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Alan @ CIT · 01-20-2008, 06:02 PM

They are both valid ways but you end up with slightly different results.

PHP Code:


			
<?php

(array) $foo;

$bar = array();



var_dump($foo);    // echo's "NULL"

var_dump($bar);    // echo's "array(0) ( )

So as you can see, typesetting the variable to an array using (array) doesn't actually create a new empty array where as array() will.

For the sake of standards, you should probably use array() when creating new arrays. Things like (array) (int) (long) etc are generally only used when converting a variable type to another:

PHP Code:


			
<?php



$foo = 16.123;

$bar = (int) $foo;



var_dump($bar);    // echo's int(16)

In this example, we have Long ($foo) which we then convert to an int by typecasting it using (int).

Alan.

01-20-2008, 06:02 PM	#2 (permalink)
Alan @ CIT Member of the Month The Frequenter Join Date: Apr 2005 Location: South UK Posts: 483 Thanks: 51	They are both valid ways but you end up with slightly different results. PHP Code: `<?php (array) $foo; $bar = array(); var_dump($foo); // echo's "NULL" var_dump($bar); // echo's "array(0) ( )` So as you can see, typesetting the variable to an array using (array) doesn't actually create a new empty array where as array() will. For the sake of standards, you should probably use array() when creating new arrays. Things like (array) (int) (long) etc are generally only used when converting a variable type to another: PHP Code: `<?php $foo = 16.123; $bar = (int) $foo; var_dump($bar); // echo's int(16)` In this example, we have Long ($foo) which we then convert to an int by typecasting it using (int). Alan.