Question

Member functions have an implicit this pointer parameter. Why does std::function accept this signature, then, where S is a simple class? (complete sample)

std::function<void(S &)> func = &S::foo;

Calling it works, too, and distinguishes objects:

S s1 = {5};
S s2 = {6};

func(s1); //prints 5
func(s2); //prints 6

What I’d normally expect is that it needs a pointer, which works as well: (complete sample)

std::function<void(S * const)> func = &S::foo;

S s1 = {5};
S s2 = {6};

func(&s1); //prints 5
func(&s2); //prints 6

Why does the first one work when I pass a reference into the member function when the implicit this parameter is a pointer?

Answer 1

std::function<SIG> can be constructed from many things that behave like functions, converting them to an appropriate std::function object.

In this case void S::foo() behaves much like a function void foo_x(S&) (as in they both require an S to call, and potentially modify S, returning nothing). Consequently std::function<void(S&)> provides a constructor for converting the member function into a function object. I.e.

std::function<void(S &)> func = &S::foo;

uses a constructor, something like std::function<void(S&)>( void(S::)() ), to create something equivalent to:

void foo_x(S & s ) { return s.foo(); }
std::function<void(S&)> func = foo_x;

Similarly,

std::function<void(S * const)> func = &S::foo;

is equivalent to

void foo_x(S * const s ) { return s->foo(); }
std::function<void(S* const )> func = foo_x;

through a constructor like std::function<void(S* const )>( void(S::)() ).