通过PHP中的引用调用内联传递

Question

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 updating the array:

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

    return null;
}

I'm using the code as followed:

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

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

b
Array ( [0] => a [2] => c )

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

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

Notice: Only variables should be passed by reference in ...
b
Array ( [0] => a [1] => b [2] => c )

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.

Update:

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

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

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

b
Array ( [0] => a [2] => c )

Answer 1

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);