Question

It is said that the arrow operator is applied recursively. But when I try to execute the following code, it prints gibberish when it is supposed to print 4.

class dummy
{
public:
    int *p;

    int operator->()
    {
        return 4;
    }
};

class screen 
{
public:
    dummy *p;

    screen(dummy *pp): p(pp){}
    dummy* operator->()
    {
        return p;
    }
};

int main()
{
    dummy *d = new dummy;
    screen s(d);
    cout<<s->p;
    delete d;
}

Answer 1

What Stanley meant by “recursive” is just that the operator is applied to every returned object until the returned type is a pointer.

Which happens here on the first try: screen::operator -> returns a pointer. Thus this is the last call to an operator -> that the compiler attempts. It then resolves the right-hand sice of the operator (p) by looking up a member in the returned pointee type (dummy) with that name.

Essentially, whenever the compiler finds the syntax aᵢ->b in code, it essentially applies the following algorithm:

1. Is aᵢ of pointer type? If so, resolve member b of *aᵢ and call (*aᵢ).b.
2. Else, try to resolve aᵢ::operator ->
   1. On success, set aᵢ₊₁ = aᵢ::operator ->(). Goto 1.
   2. On failure, emit a compile error.

I’m hard-pressed to come up with a short, meaningful example where a chain of operator -> invocations even makes sense. Probably the only real use is when you write a smart pointer class.

However, the following toy example at least compiles and yields a number. But I wouldn’t advise actually writing such code. It breaks encapsulation and makes kittens cry.

#include <iostream>

struct size {
    int width;
    int height;
    size() : width(640), height(480) { }
};

struct metrics {
    size s;
    size const* operator ->() const {
        return &s;
    }
};

struct screen {
    metrics m;
    metrics operator ->() const {
        return m;
    }
};

int main() {
    screen s;
    std::cout << s->width << "\n";
}