Posted by Chris on 05/05/05 04:09

I don't know what to tell you, that code should work (it does work,
exactly as you gave it in my PHP 5).

Chris

Fred Rathke wrote:

> Hello,
>
> how can a function get a pointer to an array? This does not work. I
> use PHP4.
>
> $t = array("test" => "unchanged");
> echo "<br>testarray unchanged:\"".$t['test']."\"";
> changearray($t);
> echo "<br>testarray hopefully changed:\"".$t['test']."\"";
>
> function changearray(&$myarray) {
> $myarray['test'] = "changed";
> }
>
> Before I tried it on my own I read this page:
> http://de2.php.net/manual/en/language.references.whatdo.php
>
> Be so nice to search for this string: "The second thing references do
> is to pass variables by-reference. This is done by making a local
> variable in a function and a variable in the calling scope reference
> to the same content. Example:"
>
> The following example I used. I only tried to do it with an array
> instead of a variable.
>
> What I need to read again to find my own mistake? I know some of php's
> commands work with an internal copy of a content.
>
> Thanks for your help and advice.
>

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