c++ - Function using alias -

c++ - Function using alias -

what next code declare;

using f1 = void(int);

i know following;

using f2 = void(*)(int); using f3 = void(&)(int);

f2 pointer function , f3 reference.

what it?

it's function type. when declare function, such as:

void func(int);

its type not pointer nor reference. above function's type void(int).

we can "prove" using type traits follows:

void func(int) {} int main() { std::cout << std::is_same<decltype(func), void(int)>::value << '\n'; std::cout << std::is_same<decltype(func), void(*)(int)>::value << '\n'; std::cout << std::is_same<decltype(func), void(&)(int)>::value << '\n'; }

live demo

the above code homecoming true first row.

is same pointer or reference?

no, function lvalue can implicitly converted function pointer per:

§4.3/1 function-to-pointer conversion [conv.func]

an lvalue of function type t can converted prvalue of type “pointer t.” result pointer function.

the relationship between function type a(args...) , reference (namely a(&)(args...)) same relationship between type t , reference (namely t&).

where's used?

it's used template parameter.

for illustration std::function takes function type stored within std::function object , can declare such object with:

std::function<void(int)> fn;

c++ c++11 using