Re: operator= function

From:
"Victor Bazarov" <v.Abazarov@comAcast.net>
Newsgroups:
comp.lang.c++
Date:
Tue, 4 Dec 2007 14:56:17 -0500
Message-ID:
<fj4bd2$p6b$1@news.datemas.de>
Kira Yamato wrote:

[..] The syntax

if (a)
{
// do something
}

is not clear what you're testing for here.


Of course not! Who in their right mind would name a variable 'a'?
But imagine that the variable is 'statusIsValid' (and its type is not
necessarily 'bool'). What prevents me from writing

    if (statusIsValid)
    {
    // do something
    }

?

 Code readability should be
highly sought for.


Sure. Don't name your variables 'a' or 'b'.

 Even with native data-types like integers, it's
better to write

       if (n != 0) ...

than just

       if (n) ...


Now, with integers it may make sense. But with a class object
I would have to define non-equality comparison to 'int' in order
for 'if (n != 0)' to compile. OTOH, if I just define operator
bool() (known as type conversion function), both 'int' _and_ my
type would work in 'if (n)' beautifully.

unless you worry about performance, at which I would think a smart
compiler would generate equivalent codes for both cases anyway.

Also, don't overload the operator bool() just to support the if-test.
You might get yourself into trouble in other situations with this
implicit conversion.


Yes, that's why some folks define 'operator void*()' instead.

My suggestion is to be explicit and have an method like
       bool isValid() const;
so that your if statement reads more explicitly

if ((a=b).isValid()) ...


Ouch. I still think that even

    if (a=b, a.isValid())

is better. But nothing beats

    a = b;
    if (a.isValid())

V
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