Reply to arrays as function arguments

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Posted by Sam Waller on 09/09/07 18:56

I'm trying do do something like this:

$a = array();
func($a);
printf("array size in main = %d\r\n", count($a));

/////////////////////////////////////


function func($_a) {
array_push($_a, 22);
printf("array size in func = %d\r\n", count($_a));
}


This prints:
array size in main = 0
array size in func = 1

It looks like the data is no longer available in the calling function. Is there a way to get this to
work so that I have the computed array available in the calling function?

thanks,
Sam

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