Question

I’m used to seeing syntax like this for function pointers

int (*pointer_name) (float, char *);
void call_function (void (*)(int), int);

In some C++03 functional libraries I see types used this way:

abc::function<void(*)(int,float)> f;

In C++11’s std::function I see the type given this way

std::function<void(int,float)> f;

There is a missing (*). Why?

The C++03 function<T> has T being an identical type to the corresponding function pointer. It’s easy to imagine the implementation.

std::function in C++11 is supported by core language enhancements. Have template argument types been extended to accomodate callability?

Answer 1

std::function (and its inspiration, boost::function) does not only store function pointers. It can also store function objects. In that sense, passing a function signature as a template parameter is similar to how a smart pointer usually take the type of the pointee as a template parameter, not a pointer type!

Contrast:

int* p; // indirection to an object of type int
std::unique_ptr<int> q; // indirection to an object of type int

with

typedef void signature_type(); // a function type

// indirection to something callable with signature_type as a signature
// i.e. f() has type void
// only work for freestanding functions however
signature_type* f;

// indirection to something callable with signature_type as a signature
// i.e. g() has type void
// not restricted to function pointers!
std::function<signature_type> g;

This is a useful convention.