Why does explode on null return 1 element?

Question

var_dump(count(explode(',',[]))); // 0
var_dump(count(explode(',',''))); // 1
var_dump(count(explode(',',null))); // 1 ????

I was expecting 0 for the last one. Can someone tell me the reason why exploding from null is 1 and not 0? (PHP 5.6.16)

Answer 1

As a workaround, use

array_filter(explode(',', $myvar));

Then you get an empty array and count() will return 0. Be aware, that it will also return an empty array, when $myvar == ''

Answer 2

explode takes a string as its second argument, by specification. You gave it a null, so before the logic of explode runs, PHP converts the null into a string. Every function operates this way: if the arguments don't match the formal specification, PHP attempts to "type juggle" until they do. If PHP can juggle, the engine motors along happily, otherwise you get a weak slap on the wrist:

PHP Warning: explode() expects parameter 2 to be string, object given in file.php on line 1

The way PHP juggles null to string is simple: null := ''. So a call with null and a call with '' are semantically equivalent because of type juggling, as seen here:

$a = explode(',', '');
$b = explode(',', null);
var_dump($a === $b); // true

$a = count(explode(',', ''));
$b = count(explode(',', null));
var_dump($a === $b); // true

So now you might ask: why does PHP return a one-element array when exploding on an empty string? That is also by specification:

If delimiter contains a value that is not contained in string ... an array containing string will be returned.