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# debug_zval_dump

(PHP 4 >= 4.2.0, PHP 5)

debug_zval_dump -- Dumps a string representation of an internal zend value to output

## Descrição

void debug_zval_dump ( mixed variable )

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

## Parâmetros

`variable`

  The variable being evaluated.

## Valores de retornado

Não há valor retornado.

## Exemplos

Exemplo 1. debug_zval_dump() example
`<?php $var1 = 'Hello World'; $var2 = ''; $var2 =& $var1; debug_zval_dump(&$var1); ?>`
O exemplo acima irá imprimir:
    string(11) "Hello World" refcount(3)

> 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:

> Exemplo 2.
> `<?php $var1 = 'Hello World'; $var2 = ''; $var2 =& $var1; debug_zval_dump($var1); // not passed by reference, this time ?>`
> O exemplo acima irá imprimir:
>     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):

> Exemplo 3.
> `<?php $var1 = 'Hello World'; debug_zval_dump($var1); ?>`
> O exemplo acima irá imprimir:
>     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.

## Veja também

var_dump()
debug_backtrace()
References Explained
Reference Counting and Aliasing (by Andi Gutmans)
References Explained (by Derick Rethans)

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