Jay Blanchard wrote:
[snip]
I've got some code from someone else I've inherited and need to sort out
some problems with. The programmer that wrote it originally was much
better than I and programmed a little over my head to say the least.
One function that I've come across that has 5 variables as input:
function($var1,$var2,!$var2,$var3->cc,$var3->bcc);
The question I have is on the 3rd input variable, what does the "!" in
front of $var2 do to that variable?
[/snip]
Hmmm. Looks like a weird mistake, to be sure. If $var2 is boolean it should
I'd say it looks odd - but I wouldn't charcterise it as a mistake without
seeing the function and the code that calls it.
the ! is the logical NOT operator - what Jay says is correct but I'd like to
add that $var2 would be negated regardless of whether the value of $var2 is a
boolean or not. how this is done internally has everything to do with the
php's auto-casting rules (i.e. the rules it uses to determine how/when to
automatically cast a variable of a given type to another type in order to
be able to use the value in a given context.
as an example, the following line fo code says " if $x is NOT set then do stuff":
if (!isset($x)) { /*do stuff*/ }
check here for more info on logical operators:
http://nl2.php.net/manual/en/language.operators.logical.php
also try running this code to get a feeling for what auto-casting does
in relation to the NOT operator:
// echo '<pre>' // uncomment this if running in a browser
class Test {}
class Test2 { var $v; function Test2($v) { $this->v = $v; } }
$a = 1;
$b = 0;
$c = true;
$d = false;
$e = "";
$f = "non-empty-string";
$g = array();
$h = array(1,2,3);
$i = new Test();
$j = new Test2("test");
var_dump(!$a, !$b, !$c, !$d, !$e, !$f, !$g, !$h, !$i, !$j);
pass the opposite of its current state. Is $var2 a boolean? It is being
passed twice here, once AS and once AS NOT.
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