NYCPHP Meetup

[Joseph Crawford]

> I'm curious to know what the "&" is for in the statement below?
> 
> -Aaron


Let me explain a bit and try to make it so you understand.  In PHP &
is the reference operator, this allows you to pass a variable by
reference rather than by value.  Here is an example

$var = 4;
func( $var );
// echo's 4
echo $var;

function func( $var ) {
    $var++;
    // echo's 5
    echo $var;
}

if you run this code, you will see how the values of $var are
different inside the function and outside the function.  The one
outside the function is 4 and the one inside is 4.  When you pass by
value php defaults to making a copy of the variable.

When you use the & (reference operator) it will pass $var by reference
rather than by value, this passes the memory address rather than
making a copy of the variable.  Let me show anoher example.

$var = 4;
func( &$var );
// echo's 5
echo $var;

function ( $var ) {
    $var++;
}

If i am correct about this (which i believe i am) when you echo $var
after the function call, $var now equals 5 because you passed the
variable by reference rather than letting PHP make a copy.  You need
to be careful passing variables/objects by reference, if you do not
intend to have the function change the value of the variable outside
of the functions scope, do not pass by reference.

I hope you understood everything i have said, if you do not, please
feel free to ask any questions reguarding my reply.


-- 
Joseph Crawford Jr.
Codebowl Solutions
codebowl at gmail.com

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