Use of Operator ->*

Question

Possible Duplicate:
How to overload the ->* operator?

What is the meaning of operator ->*?

and how it can be useful in overloading ?

See also http://stackoverflow.com/questions/2696864/are-free-operator-overloads-evil . I don't think this is a duplicate of the above, but not a really good question either. — Potatoswatter, Jan 03 '12 at 20:11
I was going to answer, but here is the jist — `operator->*` defines a binary operator just like `operator+`, `operator*`, etc. It has higher precedence than all the other binary operators so it is useful in forming member accesses, but lower than `operator[]`. It is widely considered to be obscure and may confuse users. — Potatoswatter, Jan 03 '12 at 20:18

score 1 · Answer 1 · answered Jan 03 '12 at 20:04

operator->* is for pointers to members.

struct foo{
  void bar(){}
};

int main(){
  typedef void (foo:*foo_memptr)();
  foo_memptr pfunc = &foo::bar;
  foo f;
  foo* pf = &f;
  (f.*pfunc)(); // on object or reference
  (pf->*pfunc)(); // on pointer to object
}

Overloading it is usually only useful for smart pointers, and even them don't do it because it's really complicated and the same functionality can be achieved by ((*pf).*pfunc)(). Scott Meyers even wrote a PDF on how to do it!

score 1 · Accepted Answer · edited May 23 '17 at 11:55

1

The ->* and .* operators are for accessing class members via pointers, see the following link for examples:

http://c-for-crogrammers.org.ua/ch22lev1sec6.html

You may find this SO answer useful as well.

edited May 23 '17 at 11:55

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answered Jan 03 '12 at 20:04

Andrew Clark

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Use of Operator ->*

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