return by reference and function return

userfriendly311

Is it possible to have ...

function foo ($string_name, &$offset)
{
$offset += 10;
return $some_string;
}

$offset = 0;

$new_string = foo ($old_string, $offset);

... where i get a new string returned by functioon return, and an integer returned by reference?

Anon

why would you want to return the offset?
Just pass $offset into the function by value and you'll be fine...
if not, stuff them into and array and return that...

Anon

I don't know exactly what your question is. But your notation should be possible and the result would be:

$offset = 10
and
$new_string = $some_string

hope it helps.
john

userfriendly311

I'm parsing an HTML file and it is split into "sections". I wan to process these sections and it is a repetative task. So what I want to do is Pull out the info I need (return from the function), and get the offset (points to the end of the "section" just processed). That way I can just call the function again and go to the new section.

Bobby wrote:

why would you want to return the offset?
Just pass $offset into the function by value and you'll be fine...
if not, stuff them into and array and return that...

userfriendly311

I just threw the example together to try and explain what i was trying to do.

The function parses a piece of HTML, returns some values, and I want it to return the point at which it stoppped processing.

Anon

It should be possible to do what you want....
in the function call - if you say:
pseudocode - ecuse...

$offset = 0;

In loop.....{
$newstring .= foo($old_string,&$offset)
}

as foo modifies $offset, the function will use the new value of $offset, which was modified by function foo because of pass-by-reference