php - why does casting false to an array not result in empty array

Question

I have a method return type can either be an array of things or false.

I am therefore casting the result to an array to try to ensure my code doesnt need to check for false or array.

But, php is casting false to an array containing 1 element: false:

array(1) {
  [0]=>
  bool(false)
}

WHY!

Is there a way to achieve this without the if statement?

i.e.

if ($returnValue === false) {
    return array();
} else {
    return $returnValue;
}

"the if statement. What if statement?

h2ooooooo
– h2ooooooo

2013-11-11 17:04:44 +00:00
Commented Nov 11, 2013 at 17:04 — h2ooooooo
– h2ooooooo, Commented Nov 11, 2013 at 17:04
How are you doing the casting?

Izkata
– Izkata

2013-11-11 17:05:14 +00:00
Commented Nov 11, 2013 at 17:05 — Izkata
– Izkata, Commented Nov 11, 2013 at 17:05
the casting is like (array)$returnValue

Marty Wallace
– Marty Wallace

2013-11-11 17:05:57 +00:00
Commented Nov 11, 2013 at 17:05 — Marty Wallace
– Marty Wallace, Commented Nov 11, 2013 at 17:05

h2ooooooo · Accepted Answer · 2013-11-11 17:13:43Z

6

Working as designed. The manual specifies :

For any of the types: integer, float, string, boolean and resource, converting a value to an array results in an array with a single element with index zero and the value of the scalar which was converted. In other words, (array)$scalarValue is exactly the same as array($scalarValue).

Can't you simply use a ternary operator?

return (is_array($returnValue) ? $returnValue : array());

edited Nov 11, 2013 at 17:13

answered Nov 11, 2013 at 17:08

h2ooooooo

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2 Comments

moonwave99 Over a year ago

+1 it should always be like that, whenever you expect an array as the return type.

Jake Over a year ago

Indeed. You might have an array of Booleans. Also worth noting that casting null to an array gives an empty array, not null, but which will evaluate to false in a loose conditional (with == rather than ===, or without a comparison operator).

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