PHP 5.6.0RC3 is available

debug_zval_dump

(PHP 4 >= 4.2.0, PHP 5)

debug_zval_dumpDumps a string representation of an internal zend value to output

Description

void debug_zval_dump ( mixed $variable [, mixed $... ] )

Dumps a string representation of an internal zend value to output.

Parameters

variable

The variable being evaluated.

Return Values

No value is returned.

Examples

Example #1 debug_zval_dump() example

<?php
$var1 
'Hello World';
$var2 '';

$var2 =& $var1;

debug_zval_dump(&$var1);
?>

The above example will output:

&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
?>

The above example will output:

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);
?>

The above example will output:

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.

See Also

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User Contributed Notes 2 notes

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8
AmberBlackthorn
3 years ago
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(&$foo); ?> doesn't change anything.
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-7
theriault
2 years ago
You can return a consistent refcount value for any variable (and it will work in PHP 5.3+ without problems) using the following function. 0 is returned if the variable passed has no references outside itself or the variable doesn't exist (use isset in conjunction with this function if you need to distinguish between the two).
<?php
/**
* Return the reference count of a variable.
* Returns 0 if a variable has no reference other than itself or doesn't exist.
*/
function refcount(&$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
/**
* Return whether a and b reference the same memory
*/
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;
?>
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