[RESOLVED] A series of function questions, one at a time.

bluetonic

Given this:
$a=array("Cat","Dog","Horse", "Cow");
print_r($a);
echo '<br />';
echo $a[0]
outputs:
Array ( [0] => Dog [1] => Cat [2] => Horse )
Cat

So:
$a=array("Cat","Dog","Horse", "Cow");
print_r(array_chunk($a,2));
outputs:
Array (
[0] => Array ( [0] => Cat [1] => Dog )
[1] => Array ( [0] => Horse [1] => Cow )
)
Why doesn't echo $a[0][0]; produce Cat?

Is it a limitation of the function or is it that I don't understand the nature of arrays?

laserlight

Note that array_chunk returns the result array rather than modifies the array that was passed to it.

bluetonic

Can you explain the difference between "returning the result" vs. "modifying" the array? I kinda understand it, but can't figure out what purpose the function serves if its not changing the essence of the array.

laserlight

bluetonic wrote:
Can you explain the difference between "returning the result" vs. "modifying" the array?

For the former, you get a modified copy of the original array. For the latter, the original array itself is modified.

CoderDan

to get the result you expect, you'd need to do:

$a=array("Cat","Dog","Horse", "Cow");
$a = array_chunk($a,2);
print_r($a);
// etc

bluetonic

"Returning the result" of an array only alters the "look" but not the "value structure" (or whatever the proper term is) of the array? How do I know if a function modifies the shape or the value of an array?

One other thing:
Can you explain to me why using $a the second time doesn't "redefine" its value?
$a=array("Cat","Dog","Horse", "Cow");
$a = array_chunk($a,2);

CoderDan

using $a a second time does redefine it's value, your original method did not, which is why you didn't get the result you expected.
You can check the function reference at php.net for which functions modify the content of its arguments, it's the minority.
essentially, most (normal) functions take argument(s) and return a result based off those argument(s). This means for functions like the array functions, that manipulate arrays, if you want to change your array to the result, you assign it. some other functions use a 'pass by reference' method, like array_pop, that takes an array as an argument, modifies it, and returns something else - ie it takes the array, removes the last element from it, and returns it.
(to check for pass my reference for functions that modify their arguments - the & is the key for this. ie:

function reference ( &$arg )
{
  $old_value = $arg;
  $arg = 'new value';
  return $old_value;
}
// a rather pointless function - will cause this:
$a = 'old value';
$b = reference( $a );
print $a;
print $b;
/* would have the output:
new value
old value
*/
// where as:
function noReference ( $arg )
{
  $old_value = $arg;
  $arg = 'new value';
  return $old_value;
}
// an even more pointless function - will cause this:
$a = 'old value';
$b = noReference( $a );
print $a;
print $b;
/* would have the output:
old value
old value
*/

I hope this helps explain

bluetonic

So any attempt to use the "new" array is pointless because it is actually just the old array "in disguise." The only way to use the "new array" is to redefine it with something like "$a = array_chunk($a,2); "

Is "pointless" too strong a word? I mean, once you have run the array through the function what can you do with it (assume you don't just want to echo the results)?

CoderDan

I meant the functions I made up had no point. all they did was helped illustrate the effects 😛 I didn't want to try to come up with useful functions as they would have detracted from the illustration.

and yeah, you can redefine it with $a = fun( $a ), or if you want to maintain the original, you can get a new var with $b = fun( $a ) then use $b for the modified use, while holding $a untouched for other uses... or you can simply do something directly with the fun's output without assigning it anywhere as you did with print_r( fun( $a ) ) that doesn't modify $a, but uses the fun's result as the argument for print_r. hell, you can even do $b = fun1( fun2( $a ) ) to assign $b to the result of fun1 which is in turn given the result of fun2 when passed $a.. we can go on forever 😛
The thing to watch for is when a fun modifies a variable, you don't want to use it as an argument usually - for example, stuff like: array_join( array_pop( $a ), array_pop( $a ) ); probably will have some odd results, and not what you're intending.