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

void debug_zval_dump ( mixed $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.