as of C++ 17, the answer is

std::invoke_result_t<T>

Upvotes: 15

skypjack
skypjack

Reputation: 50568

A possible solution is using a function declaration that extracts the return type as well as all the parameters. You don't have even to define it.
It follows a minimal, working example:

#include<typeinfo>
#include<cstdio>

template<typename R, typename... A>
R ret(R(*)(A...));

template<typename C, typename R, typename... A>
R ret(R(C::*)(A...));

template<typename T>
void print_name(T f)
{
    printf("%s\n", typeid(decltype(ret(f))).name());
}

void f_void() {}
int f_int(int x) { return 0; }
float f_float(int x, int y) { return 0.f; }
struct X { int f(int x, float y) { return 0; } };

int main()
{
    print_name(f_void);
    print_name(f_int);
    print_name(f_float);
    print_name(&X::f);
    return 0;
}

As you can see, the declarations provided for ret has the same return type of the submitted function or member function.
A decltype does the rest.

Upvotes: 29

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