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PHP

Create a variable called $checker and assign it a new PalprimeChecker object as the value.

Wrapping Up The Project

This third-party library gives you access to create a PalprimeChecker object. Create a variable called $checker and assign it a new PalprimeChecker object as the value.

please explain this question.

3 Answers

Robert Walker
Robert Walker
17,146 Points

Hi Nostro,

If you are confused I suggest just watching the videos again just to help clear things up, even if you have to watch it a couple of times. You could also check the PHP documentation to better understand what is happening.

In short:

PalprimeChecker is a class and to make a new object you need to do:

<?php

$checker = new PalprimeChecker();

?>

As you can see, you must first create a variable and then use the operand "new" followed by the class name and ();

The code below will pass the task but if you are still unsure about what is going on it is always best to go back over it.

<?php
include('class.palprimechecker.php');


$checker = new PalprimeChecker();


echo "The number ## ";
echo "(is|is not)";
echo " a palprime.";

?>

Another resource you could use to understand it better is: http://teamtreehouse.com/library/objectoriented-php-basics

Hope this helps.

This code did not pass in my browser

$checker = new PalprimeChecker();

why does include need parenthesis?