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:
Пример 2.
<?php
$var1 = 'Hello World';
$var2 = '';
$var2 =& $var1;
debug_zval_dump($var1); // not passed by reference, this time
?>
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Результат выполнения данного примера: string(11) "Hello World" refcount(1) |
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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):
Пример 3.
<?php
$var1 = 'Hello World';
debug_zval_dump($var1);
?>
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Результат выполнения данного примера: string(11) "Hello World" refcount(2) |
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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.