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Re^2: Warning while using ternary operator

by kcott (Abbot)
on Dec 16, 2013 at 15:19 UTC ( #1067344=note: print w/ replies, xml ) Need Help??


in reply to Re: Warning while using ternary operator
in thread Warning while using ternary operator

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'"; }

Output:

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

-- Ken


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Re^3: Warning while using ternary operator
by AnomalousMonk (Monsignor) on Dec 16, 2013 at 23:15 UTC

    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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