Member access operator and indirection operator difference

Question

When overloading, indirection operator (*obj) must return reference, but member access operator (obj->) must return pointer.

struct my_ptr
{
    some_type & operator* () { return *ptr; } // returns reference
    some_type * operator->() { return  ptr; } // returns pointer

    some_type *ptr;
};

Why is there such difference?

Answers

You can find some answers at the link above. But they doesn't completely answer why C++ was designed in such way that operator* and operator-> return different types.

https://stackoverflow.com/a/28435758/5447906

I guess this was just the only way they could think of to implement it and it turned out a bit hackish.

operator-> allows to recursively travers through multiple objects until it will meet a real pointer:

struct A {
    int foo, bar;
};

struct B {
    A a;
    A *operator->() { return &a; }
};

struct C {
    B b;
    B operator->() { return b; }
};

struct D {
    C c;
    C operator->() { return c; }
};

D d;
d->bar; // Provides access to A::bar.

Someone might consider this as a useful feature. But I personally do not. Consistency and obviousness would be a bigger advantage for me.