assignment operator (=) within a function's parameter list? - php

Question

I'm using the following code from PHPBuilder.com to handle user privileges on my site:
/**
* Correct the variables stored in array.
* #param integer $mask Integer of the bit
* #return array
*/
function bitMask($mask = 0) {
if(!is_numeric($mask)) {
return array();
}
$return = array();
while ($mask > 0) {
for($i = 0, $n = 0; $i <= $mask; $i = 1 * pow(2, $n), $n++) {
$end = $i;
}
$return[] = $end;
$mask = $mask - $end;
}
sort($return);
return $return;
}
and I'm a bit baffled by the "= 0" part of ($mask = 0) in the function parameter list. What does that do?

Answer 1

It means that if you call the function like this:
$a = bitMask();
Then $mask will be set to 0.
It is how you set default values to parameter in functions.
Example:
function example($a=0){
echo "a = $a";
}
example(10);
example();
Output:
a = 10
a = 0
If $a did not have a default value set, then calling the function like example() would give a warning.
reference: http://php.net/manual/en/functions.arguments.php (Default argument values)

Answer 2

That's the default value of $mask if no arguments are passed. This also prevents a warning from being generated when the parameter is omitted.

Answer 3

Michael's response is correct. To add to it, take note that the assignment will not have an affect on the original variable modified. Here is his code with a few more assignments / echos to illustrate this:
function example($a=0){
echo "Entering function: a = $a\n";
$a = 3;
echo "End of function: a = $a\n";
}
$a = 7;
example(10);
echo "Outside of Function: a = $a\n";
Outputs
Entering function: a = 10
End of function: a = 3
Outside of Function: a = 7