I don't understand why the following code compile and works:
template<typename Predicate>
void foo(Predicate p) {
}
bool g(int n) {
}
void user(int n) {
foo(g);
}
foo is supposed to get a function object that will run on a data structure but I made the method simpler, because what I don't understand is how can this works? A method isn't an object. The normal way to do it is to create a new class, override operator() and then send an instance of that class.
Well, in this case the Predicate parameter is substituted by a function pointer of type bool (*func) (int). Nothing wrong with that...
The Predicate template argument can be almost any type. So you can use it for function pointers and classes as well as the basic types.
If you use the function argument p as a function, then it can be anything that is callable, like a function pointer, an object whose class have an operator() member function, a pointer to a static member function, a std::bind object, a std::function object or a lambda expression.
It can't be a pointer to a member function though, because to call a pointer to a member function you need an instance to call it on. For this use std::bind.
std::cout << __PRETTY_FUNCTION__ << '\n';infooand see whatpreally is. See it live.std::for_eachandstd::copy_if