PHP code Output

Question

I was studying for my finals and I came across this question:

write the output after running this code.

<?php
function swap($x, $y) 
{
  $x = $x + 1;
  $y = $y + 2;
  return $x * $y;
}

$a = 3;
$b = swap($a, $a);
print "$a, $b";
$b = swap(&$a, &$a);
print "$a, $b";
?>

I understand exactly what this code does, however after I ran it I got a completely different answer to what I answered with and I really don't understand the output. The output I got was 3, 206, 36.

Could someone explain the output to me?

P.S. In PHP 5.3 passing references to a function call is deprecated. Meaning don't do $b = swap(&$a, &$a);. If you want to pass references, set that up in the function signature. function swap(&$x, &$y){. Docs: php.net/manual/en/language.references.pass.php — gen_Eric
– gen_Eric, Commented Apr 26, 2012 at 18:34

Aurelio De Rosa · Accepted Answer · 2012-04-19 07:32:30Z

5

What you get is actually 3, 20, 6, 36 which is the correct answer. If you don't understand why you got "206" instead of "20" and "6" it is just because you have not a space after the first print. That's it.

answered Apr 19, 2012 at 7:32

Aurelio De Rosa

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2 Comments

Pro-grammer Over a year ago

Thanks, how does each value come into place? Oh the space means the 20 and 6 are not separated thanks :)

Aurelio De Rosa Over a year ago

@F.Dot The first call pass the params by value, the second one by reference. If you need more reference, search it. You'll find a lot of information.

Rukmi Patel · Accepted Answer · 2012-04-19 07:34:32Z

1

first print statement print 3, 20 second print statement print 6,36

first dont get confused in this .. when you pass value by ref, it changes the original value.. thats why it give second output as 6,36

answered Apr 19, 2012 at 7:34

Rukmi Patel

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Comments

Neeraj · Accepted Answer · 2012-04-19 07:39:44Z

1

Replace Your function with below one and you will understand why the output is coming like that

function swap($x, $y) 
{
    echo "<BR><BR>inside function<BR>:";
    echo "<BR>Original x : ".$x;
    echo "<BR>Original y : ".$x;
    $x = $x + 1;
    $y = $y + 2;
    echo "<BR>After x : ".$x;
    echo "<BR>After y : ".$y;
    echo "<BR>^^^^^^^^^^^^^<BR";
    return $x * $y;
}

Second time value of x and y become 6 ,6 because you are using call by reference. So once $x = $x + 1 is executed then it became 4 and just after that $y = $y + 1 is executed then it made $x = 6 (4+2). so the product became 36

answered Apr 19, 2012 at 7:39

Neeraj

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Comments

Jack · Accepted Answer · 2012-04-19 07:47:10Z

1

The original value is $a = 3

The function is:

function swap($x, $y) 
{
  $x = $x + 1;
  $y = $y + 2;
  return $x * $y;
}

Since you are passing by reference, the function will modify the original value of $a. In the function, $a is passed as parameter of $x and $y so $x becomes 4. Since it is a referenced variable, $a has been modified to 4. Now $y also points to the same variable $a so now $y is actually 4. 4 + 2 makes $y = 6. $x also is 6 because both $x and $y point to the same variable $a which they both modified. So $a becomes 6 and $b becomes 36.

answered Apr 19, 2012 at 7:47

Jack

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