Need clarification on C++ template format

Question

I'm going through some template sample code and there's one thing I don't get.

Take a template methode:

template<class Seq, class T, class R>
void apply(Seq &sq, R(T::*f)() const) {
    typename Seq::iterator it = sq.begin();
    while(sq.end() != it) {
        ((*it++)->*f)();
    }
}

A sample class:

class MyClass {    
public:
    MyClass() {}
    void doSomething() const {
        std::cout << "doing stuff..." << std::endl;
    }
};

And the test code:

void testMyClass() {
    vector<MyClass*> v;
    for(size_t i = 0; i < 5; ++i) {
        v.push_back(new MyClass());
    }
    // call a member methode on all elements in container
    apply(v, &MyClass::doSomething);
}

I would be grateful if someone could explain me what is that class R for, as defined in the template definition?

Answer 1

class R in the template is used to deduce the return type of the function. In your case, it is deduced to be of type void .

Answer 2

class R refers to the return type of the function pointer being passed to the function apply . It is automatically deduced from the actually passed function pointer type, so you never really need to care about it when calling apply .

The implementation of apply discards the return value of the function, so you could simply force the passed function to return void :

template<class Seq, class T>
void apply(Seq &sq, void(T::*f)() const) {
    typename Seq::iterator it = sq.begin();
    while(sq.end() != it) {
        ((*it++)->*f)();
    }
}

However, now you restrict the call site to only pass such function pointers. Sadly, a pointer to a function which returns something isn't implcitly convertible to one which doesn't, although it would be pretty "intuitive".

So when you take a function pointer as an argument, and you don't care about the return type, it's better to accept "any" return type than "none".