Re: Does passing an uninitialized array variable initialize it? (PHP 5)

On Tue, 31 Jul 2007 02:38:04 +0200, Jerry Stuckle <jstucklex@xxxxxxxxxxxxx> wrote:
I take this back - the result of the operation is unpredictable. That's because:

$rank[$i] = trim($rank[$i++]);

$rank[$i++]

is evaluated as $rank[$i]. However, when

$rank[$i]

is evaluated, does it use the old or the new version of $i?

This is indeed dubious. Which of the '[' gets evaluated first? Allthough, left-associtivity might indicate the first one should be evaluated first (PHP4 behaviour).

However, for:

$array[$i] = $i++;

Precedence as defined should kick in clearly. The '[' isn't in the table just for show.

The bottom line: don't change a value and use it in the same statement.. Results are unpredictable.

And makes for more readable code indeed. However, if I read the documentation I fully expect the PHP4 result from my example, not the PHP5 one...
--
Rik Wasmus
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