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Bug #1269 is_array function defines the variable?
Submitted: 1999-03-26 11:05 UTC Modified: 1999-03-26 12:17 UTC
From: arneodo at initiative dot fr Assigned:
Status: Closed Package: Misbehaving function
PHP Version: 3.0.6 OS: NT4 SP3
Private report: No CVE-ID: None
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From: arneodo at initiative dot fr
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 [1999-03-26 11:05 UTC] arneodo at initiative dot fr 
I tested this code:

if( ! isSet( $a['b'] ) ) echo "a is not set<BR>";
if( is_array( $a['b'] ) ) echo "a is an array<BR>";
if( isSet( $a['b'] ) ) echo "a is set<BR>";

and the result is:

a is not set
a is set

Conclusion:

is_array() creates the variable.

Why?
PHP Version is 3.0.6 on NT4

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 [1999-03-26 12:17 UTC] zeev 
is_array() doesn't create the variable, it is created before
the call to is_array(), and passed to it.  You need to understand
how PHP's automatic declaration of variables work.

In essence:
As soon as you reference a variable, it is automatically defined.
For example, if you pass a variable to is_array(), like you have
in your example, PHP checks out whether it exists, if it doesn't -
it creates it, and then it passes it over to is_array().  It has
to pass *something*, that's why the variable is automatically defined.
The implementation of is_array() has nothing to do with this
behavior - each and every function in PHP would behave in the same
way.
There aren't too many exceptions to this case, basically, only
three special functions may reference a variable without causing
it to be implicitly declared - those are unset(), isset() and empty().
 
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