错误消息“严格的标准：只应通过引用传递变量”

Question

$el = array_shift($instance->find(..))

The above code somehow reports the strict standards warning, but this will not:

function get_arr(){
    return array(1, 2);
}
$el = array_shift(get_arr());

So when will it report the warning anyway?

Answer 1

Consider the following code:

error_reporting(E_STRICT);
class test {
    function test_arr(&$a) {
        var_dump($a);
    }
    function get_arr() {
        return array(1, 2);
    }
}

$t = new test;
$t->test_arr($t->get_arr());

This will generate the following output:

Strict Standards: Only variables should be passed by reference in `test.php` on line 14
array(2) {
  [0]=>
  int(1)
  [1]=>
  int(2)
}

The reason? The test::get_arr() method is not a variable and under strict mode this will generate a warning. This behavior is extremely non-intuitive as the get_arr() method returns an array value.

To get around this error in strict mode, either change the signature of the method so it doesn't use a reference:

function test_arr($a) {
    var_dump($a);
}

Since you can't change the signature of array_shift you can also use an intermediate variable:

$inter = get_arr();
$el = array_shift($inter);

Answer 2

$instance->find() returns a reference to a variable.

You get the report when you are trying to use this reference as an argument to a function, without storing it in a variable first.

This helps preventing memory leaks and will probably become an error in the next PHP versions.

Your second code block would throw an error if it wrote like (note the & in the function signature):

function &get_arr(){
    return array(1, 2);
}
$el = array_shift(get_arr());

So a quick (and not so nice) fix would be:

$el = array_shift($tmp = $instance->find(..));

Basically, you do an assignment to a temporary variable first and send the variable as an argument.

Answer 3

The cause of the error is the use of the internal PHP programming data structures function, array_shift() [php.net/end].

The function takes an array as a parameter. Although an ampersand is indicated in the prototype of array_shift() in the manual", there isn't any cautionary documentation following in the extended definition of that function, nor is there any apparent explanation that the parameter is in fact passed by reference.

Perhaps this is /understood/. I did not understand, however, so it was difficult for me to detect the cause of the error.

Reproduce code:

function get_arr()
{
    return array(1, 2);
}
$array = get_arr();
$el = array_shift($array);

Answer 4

The second snippet doesn't work either and that's why.

array_shift is a modifier function, that changes its argument. Therefore it expects its parameter to be a reference, and you cannot reference something that is not a variable. See Rasmus' explanations here: Strict standards: Only variables should be passed by reference

Answer 5

This code:

$monthly_index = array_shift(unpack('H*', date('m/Y')));

Need to be changed into:

$date_time = date('m/Y');
$unpack = unpack('H*', $date_time);
array_shift($unpack);

Answer 6

Well, in obvious cases like that, you can always tell PHP to suppress messages by using "@" in front of the function.

$monthly_index = @array_shift(unpack('H*', date('m/Y')));

It may not be one of the best programming practices to suppress all errors this way, but in certain cases (like this one) it comes handy and is acceptable.

As result, I am sure your friend 'system administrator' will be pleased with a less polluted error.log .