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

debug_zval_dump ( mixed $variable , mixed ...$variables ) : void

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

参数

variable

The variable to dump.

variables

Further variables to dump.

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