Re: Is it possible to override raw pointer comparison?

"Alf P. Steinbach" <>
Sun, 24 May 2009 21:08:04 +0200
* blargg:

Ian Collins wrote:

Ash wrote:

On May 24, 6:48 am, Neelesh <> wrote:

On May 24, 8:26 am, Ash <> wrote:

Why is this ok:

    bool operator==(Foo* a, Foo& b) { return true; }

and if I have:

    Foo* f1; Foo* f2;


    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?


operator overloading is
allowed only when atleast one operand is a user defined type (either a
class or an enumeration type).


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?

Foo& is a user defined type.

(to elaborate further) and operator == has no built-in meaning for
user-defined types, thus you can provide one.

Ah, well, we have a terminology problem. Or rather, the standard has one. I
don't feel like listing and discussing here all the umpteen nowhere-defined
terms that the standard uses for various groups of types, but suffice it to say
that it's an inconsistent mess.

However, even though "user-defined type" is pretty unclear, note that an
enumeration type has a built-in meaning for ==.

I think we need a term such as "customizable type", which then is an enumeration
or class type. And that we need to be clear that "built-in type" doesn't mean
whatever (very unclear) it means in the standard. But means an actual built-in
type! <g>

Then such issues can be discussed.

I think I'll start adopting that terminology.

Cheers, & hth.,

- Alf

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