Can someone please explain to me a complicated function pointer type in C++

Question

Can anyone tell me what is the type of the parameter for the function f?

int f(void (*(int,long))(int,long)) {} 

I am getting a similar type to this in when trying to compile some variadic template heavy code (my own wrapper around std::thread)...

Answer 1

The declaration

int f(void (*(int,long))(int,long)) {} 

declares a function f returning int and taking as argument a pointer to a function that takes int, long parameters and returns a pointer to a function that returns void and takes parameters int, long. Using a typedef for the innermost function pointer, this becomes more readable:

typedef void (*fptr)(int, long); int f(fptr(int, long)); 

Or with a named parameter,

int f(fptr handler(int, long)); 

This is perfectly valid code, but it is odd to see in compiler output because it uses a special syntax rule: in a function parameter list, a function type declarator declares a function pointer parameter. That is to say,

int f(fptr   handler (int, long)); // is equivalent to int f(fptr (*handler)(int, long)); 

...and you'd expect the compiler to use the lower, general form.

Answer 2

It's a function taking a pointer to a function that takes int and long as parameters and returns a function taking int and long as parameters and returns void. Probably a lot clearer if you use a trailing return type and name the function:

int f(auto g(int, long) -> void (*)(int, long));