Re: Does passing an uninitialized array variable initialize it? (PHP 5)
- From: Jerry Stuckle <jstucklex@xxxxxxxxxxxxx>
- Date: Mon, 30 Jul 2007 21:20:15 -0400
Rik wrote:
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] = $i++;
Precedence as defined should kick in clearly. The '[' isn't in the table just for show.
The bottom line: don't change a value and use it in the same statement. Results are unpredictable.
And makes for more readable code indeed. However, if I read the documentation I fully expect the PHP4 result from my example, not the PHP5 one...
--Rik Wasmus
Rik,
The key here is, precedence defines the order of OPERATOR evaluation - but not the order of OPERAND evaluation. A subtle, but important difference in cases like this.
For instance, it defines that '++' will be evaluated before '[]', it doesn't define whether [$i] will use the old or new value of $i++. So even though [$i] is evaluated before [$i++], $i++ is evaluated first and the value of $i is unpredictable.
--
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Jerry Stuckle
JDS Computer Training Corp.
jstucklex@xxxxxxxxxxxxx
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