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PHP PHP Basics Unit Converter Manipulating Numbers

Louise Turner
Louise Turner
710 Points
Posted by Louise Turner
Louise Turner
710 Points

I believe my question is correct. My variable $floatOne is a float with a value of 1.5

Why is my test not passing?

index.php 
<?php

//Place your code below this comment
$integerOne = 1;
$integerTwo = 2;

$integerOne = $integerOne + 5;

var_dump ($integerOne);
var_dump (--$integerTwo);

$floatOne = 1.5;
var_dump ($floatOne);
echo $floatOne;
echo ' multiplied by ';
echo $integerOne;
$floatOne = $floatOne * $integerOne;
echo ' = ';
echo $floatOne;
?>

1 Answer

Jennifer Nordell
seal-mask
STAFF
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher
Jennifer Nordell
.a{fill-rule:evenodd;}techdegree
Jennifer Nordell
Treehouse Teacher

Hi there! Not quite. At this point, your $floatOne is now equal to 9 instead of 1.5. And your code actually proves it. The last thing you echo out is the value of $floatOne and the result is 9.

This is the line that is causing this:

$floatOne = $floatOne * $integerOne;

This takes the value of $floatOne (which is 1.5 at this point) and multiplies it by the value of $integerOne which is 6. The result is 9. The 9 is assigned back into $floatOne overwriting its previous value.

Here's a quote from the instructions:

After adding 5 to $integerOne, and without further changes to the value of $integerOne or $floatOne, multiply $integerOne by $floatOne and display the results.

But by writing over the value of $floatOne, you've made a "further change".

Try echoing out your results directly like this:

echo $floatOne * $integerOne;

Also, it's important to note that you have a series of var_dumps and echoes in your code which are not called for and may cause problems. Echo out only what is asked for.

Hope this helps! :sparkles: