CODEWITHSUNDEEP
phpreferencepass-by-reference
user765368
user765368

Reputation: 20356

php passing by reference difference

I have a simple question here. Is there a difference between passing a variable by reference in a function parameter like:

function do_stuff(&$a)
{
    // do stuff here...
}

and do it inside the function like:

function do_stuff($a)
{
    $var = &$a;
    // do stuff here...
}

What are the differences (if any) between using these two?. Also, can anybody give me a good tutorial that explains passing by reference? I can't seem to grasp this concept 100%.

Thank you

Upvotes: 0

Views: 114

Answers (4)

deceze
deceze

Reputation: 522625

In your first example, if you modify $a inside the function in any way, the original value outside the function will be modified as well.

In your second example, whatever you do to $a or its reference $var will not modify the original value outside the function.

Upvotes: 1

Francis Lewis
Francis Lewis

Reputation: 8990

Here's a set of examples so you can see what happens with each of your questions. I also added a third function which combines both of your questions because it will also produce a different result.

function do_stuff(&$a)
{
    $a = 5;
}

function do_stuff2($a)
{
    $var = &$a;
    $var = 3;
}

function do_stuff3(&$a)
{
    $var = &$a;
    $var = 3;
}

$a = 2;
do_stuff($a);
echo $a;
echo '<br />';

$a = 2;
do_stuff2($a);
echo $a;
echo '<br />';

$a = 2;
do_stuff3($a);
echo $a;
echo '<br />';

Upvotes: 1

Marc B
Marc B

Reputation: 360912

They're not at all equivalent. In the second version, you're creating a reference to an undefined variable $a, causing $var to point to that same null value. Anything you do to $var and $a inside the second version will not affect anything outside of the function.

In the first version, if you change $a inside the function, the new value will be present outside after the function returns.

Upvotes: 1

loganfsmyth
loganfsmyth

Reputation: 161657

In the second function, the $a passed into the function is a copy of the argument passed in, (unless $a is an object), so you are making a $var a reference to the $a inside the function but it will still be separate from the variable passed to the function.

Assuming you are using a recent version of PHP, objects are automatically passed by reference too, so that could make a difference.

Upvotes: 0

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