When declaring an Array in PHP, the index's may be created out of order...I.e
Array[1] = 1 Array[19] = 2 Array[4] = 3
My question. In creating an array like this, is the length 19 with nulls in between? If I attempted to get Array[3] would it come as undefined or throw an error? Also, how does this affect memory. Would the memory of 3 index's be taken up or 19?
Also currently a developer wrote a script with 3 arrays FailedUpdates[] FailedDeletes[] FailedInserts[]
Is it more efficient to do it this way, or do it in the case of an associative array controlling several sub arrays
"Failures" array(){
["Updates"] => array(){
[0] => 12
[1] => 41
}
["Deletes"] => array(){
[0] => 122
[1] => 414
[1] => 43
}
["Inserts"] => array(){
[0] => 12
}
}
Memory effiency isn't really something you need to worry about in PHP unless you're dealing with really huge arrays / huge numbers of variables.
An array in PHP isn't really like an array in C++ or a similar lower-level language; an array in PHP is a map. You have a list of keys (which must be unique and all of type string or integer), and a list of values corresponding to the keys. So the following is a legal array:
array(0 => 'butt', 1 => 'potato', 2 => 'tulip')
but so is
array(5 => 'i', 'barry' => 6, 19 => array(-1 => array(), 7 => 'smock'))
In both cases there are 3 entries in the array, hence 3 keys and 3 values.
In addition to the keys and values in the array, one array may be distinguished from another by the order in which the key/value pairs occur. If you define an array so that it has nonnegative integers as keys, this will often be the expected order. The order matters when you use constructs like foreach().
array[3] will be undefined/unset but not causing an error, and the array will use only memory for that 3 values - php isn't like C where you have to look at those things.
Notice: Undefined offset: 3 in /data/home/sjoerd/public_html/svnreps/test/a.php on line 3. You can avoid this by checking with isset() or array_key_exists().