Can someone help me understand why Perl does this with a Pre-Increment: 

my $i = 0;
my ($j, $k) = (++$i, ++$i); #Pre-increment
print "\$j: $j\t\$k: $k\n";

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which outputs: 
$j: 2    $k: 2

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When what I actually expected was: 
$j: 1    $k: 2

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Whereas Perl outputs the following with the following code 

my $i = 0;
my ($j, $k) = ($i++, $i++); #Pre-increment
print "\$j: $j\t\$k: $k\n";

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Which, as expected, outputs: 
$j: 0    $k: 1

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According to the Camel:

The ++ and -- operators work as in C. That is, when placed before a variable, they increment or decrement the variable before returning the value, and when placed after, they increment or decrement the variable after returning the value.
To my surprise, a similar test program in a C++ IDE netted the exact same result. Wow...OK, fair enough.

I can accept this is and it just means I'll have to expand my code a few extra lines to accommodate. I would like to know the why of it though. Why does both ++$i variables get evaluated before being fed into print whereas $i++ is evaluated in sequence? What's the logic behind that?

In reply to Pre vs Post Incrementing variables by SavannahLion

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