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PHP Tip of the Week Archive

• If Brevity: The Ternary Operator
• Easy Database Inserts
• Output Buffering and Editing Web Pages On the Fly
• Searching Arrays
• Random Thoughts
• Faster Page Transfer Using Compression
• Persistent Values in Web Forms
• Using HEREDOC Syntax with HTML
• Calculating Dates More Easily
• Using Single and Double Quotes
• Timing Your PHP Scripts
• When Nothing is Something: Zero, Empty, Null, and Negative Values
• Nesting Instincts
• Useful (for) Characters: The Increment Operator
• The "static" keyword
• Those !#@% Characters!
• The Value of PHP Classes and OOP
• The Simple Things in (programming) Life
• Just a Passing Reference
  
  
  I�ve been looking through a lot of code written by others lately, and I�ve run across this scenario quite a bit:
  
  <?php
  
  function bumpit ($var ) {
    $var++;
    return $var;
  }
  
  $a = 5;
  $a = bumpit($a);
  print "Here is var a now: $a"; //$a is "6" now
  
  ?>
  
  That is, a function is sent the content of a variable ($a), the function modifies and returns that content, and the content is reassigned to the same variable.
  
  While that�s a perfectly allowed approach, it is not the most efficient approach, and it becomes less and less efficient the larger $a is to begin with.
  
  Why? Because the content of $a is actually copied into $var in this example. So we have two variables with exactly the same content (before $var is modified in bumpit( ) ). And that�s not terribly efficient. It would be better if we could modify $a directly within the function; we can, and here's one approach:
  
  <?php
  
  function bumpit(&$var) {
    $var++;
  }
  
  $a = 5;
  bumpit($a);
  print "Here is var a now: $a"; //$a is "6" now!
  
  ?>
  
  What happened to $a?
  
  First, it was passed into bumpit( ) by reference, indicated by the �&$var�. This means $var now points to the same content that $a does. $var is NOT a copy of $a.
  
  How does that work? We have a beloved family friend whose first name is Carol. However, we all know him as Grandpa Pinkie. Two names, same person.
  
  That�s an illustration of $a and $var in our second example above: two names, same content. Passing $a into bumpit( ) by reference does this.
  
  So when we modify $var within the function in that second example ($var++;), we are actually modifying the same content that $a points to. So $a is incremented when $var is incremented�because they both point to the same content. And we don�t even have to return $var from bumpit( ) in the second example. Just modifying $var within bumpit( ) modifies $a outside of bumpit( ).
  
  Again, if $a were a couple of megabytes of info to begin with, we�ve just saved ourselves the trouble of copying that content to another variable if we pass $a to a function by reference, like bumpit(&$var).
  
  A suggestion: It is easy to overlook that extra ampersand (&) in front of a variable name when passing it into a function. So comment your code so that you � and others after you � will be alerted to what you are doing when you pass by reference.
  
  Read more here:
  
  http://www.php.net/manual/en/language.references.pass.php
  
  
• Simple Image Creation with PHP