CODEWITHSUNDEEP
phpcount
Jeremie Ges
Jeremie Ges

Reputation: 2743

Why count(false) return 1?

Do you know why <?= count(false) ?> returns 1?

Upvotes: 19

Views: 10758

Answers (4)

Pwncopter
Pwncopter

Reputation: 31

A nice way to remember this in your mind:

  • count(false) is basically the same as:
  • count ("one boolean"), and therefore there are "ONE" booleans as result.

Upvotes: 3

Gabriel
Gabriel

Reputation: 464

It looks to me like PHP is preventing one from using count() to determine if an element is an array or an object. They have dedicated functions for this (is_array(), is_object()) and it may be tempting to naively use count() and check for a false condition to determine array or object. Instead, PHP makes non-objects, non-arrays return 1 (which is truthy) so that this method cannot be be naively used in this way (since 0 is a valid, falsy result for an empty array/object).

This may be the why behind the choice of value to be returned by the function in the situation you're describing.

Upvotes: 2

Chris Laplante
Chris Laplante

Reputation: 29658

It's specified behavior:

If var is not an array or an object with implemented Countable interface, 1 will be returned.

According to http://php.net/manual/en/function.count.php

Upvotes: 27

Starx
Starx

Reputation: 79031

Because false is also a value and if the count() does not get array but a valid variable it returns true which is 1.

$result = count(null);
// $result == 0

$result = count(false);
// $result == 1

Upvotes: 10

Related Questions