Using a function pointer with a trailing return type

Question

There are some Stack Overflow users who strongly advocate always using the new C++11 trailing return type convention when writing functions, such as main()->int. I can see advantages, as it makes the notation uniform. However, when declaring a function pointer, I cannot find any way of using a trailing return form, i.e. can declare either

typedef int(*fp)(int);

or

using fp = int(*)(int);

for a function pointer taking an int and returning an int.

Is there a way of using the new trailing return syntax in declaring such a function pointer? For example, something like

using fp = (*)(int)->int;

but this doesn't compile. If not, is there a reason why the new syntax is not applicable to function pointers?

Answer 1

You have to use auto

using fp = auto (*)(int) -> int; typedef auto (*fp)(int) -> int; // alternatively 

Trailing return type syntax means you put the auto before the function name to indicate that the return type follows (or in the case of c++14, should be deduced). For function pointers, the same rule applies except that it can't deduce it (for obvious reasons).

This is true for functions as well though, your example

main() -> int {... } 

is not valid either without the preceding auto

Answer 2

You want one of these depending on whether you are using using declaration or a typedef:

using fp = auto (*)(int)->int; typedef auto (*gp)(int) -> int;