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Rich Mathis
3,626 PointsHung up on objects question 5 of 7
Here is the challenge...
PalprimeChecker objects have a method called isPalprime(). This method does not receive any arguments. It returns true if the number property contains a palprime, and it returns false if the number property does not contain a palprime. (Tip: 17 is not a palprime.) Turn the second echo statement into a conditional that displays “is” or “is not” appropriately. The conditional should call the isPalprime method of the $checker object. If the isPalprime method returns true, then echo “is”; otherwise, echo “is not”.
This is what I have and I can't figure out what the issue is.
<?php
include ('class.palprimechecker.php');
$checker = new PalprimeChecker;
$checker->number = 17;
echo "The number " . $checker->number . " ";
if ($checker->number(isPalprime)) {
echo "is";
}
else {
echo"is not";
}
echo " a palprime.";
?>
2 Answers
Bella Ratmelia
Front End Web Development Techdegree Graduate 28,947 PointsThe question said "PalprimeChecker objects have a method called isPalprime()" $checker is a PalprimeChecker object. Therefore, to call the method that belongs to that object, you can simply do it like this:
$checker->isPalprime()
Therefore your code should be:
if ( $checker->isPalprime() ) {
echo "is";
}
else {
echo"is not";
}
Hope this helps!
Rich Mathis
3,626 PointsThanks! I was staring at that for a long time and it finally makes sense.