Xorifelse Xorifelse - 2 years ago 72
PHP Question

In-lining a pass by reference call in PHP

I have a question regarding the pass-by-reference in PHP. I've searched online but couldn't see anything specific to this issue. The following function removes the key from the array and returns it's value.

function array_fetch($k, array &$a){
if(isset($a[$k]) || array_key_exists($k, $a)){
$v = $a[$k];
return $v;

return null;

function foo(){
return ['a', 'b', 'c', 'd', 'e'];

$a = foo();
echo array_fetch(1, $a);

bArray ( [0] => a [2] => c [3] => d [4] => e )

So it works like a charm and now I want to make the code a little shorter:

echo array_fetch(1, $a = foo());

Notice: Only variables should be passed by reference in ...

bArray ( [0] => a [1] => b [2] => c [3] => d [4] => e )

Am I wrong to assume that I give a variable as reference? Apparently so, cause the array is unchanged as well but I don't understand why this happens. Even if I enclose the expression with
it doesn't help.

The only thing I can think of, is that you don't pass the reference of the variable, but PHP picks it up as the name instead.

Is there a clever work around for this issue?

A viable work-around is to use a wrapper function like so:

function &ref($var){
return $var;

echo array_fetch(1, $a = &ref(['a', 'b', 'c', 'd', 'e']));

Answer Source

array_fetch(1, $a = foo()); is not

assign $a and pass $a to function.

It is

assign $a and pass the result of assign to function.

And result of assign operation is the value which is assigned.

So, array_fetch(1, $a = foo()); is equivalent to array_fetch(1, ['a', 'b', 'c']) where second argument is not a variable.

So, the only solution is still:

$a = foo();
echo array_fetch(1, $a);
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