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Start your free trialAaron Gates
1,573 PointsAdding to int
I added the following code $integerOne = 1; $integerTwo = 2; var_dump( $integerOne + 5); var_dump( $integerTwo - 1);
but am told that I have not changed $integerOne. It works in the compiler. Not sure what I am doing wrong.
<?php
//Place your code below this comment
$integerOne = 1;
$integerTwo = 2;
var_dump( $integerOne + 5);
var_dump( $integerTwo - 1);
?>
1 Answer
Amy Kelly
3,588 PointsI think it may be because you are doing it as a var_dump which means that the new value is not getting saved, which is why it is saying that you have not changed $integerOne. My solution to this is to change the actual variables:
$integerOne = 1;
$integerTwo = 2;
$integerOne += 5;
$integerTwo -= 1;
Aaron Gates
1,573 PointsAaron Gates
1,573 PointsThanks Amy, that worked.