Re: SV: SV: [PHP4BETA] unset() bug or feature?

At 19:12 03/01/2000 , Johan Ekenberg wrote:

Yes. But look at this: $array[0] = "Set";

   print count($array);                             // prints: 1
   unset ($array[0]);                               // $array

now contains 0 items.

   reset($array);                                   // This doesn't help

It doesn't help what? As far as I understand it doesn't reset the internal array pointer to $array[0]. If it would, the subsequent addition to $array[] would be placed in $array[0] and not $array[1] as is the case.

Ah no. There are really two internal pointers (of different kinds) inside a hash. One is a real pointer, that points to the current entry in the hash (the one that would be returned by the next call to each()). The other is basically just an incrementing number, that is used to give automatic indices to array elements that were inserted with implicit keys (i.e. $foo[] = $bar). This number never decreases.

   print count($array);                             // prints: 0
   $array[] = "Set";                                // Add to

$array (which contains _zero_ items!)

   print count($array);                             // prints: 1
   print "$array[0]";                               // prints: (nothing)
   print "$array[1]";                               // prints: Set

Questions: 1. Why isn't the internal array pointer reset when any or

all items are

unset()?

Why should it? It should only move if the element being unset is the one currently pointed by the internal pointer, and then too it shouldn't reset, but move to the next element. Ok, I won't argue more on that.

2. Why doesn't reset() reset the internal array pointer?

It does. What makes you think it doesn't? See comment above. $array[] = "Set" adds to $array[1] even though $array was just reset(). Shouldn't it add to $array[0] ??

I missed your earlier posts I guess, but that's the intended behavior. Zeev -- Zeev Suraski <zeev@zend.com> http://www.zend.com/