C++ function call question

Question

Hi, i have a question about the function call in the following example:

int main()
{
    int a, b;
    cin  >> a >> b >> endl;
    cout << *f(a,b);
    return 0;
}

So is *f(a,b) a valid function call?

Edit:: sorry for the errors, i fixed them now i'm a little tired

Answer 1

Whatever f is, *f(a, b) attempts to apply the indirection operator to the result of f(a, b).

If f is a function pointer and you're trying to call it, while you could do this:

(*f)(a, b)

Just doing f(a, b) is simpler.

Answer 2

I don't think there is any way to tell without seeing the definition of f. C++ requires a lot of context to know what is going on.

Answer 3

It is very tough to tell if it is valid or not without knowing what 'f' is. But if it returns something that can be dereferenced it looks fine as long as that 'dereferenced' value can be printed (an overloaded operator <<) should exist for the type.

Answer 4

The code at least could be reasonable. For it to work, f must be defined as a function that returns either of two sorts of things: either it returns a pointer, in which case the * dereferences the pointer, so whatever it was pointing at gets sent to standard output. Otherwise, f must return some user-defined type that defines operator * to return something that's compatible with cout.