Conditional aka ternary operator in perlop.

Ahhh, thank you!


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"Ask not what you can do for your country. Ask what's for lunch."
-- Orson Welles

<condition> ? <do if condition was true> : <do if it wasn't>
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$x == 1 ? print "x was '1'" : print "x was something else";
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$x == 1 ? print "x was '1'" : print "x was something else";

While that's technically correct it doesn't take advantage of the fact that the ternary operator has a return value; whatever expression is evaluate after the ?.

See another adaptation that takes advantage of this feature.

print( "x was " . $x == 1 ? "'1'" : "someting else" );
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Dave

take advantage of the fact that the ternary operator has a return value

How true; Perl is a bit unorthogonal in this respect, in that there is (at least) one case where you can achieve this effect with a plain 'if':

sub lessthan {
  if($_[0] < $_[1]) {1} else {0}
}
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Some languages (old examples: Algol60 and Algol68) are more regular in this respect...

BTW, does someone know by chance which was the first language which used ?: to represent "ternary if"? Of course you had this already in LISP or BCPL, but with a different syntax. The earliest occurance of ?: I know of, is in C, but I wouldn't be surprised if it was used in even older languages already.

-- 
Ronald Fischer <ynnor@mm.st>

Thanks for the explanation! I thought it might be something like that.


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"Ask not what you can do for your country. Ask what's for lunch."
-- Orson Welles

Conditional Operator

Ternary "?:" is the conditional operator, just as in C. It works much 
+like an if-then-else. If the argument before the ? is true, the argum
+ent before the : is returned, otherwise the argument after the : is r
+eturned. For example:

    printf "I have %d dog%s.\n", $n,
            ($n == 1) ? '' : "s";

Scalar or list context propagates downward into the 2nd or 3rd argumen
+t, whichever is selected.

    $a = $ok ? $b : $c;  # get a scalar
    @a = $ok ? @b : @c;  # get an array
    $a = $ok ? @b : @c;  # oops, that's just a count!

The operator may be assigned to if both the 2nd and 3rd arguments are 
+legal lvalues (meaning that you can assign to them):

    ($a_or_b ? $a : $b) = $c;

Because this operator produces an assignable result, using assignments
+ without parentheses will get you in trouble. For example, this:

    $a % 2 ? $a += 10 : $a += 2

Really means this:

    (($a % 2) ? ($a += 10) : $a) += 2

Rather than this:

    ($a % 2) ? ($a += 10) : ($a += 2)

That should probably be written more simply as:

    $a += ($a % 2) ? 10 : 2;
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(Alas, if only google used regexps.)

(Google code search does.)