Simplest way to determine return type of function

Question

Given a very simple, but lengthy function, such as:

int foo(int a, int b, int c, int d) {
    return 1;
}

// using ReturnTypeOfFoo = ???

W

Answer 1

I don't know if is the simplest way (if you can use C++17 surely isn't: see NathanOliver's answer) but... what about declaring a function as follows:

template <typename R, typename ... Args>
R getRetType (R(*)(Args...));

and using decltype()?

using ReturnTypeOfFoo = decltype( getRetType(&foo) );

Observe that getRetType() is only declared and not defined because is called only a decltype(), so only the returned type is relevant.

Answer 2

Most simple and concise is probably:

template <typename R, typename... Args>
R return_type_of(R(*)(Args...));

using ReturnTypeOfFoo = decltype(return_type_of(foo));

Note that this won't work for function objects or pointers to member functions. Just functions, that aren't overloaded or templates, or noexcept.

But this can be extended to support all of those cases, if so desired, by adding more overloads of return_type_of.

Answer 3

You can leverage std::function here which will give you an alias for the functions return type. This does require C++17 support, since it relies on class template argument deduction, but it will work with any callable type:

using ReturnTypeOfFoo = decltype(std::function{foo})::result_type;

---

We can make this a little more generic like

template<typename Callable>
using return_type_of_t = 
    typename decltype(std::function{std::declval<Callable>()})::result_type;

which then lets you use it like

int foo(int a, int b, int c, int d) {
    return 1;
}

auto bar = [](){ return 1; };

struct baz_ 
{ 
    double operator()(){ return 0; } 
} baz;

using ReturnTypeOfFoo = return_type_of_t<decltype(foo)>;
using ReturnTypeOfBar = return_type_of_t<decltype(bar)>;
using ReturnTypeOfBaz = return_type_of_t<decltype(baz)>;