How does the template parameter of std::function work? (implementation)

Question

In Bjarne Stroustrup\'s home page (C++11 FAQ):

struct X { int foo(int); };

std::function f;
f = &X::foo; //pointer to membe         


        

Answer 1

To answer the question in the title. The parameter that std::function uses is a nice trick to pass many type parameters as a single template parameter. Those arguments being the argument types and the return type of a function.

It turns out that std::function tries to type-erase a general functor but that is just coincidence.

As a matter of fact, once upon a time there were compilers that wouldn't accept such tricks and the boost::function precursor had a portable syntax by which all the parameters could be passed separately:

Preferred syntax

boost::function sum_avg;

Portable syntax

boost::function4 sum_avg;

https://www.boost.org/doc/libs/1_68_0/doc/html/function/tutorial.html#id-1.3.16.5.4

So that's how the template parameters of std::function work, at the end it is just a trick to make a lot of parameters look like a function call. Function pointers to that type of function are not necessarily involved in the class.