Why member-of-pointer at operator() doesn't work?

Question

Why work the function call operator not together with a member-of-pointer operator?

struct F
{
    void operator()(){}
};

int main()
{
    F * f = new F;
    f->();   // Why this doesn't compile?
}

Answer 1

I guess the C++ gods simply thought the syntax was too obscure. The correct syntax is either:

f->operator()();

Or:

(*f)()

If the -> operator isn't overloaded then f-> is roughly equivalent to (*f). , (*f).() doesn't compile either, you would have to use:

(*f).operator()();

Answer 2

您应该将其称为f->operator()()或(*f)() 。

Answer 3

I don't think there's any specific reason.

Both -> and . expect the name of a member to follow, and you haven't provided one.

There's just no need to create a special case in the grammar to permit that. It's complicated enough as it is!

(*f)() is the way to go.

Sorry if that's brief and unfulfilling, but if you're looking for some deep meaning here I think you're bound to be disappointed!