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Re: Warning while using ternary operator

by Anonymous Monk
on Dec 16, 2013 at 13:42 UTC ( #1067329=note: print w/replies, xml ) Need Help??

in reply to Warning while using ternary operator

The ternary operator is not a shorthand way of writing an if block. It's a shorthand way of writing an if block that returns a value.
if ($expression) { $var = "foo"; } else { $var = "bar"; } # WRONG $expression ? $var = "foo" : $var = "bar"; # correct $var = $expression ? "foo" : "bar";

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Re^2: Warning while using ternary operator
by kcott (Chancellor) on Dec 16, 2013 at 15:19 UTC

    This is bad information.

    The expression

    $expression ? $var = "foo" : $var = "bar";

    is only "# WRONG" because of operator precedence (see perlop: Operator Precedence and Associativity). Here's how it is parsed:

    $ perl -MO=Deparse,-p -e '$expression ? $var = "foo" : $var = "bar";' (($expression ? ($var = 'foo') : $var) = 'bar'); -e syntax OK

    This is explained in perlop: Conditional Operator. Correcting the precedence with parentheses:

    $ perl -MO=Deparse,-p -e '$expression ? ($var = "foo") : ($var = "bar" +);' ($expression ? ($var = 'foo') : ($var = 'bar')); -e syntax OK

    I do not dispute that '$var = $expression ? "foo" : "bar";' is a better way to write '$expression ? ($var = "foo") : ($var = "bar");'. In fact, the documentation also recommends this.

    Once corrected as indicated, here's how your "# WRONG" code behaves:

    #!/usr/bin/env perl -l use strict; use warnings; for my $expression (0, 1) { my $var = ''; $expression ? ($var = 'foo') : ($var = 'bar'); print "\$expression = '$expression'; \$var = '$var'"; }


    $expression = '0'; $var = 'bar' $expression = '1'; $var = 'foo'

    -- Ken

      But what does the syntactically correct statement
          $expression ? $var = "foo" : $var = "bar";
      actually do?

      >perl -wMstrict -le "for my $expression (0, 1) { my $var = ''; $expression ? $var = 'foo' : $var = 'bar'; print qq{if \$expression is $expression, \$var is '$var'} } " if $expression is 0, $var is 'bar' if $expression is 1, $var is 'bar'

      It seems the statement is just an expensive equivalent to the statement
          $var = 'bar';
      The reason why is left as a exercise for simonz (after the recommended perusal of perlop :)

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