# 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 `$variable` [, mixed `$...` ] ) : void

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)