I cannot understand why the answer is six? It is six...but why?

Hey Benjamin!

I assume you're talking about this question?

<?php

function sum_two_numbers($one,$two) {

    $sum = $one + $two;

    return $sum;

}

sum_two_numbers(1,5);

?>

The sum_two_numbers function takes 2 parameters representing 2 different numbers. In this case, $one contains an integer with the value 1 and $two contains an integer with the value 5. The function adds the two numbers, so 1 + 5, returning 6.

Hey Aimee, I'm actually talking about a different one. The one you posted is a trick question because there is no 'echo' command ;) but I fully understand that the return value would be 6 on that. However, I do not understand why the return value is 6 on this one:

<?php

$numbers = array(1,2,3,4);

$total = count($numbers);

$sum = 0;

$output = "";

$i = 0;

foreach($numbers as $number) {

    $i = $i + 1;

    if ($i < $total) {

        $sum = $sum + $number;

    }

}

echo $sum;

?>

What is the value of $sum?

Hi Andrew, That helps a ton. I edited your answer to what makes more sense to me. Does this still seem correct or am I not getting it still?

(Starting with $total = 4, $i = 0, $sum = 0): 1.

$i is incremented - so $i = 1 $i is still less than 4, so the first value in the array(1) is added to the sum where:

$sum = 0 + 1 (first-value in the array).

$i is incremented - so $i = 2 $i is still less than 4, so the second value in the array(2) is added to the sum where:

$sum = 1 + 2 (second-value in the array).

$i is incremented - so $i = 3 $i is still less than 4, so the third value in the array(3) is added to the sum where:

$sum = 3 + 3 (third-value in the array).

$i is incremented - so $i = 4 $i is no longer less than 4, so the last number in the array is never actually added to the sum and $sum is still 6