php: property of custom type

Question

I am trying to write the following code, though it is wrong it will probably demonstrate best why I am trying to do.

class myClass
{
    private $name = "";
    private $startAddress = new myAddress();  // this is the issue
    private $endAddress = new myAddress();    // this is the issue
}

How can I accomplish this properly?

Thank You

Mike Lewis · Accepted Answer · 2011-04-05 23:57:45Z

5

 <?php
class myClass {
   private $endAddress;
   public function __construct() {
       $this->endAddress = new myAddress();
   }
 }

Take advantage of the constructor, which is called every time you create a new object.

edited Apr 5, 2011 at 23:57

answered Apr 5, 2011 at 23:50

Mike Lewis

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3 Comments

KingCrunch Over a year ago

@alex: If you omit the visibility, public is assumed.

alex Over a year ago

@KingCrunch I like to show it. Otherwise it feels like I'm still using PHP4 :P

user401090 Over a year ago

Thank you all very much! So speedy with the response! This was a big help. Thanks again! :)

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