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Start your free trialJustin Moore
718 PointsHow do I assign the variable "$checker" a new PalprimeChecker object?
What does this mean? What does the syntax look like? I can't figure out what the question is asking
<?php
include("class.palprimechecker.php");
echo "The number ## ";
echo "(is|is not)";
echo " a palprime.";
?>
2 Answers
Siddhant Mehta
Courses Plus Student 478 PointsYou simply need to create instance of class here i.e. object.
$checker = new PalprimeChecker();
Kaisma Penn-Titley
PHP Development Techdegree Graduate 20,208 PointsThis worked, thanks!
Jason Anello
Courses Plus Student 94,610 PointsJason Anello
Courses Plus Student 94,610 Pointschanged comment to answer