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PHP Build a Simple PHP Application Wrapping Up The Project Objects

Nthulane Makgato
Nthulane Makgato
Courses Plus Student 19,602 Points

Wrapping up the website task 5/7

If there ever was a code challenge that rubbed me up the wrong way, this would be it. here is the question:

"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”."

$checker = new PalprimeChecker();

$checker -> number = 17;

echo "The number " . $number;

if($checker-> isPalprime()) {
    echo "is ";
} else {
    echo "is not ";
} else {
    echo "a palprime. ";

This is my best effort yet it doesn't pass.

  1. What I want to know is how can "isPalprime" not receive arguments yet check if a number is a palprime?

  2. how is "isPalprime" checking if $checker is a prime through "($checker->isPalprime)" because it looks like $checker is assigning a property.

  3. How does "isPalprime" extract 17 from $checker since $number is just a property of the object $checker.

Any help would be appreciated.

1 Answer

Coco Jackowski
Coco Jackowski
12,914 Points

1) That "->" lets you get or set object properties just like any other variable, but it also lets you call methods, so $checker->isPalprime() is calling the isPalprime() method of $checker.

2) The isPalprime() method is a method of the object $checker, and is therefore able to 'see' the value of number, which is just another property of $checker

Maybe there is more to your code, but this line

echo "The number " . $number;

seems to include a variable that has no value and was never initialized.