Re: Opposite of ! operator.

"" <>
Sat, 22 Nov 2008 20:26:35 -0800 (PST)
On Nov 22, 2:38 am, Alan Johnson <> wrote: wrote:

If I have:

class Something {
  bool operator ! () const;

Then I can do:

Something s(...);
if (!s) ...;

My question is, what operator do I need to overload if I want to be
able to define the behavior of this instead:

Something s(...);
if (s) ...; // <-- notice no !

Sorry if the post title was stupid, I couldn't think of a better one.


As others have mentioned the straightforward approach is to implement
operator bool or operator void *. Each of these has some type
conversion drawbacks that aren't immediately obvious. Safe solutions
exist but it gets complicated quickly. See:


Thanks guys, that was some interesting info. I've gone with operator

I have another question. What is the purpose of being able to overload
the unary ! operator? It seems like it would have been less ambiguous
if the rule was convert to bool implicitly then use ! on the resulting
bool, rather than making ! available as an overloadable operator. The
reason I ask is it seems like it's clearer in all cases to just
implement operator bool and not implement operator !.

For example, here is something weird:

class Ambiguous {
    operator bool () const { return true; }
    bool operator ! () const { return true; }

And then I can do:

  Ambiguous a;
  assert(a && !a); // doesn't read well, does it?

Another weird thing is that it seems that it will use the bool
conversion operator by default if you don't have a ! operator defined:

class A {
    operator bool () const { cout << "hi"; return true; }

Evaluating "!A();" would print "hi".

So I guess my question is: When would you ever want to override ! but
not bool?


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