debug_zval_dump
(PHP 4 >= 4.2.0, PHP 5, PHP 7)
debug_zval_dump — Dumps a string representation of an internal zend value to output
说明
$variable
[, mixed $...
] )Dumps a string representation of an internal zend value to output.
参数
-
variable -
The variable being evaluated.
返回值
没有返回值。
范例
Example #1 debug_zval_dump() example
<?php
$var1 = 'Hello World' ;
$var2 = '' ;
$var2 =& $var1 ;
debug_zval_dump (& $var1 );
?> 以上例程会输出：
&string(11) "Hello World" refcount(3)
Note: Beware the refcount
The refcount value returned by this function is non-obvious in certain circumstances. For example, a developer might expect the above example to indicate a refcount of 2. The third reference is created when actually calling debug_zval_dump() .
This behavior is further compounded when a variable is not passed to debug_zval_dump() by reference. To illustrate, consider a slightly modified version of the above example:
<?php
$var1 = 'Hello World' ;
$var2 = '' ;
$var2 =& $var1 ;
debug_zval_dump ( $var1 ); // not passed by reference, this time
?>以上例程会输出：
string(11) "Hello World" refcount(1)Why refcount(1)? Because a copy of $var1 is being made, when the function is called.
This function becomes even more confusing when a variable with a refcount of 1 is passed (by copy/value):
<?php
$var1 = 'Hello World' ;
debug_zval_dump ( $var1 );
?>以上例程会输出：
string(11) "Hello World" refcount(2)A refcount of 2, here, is extremely non-obvious. Especially considering the above examples. So what's happening?
When a variable has a single reference (as did $var1 before it was used as an argument to debug_zval_dump() ), PHP's engine optimizes the manner in which it is passed to a function. Internally, PHP treats $var1 like a reference (in that the refcount is increased for the scope of this function), with the caveat that if the passed reference happens to be written to, a copy is made, but only at the moment of writing. This is known as "copy on write."
So, if debug_zval_dump() happened to write to its sole parameter (and it doesn't), then a copy would be made. Until then, the parameter remains a reference, causing the refcount to be incremented to 2 for the scope of the function call.
参见
- var_dump() - 打印变量的相关信息
- debug_backtrace() - 产生一条回溯跟踪(backtrace)
- References Explained
- » References Explained (by Derick Rethans)
用户评论:
[#1] theriault [2012-06-23 08:43:25]
You can return a consistent refcount value for any variable (and it will work in PHP 5.3+ without problems) using the following function. 0 is returned if the variable passed has no references outside itself or the variable doesn't exist (use isset in conjunction with this function if you need to distinguish between the two).
<?php
function refcount(&$var) {
ob_start();
debug_zval_dump(array(&$var));
return preg_replace("/^.+?refcount\((\d+)\).+$/ms", '$1', substr(ob_get_clean(), 24), 1) - 4;
}
# Example
$a = 34;
refcount($a) == 0;
$b = &$a;
refcount($a) == 1;
?>
Using the above function as a dependency, you can determine if two variables reference the same space in memory using another small function:
<?php
function reference(&$a, &$b) {
$d = refcount($b);
$e = &$a;
return refcount($b) != $d;
}
# Example
$a = 3;
$b = 5;
$c = &$a;
$d = &$c;
reference($a, $b) == false;
reference($a, $c) == true;
reference($a, $d) == true;
?>
[#2] AmberBlackthorn [2010-10-23 05:35:51]
The add of "Call-time pass-by-reference" as E_DEPRECATED makes it impossible to get the real refcount without getting an error, since
<?php error_reporting(E_ALL ^ E_DEPRECATED); ?>
and even
<?php @debug_zval_dump(&$foo); ?>
doesn't change anything.