Reply to Re: why this error?

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Posted by farrishj@gmail.com on 05/28/07 21:55

On May 28, 4:36 pm, vinnie <centro.ga...@gmail.com> wrote:
> This is my first "class" algorithm: when i execute it, i get this
> error:
>
> <<Fatal error: Call to undefined function: sum() in d:\class.php on
> line 20>>
>
> what's wrong? The class has the function, so why it says undefined?
>
> <?php
> print("first time class");
> class somma
> {
> var $uno=10;
> var $due=15;
> var $tre=15;
> function sum($uno, $due, $tre)
> {
> $totale = $uno + $due + $tre;
> return($totale);
> }
> function sum_2()
> {
> $totale = $this->sum($this->uno, $this->due, $this->tre);
> return $totale;
> }
> }
> $sum=sum(34,40);
> print("$sum");
> ?>

When using objects, you have to "create" them using the "new" keyword.
So consider a class a demarcation of what makes up an object, with
references to that demarcation being stored in the $variable that was
used to create the object.

Consider:

<code>
<h4>Somma object test class</h4>
<p>Print a sum from the Somma class</p>
<pre>
<?php

class Somma {
var $totale = 0;
var $tre = 15;
// Php4 constructor, "makes" object and
// returns a reference to the object in
// memory, to be stored in the $variable
// that is = new Somma($uno, $due);
function Somma($uno, $due) {
$this->totale = $uno + $due + $this->tre;
}
}
// $variable stores reference to object in
// memory
$variable = new Somma(34,40);
echo $variable->totale . "\n\n";
// Let's look at the object
print_r($variable);

?>
</pre>
</code>

Note the use of $this-> and $variable in the different scopes ($this->
is used inside the class, $variable is used to refer to the object's
methods and properties, which are called $variable->Somma(85,98);

[Back to original message]


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