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HomeBackend DevelopmentPHP Tutorialweakreference Thoughts caused by PHP's call_user_func passing reference

The question was raised
Netizen bercmisir left a message in the hospital, regarding the documentation of the call_user_func function in the PHP manual, which is roughly as follows:
http://php.net/manual/en/function.call-user-func.php
Under parameter There is such a sentence:
Note: Note that the parameters for call_user_func() are not passed by reference.
A simple translation means that the parameters of this function cannot be passed by reference.
Another example:

Copy the code The code is as follows:


error_reporting(E_ALL);
function increment(&$var)
{
$var++;
}
$a = 0;
call_user_func( 'increment', $a);
echo $a."n";
call_user_func_array('increment', array(&$a)); // You can use this instead before PHP 5.3
echo $a."n" ;
?>


The output is:
0
1
The problem with netizen bercmisir is:
call_user_func('increment', $a); the output is 0, and call_user_func('increment', &$a); But the output is 1, which clearly says that it cannot be passed by reference.
Looking for the root cause
Then we will further look for the root cause. The information of this Note is actually the final processing result of the bug http://bugs.php.net/bug.php?id=24931.
And in call_user_func('increment', &$a); although the result of 1 is output, under normal circumstances, there will be a warning message: Deprecated: Call-time pass-by-reference has been deprecated.
What is the reason for this?
Let’s look at an example first:

Copy the code The code is as follows:


error_reporting(E_ALL);
function increment(&$var)
{
$var++;
}
$x = 1;
increment( $x);
echo $x;
?>


The result is 2, and there is no warning message like expected to be a reference, value given. On the contrary, if the 8th line of code is modified to &$x, will get a deprecation warning. It can be verified that in fact, during the transfer process in PHP, the variable will decide whether to transfer a reference or a value based on whether the formal parameter requires a reference or a value, and does not need to be transferred explicitly (on the contrary, explicit transfer will be abolished soon) ).
Continue to delve deeper
http://www.php.net/manual/en/language.references.pass.php
In the PHP manual, in the section introducing the passing of references, there is a Note in the middle that says: in function calls There is no need to pass a reference (that is, the explicit call mentioned in the previous section). In 5.3, if an explicit call is made, a deprecation warning will appear.
Analyze the source code
Some people say: When writing in php, everything is a reference.
Check the php source code, there is a function definition zend_do_pass_param in line 1579 of ./Zend/zend_compile.c. (php5.2.13)
There is such a judgment:
if (original_op == ZEND_SEND_REF && !CG(allow_call_time_pass_reference)) {Print abolition warning. }
The general meaning is that if a reference is passed and the allow_call_time_pass_reference of php.ini is No, a warning will be printed.
Looking at where zend_do_pass_param is used, we can find that in the parser stage, whether it is var, value or reference is passed based on the definition of the elements in the parameter ZVAL structure. (php5.2.13 ./Zend/zend_language_parser.y/c 451/3593)
Conclusion
References are actually similar to file hard links in Linux, but they are different from pointers in C language. During the parser phase, php will use the context according to the context. It is up to the environment to decide whether to pass a reference or a value. The call_user_function mentioned in this article does not determine whether it is a reference or a value. Therefore, the previous example call_user_function does not work when passing by value, but the correct result is obtained when passing by reference (but there is actually a deprecation warning).

The above introduces the thoughts about weakreference caused by PHP's call_user_func passing reference, including the content of weakreference. I hope it will be helpful to friends who are interested in PHP tutorials.

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