PHP: Working with functions Question 4 of 9. Tell me why i'm wrong!

Question

I got this question right but I only marginally understand why so I'm hoping someone can clear this up.

Why is does the below code output 6? I thought it was 3. I put my logic in comments. What did I get wrong? I'm guessing it has something to do with the $output but I'm not sure what.

<?php

$numbers = array(1,2,3,4); // creates an array of numbers

$total = count($numbers); //counts the numbers starting from 0 So the final count is actually 3. 

$sum = 0; //creates a variable called $sum and assigns it the number 0. 

$output = ""; // creates a varialbe called $output and assigns it a blank string 

$i = 0; // creates a variable called $i and assigns it a number 0. 

foreach($numbers as $number) { // loops through the $numbers variable, one at a time and calls each one a $number

    $i = $i + 1; // takes the $i variable and adds 1 (which seems to end up at 1 since 0+1=1. )

    if ($i < $total) { // if the result of i + 1 (which according to my logic is 1) is less than the total of $numbers (which i believe #numbers is 3) so it is less... then

        $sum = $sum + $number; // add the $sum ( I believe is 0 since nothing changed that variable) to the total of $number (which I still believe is 3.)

    }

}

echo $sum; // according to my logic is 3. But it's 6. Why?

?>

Answer 1 · 2014-05-18T20:01:47Z

May 18, 2014 8:01pm

Here is the correction your logic.

<?php

$numbers = array(1,2,3,4); // creates an array of numbers

$total = count($numbers); //this function doesn't count the numbers starting at 0, it actually counts the number of elements in the array, so starting at 0 it counts, 0, 1, 2, 3 which means the total here is equal to 4 not 3. 

$sum = 0; //creates a variable called $sum and assigns it the number 0. 

$output = ""; // creates a variable called $output and assigns it a blank string 

$i = 0; // creates a variable called $i and assigns it a number 0. 

foreach($numbers as $number) { // loops through the $numbers variable, one at a time and calls each one a $number this is true, which means because there are 4 elements in the array, this loop will cycle through 4 times.

    $i = $i + 1; //The first time $i will be set to 1, the second time through 2, the third time 3, and the fourth time 4

    if ($i < $total) { //this if statement will pass the first three times through, but will fail on the 4th because $i will be for and $total is 4 so $i is not less than $total

        $sum = $sum + $number;//Given that this statements runs three times the result each of the three times will be
//1 = 0 + 1 1 is the new value of sum, the old value was 0 and we added 1 to it because the first time through our for loop we use the first array element which is one
//3=1+2 this time we add 2 to the old value because we are on the second array element
//6=3+3 this time we add three because we use the 3rd array element,

    }

}

echo $sum; //the value will be 6 as shown above. We never add 4 to the string because of the failed if statement $i<$total which shows the correct number and how the logic runs through this code.

?>

I hope this helps

Answer 2 · 2014-05-18T20:02:50Z

May 18, 2014 8:02pm

You have 4 elements in array so $total = count($numbers) will return 4 in this case your foreach will be executed 3 times

i=1 number=1 if ($i < $total) is true sum = 1

i=2 number=2 if ($i < $total) is true sum = 1 + 2 = 3

i=3 number=3 if ($i < $total) is true sum = 3 + 3 = 6

i=4 number=4 if ($i < $total) is false

Gareth was 1 min faster :)

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PHP: Working with functions Question 4 of 9. Tell me why i'm wrong!

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Gareth Borcherds

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Mike Wojtkowski

Mike Wojtkowski