Re: Is it possible to override raw pointer comparison?

From:
"Bo Persson" <bop@gmb.dk>
Newsgroups:
comp.lang.c++
Date:
Sun, 24 May 2009 13:22:10 +0200
Message-ID:
<77sounF1jcjm1U1@mid.individual.net>
Ash wrote:

On May 24, 6:48 am, Neelesh <neelesh.bo...@gmail.com> wrote:

On May 24, 8:26 am, Ash <ashiru...@googlemail.com> wrote:

Why is this ok:
bool operator==(Foo* a, Foo& b) { return true; }

and if I have:
Foo* f1; Foo* f2;
then
f1 == *f2
is valid and calls the above function.

but the following (to try to override equality when called using
two raw pointers: f1 == f2):
bool operator==(const Foo* a, const Foo* b) { return true; }
doesn't compile?
(g++ tells me "test2.cpp:9: error: ?bool operator==(const Foo*,
const Foo*)? must have an argument of class or enumerated type")

Is it not possible to override the comparison of two pointers (f1
== f2) or have I simply not got the exact signature right?


g++ is telling exactly what the issue is: operator overloading is
allowed only when atleast one operand is a user defined type
(either a class or an enumeration type). A "pointer" is not a user
defined type (even if it is a pointer to a class type) and hence
it is not possible to overload any operator that takes two
pointers, eg. operator== or operator= or operator- etc.


I suspected as much, but in which case I don't udnerstand why the
first case:
bool operator==(Foo* a, Foo& b) { return true; }
is ok? Where's the logic of the C++ standards body in allowing this,
which is an operation on a pointer and not when both operands are?


The logic behind this is that the comparison of two pointers is
already defined by the language. You cannot add another one.

To overload a function or operator, the set of parameters must be
different from those of existing functions and operators with the same
name.

Bo Persson

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