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7,067 PointsCreate 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
17,146 PointsHi 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');
$checker = new PalprimeChecker();
echo "The number ## ";
echo "(is|is not)";
echo " a palprime.";
?>
Another resource you could use to understand it better is: http://teamtreehouse.com/library/objectoriented-php-basics
Hope this helps.
MUZ140286 Ronald Mutsikwi
14,788 Points$checker = new PalprimeChecker();
koltton fox
2,576 Pointswhy does include need parenthesis?
koltton fox
2,576 Pointskoltton fox
2,576 PointsThis code did not pass in my browser