Daevid Vincent wrote:
<?PHP
$num = 123;
$num = $num++;
print $num; //this prints 123 and not 124 ?!!
$num = 123;
$num = ++$num;
print $num; //this prints 124 as expected
$num = 123;
$num++;
print $num; //this prints 124 as expected
$num = 123;
print $num++; //this prints 123 and not 124 ?!!
print $num++; //this NOW prints 124
?>
So then I read the manual because I think I'm loosing my mind and perhaps
it's backwards day and nobody told me:
http://us3.php.net/manual/en/language.operators.increment.php
I'm baffled as to the reasoning behind:
"$a++ :: Post-increment :: Returns $a, then increments $a by one."
Why would you EVER want $num = $num++; to give you back the value you
already had? Even if we did $foo = $bar++; I would still logically (and
common sensely) expect $foo to be the increment of $bar!
It also seems counter-intuitive, as I was always lead to believe everything
is processed on the right of an equals sign and then assigned back to the
left side of the equals sign. In this case, it does a mixture of both.
Can someone PLEASE explain why the developers of PHP chose this seemingly
whacky logic?
This logic is almost universal in modern programming languages. Some
people do the following:
<?php
$index = 1;
foreach( @items as @item )
{
echo '<tr>'
.'<td>'.($index++).'</td>'
.'<td>'.$item.'</td>'
.'</tr>';
}
?>
Some people consider it weird to start at 0 for the above index, because
the index is only used when it is 1 or more. But the equivalent of the
above is the following:
<?php
$index = 0;
foreach( @items as @item )
{
echo '<tr>'
.'<td>'.(++$index).'</td>'
.'<td>'.$item.'</td>'
.'</tr>';
}
?>
Cheers,
Rob.
--
http://www.interjinn.com
Application and Templating Framework for PHP
--
PHP General Mailing List (http://www.php.net/)
To unsubscribe, visit: http://www.php.net/unsub.php
