Posted by Bit Byte on 10/10/06 19:35

Colin Fine wrote:

> Robin wrote:
>
>> Bit Byte wrote:
>>
>>> 1). Is the function signature correct (valid syntax?)
>>
>>
>> Depends on your PHP-Version, i.e.:
>>
>> PHP 4:
>> function foo(&$arr){}
>>
>> PHP 4 knows nothing about visibilty and always passes by value, hence
>> the '&' to signal a pass-by-reference. Nice text about using objects
>> and references in PHP 4:
>> http://www.obdev.at/developers/articles/00002.html
>>
>> PHP 5:
>> public function foo(array $arr){}
>>
>> PHP 5 always passes by reference, so no '&' is needed.
>>
>> Cheers
>> Robin
>
>
> I don't believe this is correct.
>
> http://www.php.net/manual/en/functions.arguments.php says:
>
> "By default, function arguments are passed by value (so that if you
> change the value of the argument within the function, it does not get
> changed outside of the function). If you wish to allow a function to
> modify its arguments, you must pass them by reference.
>
> "If you want an argument to a function to always be passed by reference,
> you can prepend an ampersand (&) to the argument name in the function
> definition:"
>
> I think that what you are thinking of is that in PHP5 *objects* are
> always passed by reference, but arrays are not objects.
>
> Therefore the & *is* still needed in PHP5.
>
> Furthermore, I can't find where it says so, but I'm pretty sure that the
> PHP5 extension that allows you to specify a type for a function argument
> is limited to objects.
>
> If I am right, then :
>
> for an array in PHP5
>
> public function foo(&$arr) {
>
> but if you use an object class MyArray
>
> public function foo (MyArray $arr) {
>
> Colin
>
>
>

Thanks all. This is a very detailed (and very useful) reply.

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