Question

$origarray1 = array(2.4, 2.6, 3.5);
$origarray2 = array(2.4, 2.6, 3.5);

print_r(array_map('floor', $origarray1)); //$origarray1 stays the same

//changes $origarray2
function callback_array_walk(&$v, $k) { $v = floor($v); }
array_walk($origarray2, callback_array_walk(&$v, $k)); 
print_r($origarray2);

//this is a more proper use of array_walk
function callback_array_walk2($v, $k) { echo "$k => $v", "\n"; }
array_walk($origarray1, callback_array_walk2($v, $k));

//array_map accepts several arrays
function callback_arraymap($a, $b) { return $a * $b; }
print_r(array_map(callback_arraymap($a, $b), $origarray1, $origarray2));

//select only elements that are > 2.5
function callback_array_filter($a) { return $a > 2.5; }
print_r(array_filter($origarray1,callback_array_filter($a)));

I got error here

array_walk() [function.array-walk]: Wrong syntax for function name in ….

What I misunderstand of these three type(using,calling,passing )?

Answer 1

Try passing the callback function to array_walk as a string without any arguments:

function callback_array_walk(&$v, $k) { $v = floor($v); }
array_walk($origarray2, 'callback_array_walk');
                        ^^^^^^^^^^^^^^^^^^^^^

Similarly for array_map and array_filter functions pass the callback function as a string.