Help with an Array Structure..

[Arakon]

I'm just starting out with PHP, although I've had some experience with Java and C++. My Question I'm sure is Novice level, and probably really stupid.... but..

WHy doesn't this code work?

<?php 

//Variable Declaration
$myArray   = array(); 


addArrayElement("foo");
testArray($myArray);
addArrayElement("bar");
testArray($myArray);

/*
	function designed to add elements to an array
*/
	function addArrayElement($element){
		$i = count($myArray);
		$myArray[$i] = $element;
	}//End function addArrayElement($element)

/*
	function designed to test an array
*/
	function testArray($thisArray){
		print_r($thisArray);	
	}//End function testArray

?>

Thanks in Advance....

[LordShryku]

It doesn't work because addArrayElement doesn't know what the array is. Counting it is fairly useless to. In fact, using functions for these is un-needed, but I'll assume you're testing something, or this is part of something bigger. Try this instead:

<?php 
$myArray   = array(); 

$myArray = addArrayElement("foo",$myArray);
testArray($myArray);
$myArray = addArrayElement("bar",$myArray);
testArray($myArray);

function addArrayElement($element, $array){
   $array[] = $element;
   return $array;
}
function testArray($thisArray){
   print_r($thisArray);    

}
?> 

[leatherback]

Hi,

expanded your code a bit (So you can define which array to test) and added the needed (?) return structure:

<?php  

//Variable Declaration 
$myArray   = array();  


$myArray = addArrayElement("foo", $myArray); 
testArray($myArray); 
$myArray = addArrayElement("bar", $myArray); 
testArray($myArray); 

/* 
    function designed to add elements to an array 
*/ 
    function addArrayElement($element, $inArray){ 
        $i = count($inArray); 
        $inArray[$i] = $element; 
        return $inArray;
    }//End function addArrayElement($element) 

/* 
    function designed to test an array 
*/ 
    function testArray($thisArray){ 
        print_r($thisArray);     

    }//End function testArray 

?> 

[Arakon]

Ah I see, so the function cannot access a variable that isn't expressly passed to it? I was under the impression that a variable declared outside a function would be treated as a Global... hmm. Thanks.

Guess I'll have lookup on how to create/deal/manipulate Globals.

The point of the excercise was to practice OO coding on PHP. Its just a test to learn how it works differently than Java. Also Arrays are a pretty basic structure and VERY common. SO its good to know how to address them. Also making simple functions helps me track down errors easier with DEBUG statements and not have to rewrite the wheel everytime I want to use one. Just inherit code from one Project to another.

[ahundiak]

Slight variation:

<?php
    //Variable Declaration
    $myArray   = array();
    addArrayElement("foo", $myArray);
    testArray($myArray);
    addArrayElement("bar", $myArray);
    testArray($myArray);

/*
    function designed to add elements to an array
*/
/* *** NOTICE THE & *** */
function addArrayElement($element, &$inArray){
    $i = count($inArray);
    $inArray[$i] = $element;
}//End function addArrayElement($element)

/*
    function designed to test an array
*/
function testArray($thisArray){
    print_r($thisArray);
}//End function testArray

?>

Using the & in the function declaration causes a reference to the array to be passed instead of a copy of the array. This in turn enables you to directly update the content of the array.

[Arakon]

COOOL!

Now that is some majorly useful information!

Thanks!

the Power of '&' has been revealed to me, and I revel in its splendor!

I feel I should be making some religious trek of some sort now.

[Shrike]

Or if you want to use OO, then use it 🙂

class Element {

  var $myArray = array();

  // Constructor
  function element(){
    $this->addElement("foo");
    $this->testArray();
    $this->addElement("bar");
    $this->testArray();
  }

  function addElement($element){
    $this->myArray[] = $element
  }

  function testArray(){
    print_r($this->myArray);
  }
}

$element = new Element();

Passing variables by reference is kinda similar to using objects, in a roundabout way.

Sorry for any typos 😉

[toplay]

I just wanted to comment on your remark of: "I was under the impression that a variable declared outside a function would be treated as a Global"

A variable declared outside a function is potentially Global, however, you need to specify the variable(s) desired to access with a Global command as in:

$DEBUG = TRUE;

function debugOut($msg);
{
    global $DEBUG;

 if ($DEBUG) echo $msg, '<br>';

 return;
}

debugOut('Must remember to use global in a function');

The above code will produce an output of the debug message. If you remove the global command within the function, then the function won't recognize the $DEBUG variable's value and you may get unpredictable results.

Take care.

[superwormy]

Originally posted by Arakon
Ah I see, so the function cannot access a variable that isn't expressly passed to it? I was under the impression that a variable declared outside a function would be treated as a Global... hmm. Thanks.

Guess I'll have lookup on how to create/deal/manipulate Globals.



The point of the excercise was to practice OO coding on PHP. Its just a test to learn how it works differently than Java...

Interesting... variable scoping and return; statements works exactly the same in Java and C++ as it does in PHP ( minus C pointers and that crap ) which makes me wonder why you would be under this impression... but regardless, go read up on scoping in ANY programming book to get the idea.

Also, what you're doing isn't OO coding at all. Its procedural, you're writing FUNCTIONS, not OBJECTS...

Also for the record, there are very few cases you should really use references, they go against the idea of returning values and variable scoping which makes modern languages ( read: C, Java, PHP ) so great. global $var; delcarations and goto; statements fall under that same problem many times. References are there for a reason, and they should only be used in those cases when its really required. Otherwise, return $values; instead.

[ahundiak]

variable scoping and return; statements works exactly the same in Java and C++ as it does in PHP

For C/C++ variables declared outside functions are global and can be accessed inside of functions or methods by default. Very different from PHP which requires use of the global keyword (or $GLOBALS) to get at them from functions/methods.

In Java of course everything is associated with an object so the notion of global variables really does not exist.

BTW, OOP does not require using objects and methods.

[superwormy]

Welll... I suppose, but only variables declared outside of the main() function are global, and since PHP doesn't have a main() function...

How exactly does Object Oriented Programming NOT require objects...? Even then name has 'Object' in it... prorgramming using functions is usually refered to as function or procedural programming...