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Todd Tate
540 PointsAm I right or is this a bug?
I just started paying for this and have been stuck on this LAST question on this quiz section.
I want to know if this is a bug
Challenge:
Create a New Float Variable named $floatOne with a value of 1.5. Without changing the value of $integerOne or $floatOne, multiply $integerOne by $floatOne and display the results.
My answer:
<?php //Place your code below this comment $integerOne = 1; $integerTwo = 2; $integerOne = $integerOne +5; var_dump ($integerOne); $integerTwo = $integerTwo -1; var_dump ($integerTwo); $floatOne = 1.5; var_dump ($floatOne * $integerOne); echo ($floatone); ?>
Thanks!
3 Answers
Jennifer Nordell
Treehouse TeacherHi there! It's not a bug. You're using var_dump to display all information regarding that variable including both the type and the value. It's wanting you to use echo to display the value of the variable to the page. Give it another shot with this hint in mind!
edited for additional note
As a side note: it's important when doing these challenges to try not to do anything the challenge doesn't explicitly ask for. Even if functional, it can cause the challenge to fail as it no longer meets the strict requirements.
Todd Tate
540 PointsThanks!
Wow I've tried that and the value of $floatOne is 9, so I've tried -7.5 and that don't work either. I;ve watched videos multiples times lol this below still will not work.
<?php //Place your code below this comment $integerOne = 1; $integerTwo = 2; $integerOne = $integerOne +5; var_dump ($integerOne); $integerTwo = $integerTwo -1; var_dump ($integerTwo); $floatOne = 1.5; $floatOne = $floatOne * $integerOne; echo ($floatone); ?>
Jennifer Nordell
Treehouse TeacherHi again, Todd Tate! First, you still have var_dump scattered throughout your code, which the challenge doesn't require nor request. But the challenge explicitly states to not change the value of $floatOne. You have written this line:
$floatOne = $floatOne * $integerOne;
This line takes 1.5 and multiplies it by 6 then assigns the value back into $floatOne thus changing the original value from 1.5 to 9.
Here was my solution:
<?php
//Place your code below this comment
$integerOne = 1;
$integerTwo = 2;
$integerOne += 5;
$integerTwo -= 1;
$floatOne = 1.5;
echo $integerOne * $floatOne;
?>
Here we set up $integerOne and $integerTwo with the initial values. I used the unary operators to add 5 and subtract one from them respectively. Then we create $floatOne and set it equal to 1.5. Finally, without changing any values, we echo out the result of the multiplication of $integerOne and $floatOne.
Hope this clarifies things!
Todd Tate
540 PointsDo you work for Treehouse?
Jennifer Nordell
Treehouse TeacherNo 