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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

In reply to Re^2: Warning while using ternary operator by kcott
in thread Warning while using ternary operator by simonz

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