Solved

Trying to make sense of $stack[count($stack)] = &$tree;

Posted on 2008-10-20
4
224 Views
Last Modified: 2012-05-05
I'm trying to roll an XML file into a PHP array using Expat Parser, and I'm relatively new to PHP.  I came across the following code: http://us.php.net  /xml_set_element_handler.  (The sample code is about 1/3 the way down the page, author: "rubentrancoso at gmail dot com".)

I'm struggling with the last line of code in the following code block:

$file = "flow/flow.xml";
$depth = 0;
$tree = array();
$tree['name'] = "root";
$stack[count($stack)] = &$tree;

1) As I understand it, count returns the number of elements in the array, but there are no elements in the array, because $stack was just declared.  Right?
2) What's up with &$tree?  

(I have also attached the full sample of the code.)  
<?php
 
$file = "flow/flow.xml";
$depth = 0;
$tree = array();
$tree['name'] = "root";
$stack[count($stack)] = &$tree;
 
function startElement($parser, $name, $attrs) {
   global $depth;
   global $stack;
   global $tree;
  
   $element = array();
   $element['name'] = $name;
   foreach ($attrs as $key => $value) {
        //echo $key."=".$value;
        $element[$key]=$value;
    }
 
   $last = &$stack[count($stack)-1];
   $last[count($last)-1] = &$element;
   $stack[count($stack)] = &$element;
 
   $depth++;
}
 
function endElement($parser, $name) {
   global $depth;
   global $stack;
 
   array_pop($stack);
   $depth--;
}
 
$xml_parser = xml_parser_create();
xml_set_element_handler($xml_parser, "startElement", "endElement");
if (!($fp = fopen($file, "r"))) {
   die("could not open XML input");
}
 
while ($data = fread($fp, 4096)) {
   if (!xml_parse($xml_parser, $data, feof($fp))) {
       die(sprintf("XML error: %s at line %d",
                   xml_error_string(xml_get_error_code($xml_parser)),
                   xml_get_current_line_number($xml_parser)));
   }
}
xml_parser_free($xml_parser);
$tree = $stack[0][0];
echo "<pre>";
print_r($tree);
echo "</pre>";
?>

Open in new window

0
Comment
Question by:jdana
[X]
Welcome to Experts Exchange

Add your voice to the tech community where 5M+ people just like you are talking about what matters.

  • Help others & share knowledge
  • Earn cash & points
  • Learn & ask questions
  • 2
  • 2
4 Comments
 
LVL 15

Expert Comment

by:MMDeveloper
ID: 22758179
stack[count($stack)] is the same as

$stack[] =

it simply says that the following value will be appended to the end of the array.


&$tree is a reference pointer per say. That means that whatever changes are made to the value in $stack[count($stack)] are replicated automatically to the original $tree value.


If I set this

$a = 5;
$b = &$a;

$b = 0;

now both $b and $a = zero. What happens to $b also happens to $a automatically.
0
 

Author Comment

by:jdana
ID: 22758813
Does $b = &$a; set $b equal to $a BY REFERENCE as opposed to BY VALUE?
0
 
LVL 15

Accepted Solution

by:
MMDeveloper earned 250 total points
ID: 22758924
you are correct.
0
 

Author Closing Comment

by:jdana
ID: 31507826
Thanks!
0

Featured Post

Free Tool: IP Lookup

Get more info about an IP address or domain name, such as organization, abuse contacts and geolocation.

One of a set of tools we are providing to everyone as a way of saying thank you for being a part of the community.

Question has a verified solution.

If you are experiencing a similar issue, please ask a related question

These days socially coordinated efforts have turned into a critical requirement for enterprises.
Password hashing is better than message digests or encryption, and you should be using it instead of message digests or encryption.  Find out why and how in this article, which supplements the original article on PHP Client Registration, Login, Logo‚Ķ
The viewer will learn how to count occurrences of each item in an array.
The viewer will learn how to look for a specific file type in a local or remote server directory using PHP.

626 members asked questions and received personalized solutions in the past 7 days.

Join the community of 500,000 technology professionals and ask your questions.

Join & Ask a Question