Better expression for $x = !!$y||0

[borisz]

Hi, I search for a better/shorter/more_readable way to set a var $x to exactly 0 or 1 depending on another var.
For example:

if ( $y ) {
  $x = 1;
  ...
} else {
 $x = 0;
}

I rewrote this to:

if ( $x = !!$y || 0 ) {
 ...
}

[pelagic]

$x = $y ? 1 : 0;
and: timtowtdi ...

[salva]

and: timtowtdi ...

$x = int !!$y;

[merlyn]

$x = int !!$y;

There's no promise in the docs that the result of "!" will be "1" and "undef". It could be "37" for "a true value".

[reasonablekeith]

quote:

I rewrote this to:

if ( $x = !!$y || 0 ) {
 ...
}

...

I'm sorry, but that's just awful. I'd never guess what you were actually trying to do here if I stumbled across this code, regardless of how correct it might be.

In my opinion there's not much wrong with your first example, and although the conditional operator does add some brevity, the if else construct is almost certainly more readable to more people.

[Anonymous Monk]

I'd go for $y ? 1 : 0. Other suggestions, like !!$y || 0 or 1 - !$y may not return 0 or 1 if $y has overload magic.

It also doesn't require the reader to know that negating a false value returns 1 (and not some other true value).

[ysth]

Shorter: $x = 1-!$y

Better, IMO, but much depends on what the reader knows about x and y from the surrounding code:

  $x = 0;
  if ($y) {
     $x = 1;
     ...
  }

[merlyn]

Like others in this thread, you too are presuming the sanity of the ! return. In fact, ! can return 37 for true. There's no promise in the Perl docs otherwise.

[ysth]

I said shorter, I didn't say accurate :). But really, if you presume an overloaded ! that returns something other than perl's true and false values (and that's presuming an awful lot) there's nothing you can do to $y besides examine it in boolean context with if or ?:.

[salva]

for maximum readibility and if you are doing this kind of conversion in more than one place, you can also create a new "booleanize" operator:

sub booleanize ($) { shift ? 1 : 0 }

$x = booleanize $y;

[prasadbabu]

One way.

$y?$x=0:$y=1;

update:$x=$y?0:1 is correct as ambs mentioned.

[ambs]

Not properly this (typo?). The other reply is correct ($x=$y?0:1)

[ghenry]

See the Conditional Operator, to understand the above better.

[borisz]

Thanks all, sure I know about the conditional. It just look unnatural in a if statement to me. Thats why I used || in the first place. But since all answers are more or less the same. I shouldn't be so picky.

[ikegami]

Since you don't like seeing the conditional operator in the if, why don't you move it to the outside?

$x = $y ? 1 : 0;

if ($x) {
   ...
}

It has the added advantage of removing an assignement from an if's condition, something I find quite unnatural and unreadable.