Can ‘auto’ be used in a function declaration?

Question

Background:

I found this handy random number generator and wanted to make a header file for it: http://www.cplusplus.com/reference/random/

std::default_random_engine generator;
std::uniform_int_distribution<int> distribution(1,6);
auto dice = std::bind ( distribution, generator );
int wisdom = dice()+dice()+dice();

However, in C++11, a function declaration with return type ‘auto’ requires a trailing return type so the compiler can decide what the type is. E.g.:

auto foo(int a, int b) -> decltype(a*b);

Problem:

It appears like my header would need to be almost as long as the function itself to determine the type:

std::default_random_engine generator;
std::uniform_int_distribution<int> distribution(1,6);
auto roll() -> decltype(distribution(generator));

Question:

Is there a way around determining the full return type for a function declaration (in a header) that uses the ‘auto’ type? If not, what should my dice() header look like?

Answer 1

Since you use int as the template type for std::uniform_int_distribution, the return type of distribution(generator) is int. Unless the real code is templated as well, then the return type could be hard-coded to int.

And if the real code is templated then you can use the result_type member of std::uniform_int_distribution:

template<typename T>
typename std::uniform_int_distribution<T>::result_type roll();

Or simply the template type itself:

template<typename T>
T roll();