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###### Lucas Santos

19,315 Points# Can someone please explain this code

Im stuck on this quiz and I believe it's asking me what this code would output on the screen.

```
<?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) {
$output = $number . $output;
}
}
echo $output;
?>
```

## 3 Answers

###### Dino Paškvan

Courses Plus Student 44,107 PointsWhen this particular piece of code starts executing, there are four variables initialized at the top. First, you can ignore `$sum`

because it's not used elsewhere in the code. `$i`

is `0`

, `$total`

is `4`

because there are 4 elements in the `$numbers`

array, and `$output`

is an empty string.

On the first iteration of `foreach`

, `$i`

gets immediately increased to `1`

. Then we check if it's smaller that `$total`

. Because it is, the current number from the `$numbers`

array (in this case it's the first element — `1`

) gets concatenated with the `$output`

and the resulting string is assigned as the new value of the `$output`

variable, which now equals to `"1"`

.

On the second iteration, `$i`

is increased to `2`

. It's still smaller than `$total`

, so now the second element of the array (`2`

) is concatenated with `$output`

(`"1"`

) and the resulting string (`"21"`

) is assigned as the new value of `$output`

.

The same is true for the third iteration, only now the resulting string is `"321"`

.

In the fourth iteration, `$i`

is now `4`

and as it's not smaller than `$total`

, no value re-assignment of `$output`

happens.

So, the final output is the resulting string from the third iteration: `321`

.

###### Guillaume Maka

Courses Plus Student 10,224 Points```
<?php
// declare an array of 4 numbers
$numbers = array(1,2,3,4);
// assign the length of $number array to $total -> 4
$total = count($numbers);
// declare a variable $sum and initialize it to 0
$sum = 0;
// declare a variable $output and initialize it to an empty string
$output = "";
// declare a variable $i and initialize it to 0
$i = 0;
// do a foreach loop to grab each element in the array
foreach($numbers as $number) {
// track how many element we have already take
// ex: at the first iteration, $i = 0 + 1
$i = $i + 1;
// check we don't exceed the total sze of the array
// ex: at the first iteration, $i = 1 , $total = 4, so 1 < 4 then enter into the if statement
if ($i < $total) {
// concatenate the given $number to the $output
// ex: at the first iteration, $number = 1, $ouptut = ''
$output = $number . $output;
// $output = '1'
}
}
// print the output to the screen
// $output = '321'
echo $output;
?>
```

Hope that will help you.

###### Lucas Santos

19,315 PointsThanks a lot this helps!

###### Lucas Santos

19,315 PointsThanks a lot this helps!

## Lucas Santos

19,315 Points## Lucas Santos

19,315 Pointscrystal clear

## Lucas Santos

19,315 Points## Lucas Santos

19,315 PointsHey so when ever you concatenate a number or value to an empty variable it will always add it to the left meaning the beginning of the empty string rite? I would think it would be like reading and it adds the numbers to the rite like 123 instead of 321 but I guess not.

## Guillaume Maka

Courses Plus Student 10,224 Points## Guillaume Maka

Courses Plus Student 10,224 PointsTo get 123 you just need to do the opposite

`$output = $output . $number;`

That was your question ?

## Lucas Santos

19,315 Points## Lucas Santos

19,315 PointsI see ok got it thanks!