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Start your free trialDwayne Munro
Front End Web Development Techdegree Student 13,108 PointsThe second echo statement was set up to display the number of letters, but it always showed "2", and we have additional
What is wrong with my code
<?php
$letters = array();
$letters[] = "A";
$letters[] = "C";
$letters[] = "M";
$letters[] = "E";
echo "Today's challenge is brought to you by the ";
$count_letters = function() {
echo count($letters);
}
echo " letters: ";
echo "AC";
echo ".";
?>
1 Answer
Serhii Lihus
6,352 PointsHi there Dwayne!
First of all. Your code wont output result because of two things:
1) You've define closure, but didn't call it.
$count_letters = function() {
echo count($letters);
}
$count_letters(); // but this call wont output the result and if you read further you'll understand why ;)
2) You need to remember about local and global scope. In your case it's local, and you can only pass your array of letters by reference, or more precisely through a parameter. Your function won't see the array outside of it's scope.
Here is the working code sample:
$letters = array();
$letters[] = "A";
$letters[] = "C";
$letters[] = "M";
$letters[] = "E";
echo "Today's challenge is brought to you by the ";
$count_letters = function($letters) {
echo count($letters);
};
$count_letters($letters); // this call will output the right result ;)
echo " letters: ";
echo "AC";
echo ".";
Additional info about closures you can read here where you can read in detail about clouser.
Best regards.