Type check in all array elements

Is there any simple way of checking if all elements of an array are instances of a specific type without loopin开发者_JAVA技巧g all elements? Or at least an easy way to get all elements of type X from an array.

$s = array("abd","10","10.1");
$s = array_map( gettype , $s);
$t = array_unique($s) ;
if ( count($t) == 1 && $t[0]=="string" ){
    print "ok\n";
}

You cannot achieve this without checking all the array elements, but you can use built-in array functions to help you.

You can use array_filter to return an array. You need to supply your own callback function as the second argument to check for a specific type. This will check if the numbers of the array are even.

function even($var){
  return(!($var & 1));
}

// assuming $yourArr is an array containing integers.
$newArray = array_filter($yourArr, "even");
// will return an array with only even integers.

As per VolkerK's comment, as of PHP 5.3+ you can also pass in an anonymous function as your second argument. This is the equivalent as to the example above.

$newArray = array_filter($yourArr, function($x) { return 0===$x%2; } );

Is there any simple way of checking if all elements of an array [something something something] without looping all elements?

No. You can't check all the elements of an array without checking all the elements of the array.

Though you can use array_walk to save yourself writing the boilerplate yourself.

You can also combine array_walk with create_function and use an anonymous function to filter the array. Something alon the lines of:

$filtered_array = array_filter($array, create_function('$e', 'return is_int($e)'))