Why does count (false) return 1?

Question

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

Answer 1

Specified behavior:

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

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

Answer 2

Because false also a value, and if count () does not receive an array, but a real variable, it returns true , which is 1 .

 $result = count(null); // $result == 0 $result = count(false); // $result == 1 

Answer 3

It seems to me that PHP does not allow the use of count() to determine if an element is an array or an object. They have special functions for this ( is_array() , is_object() ), and it may be tempting to use count() naively and check the false condition to define an array or object. Instead, PHP does non-objects, non-arrays returns 1 (which is true), so this method cannot be naively used that way (since 0 is a valid, false result for an empty array / object).

This may be the reason for choosing the value returned by the function in the described situation.

Answer 4

A good way to remember this in your mind:

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