How does x|0 floor the number in JavaScript?

Question

In the accepted answer on my earlier question ( What is the fastest way to generate a random integer in javascript? ), I was wondering how a number loses its decimals via the symbol | .

For example:

var x = 5.12042;
x = x|0;

How does that floor the number to 5?

Some more examples:

console.log( 104.249834 | 0 ); //104
console.log( 9.999999 | 0 );   // 9

Answer 1

Because, according to the ECMAScript specifications, bitwise operators operators call ToInt32 on each expression to be evaluated.

See 11.10 Binary Bitwise Operators:

The production A : A @B, where @ is one of the bitwise operators in the productions above, is evaluated as follows:

1. Evaluate A.

2. Call GetValue(Result(1)).

3. Evaluate B.

4. Call GetValue(Result(3)).

5. Call ToInt32(Result(2)).

6. Call ToInt32(Result(4)).

7. Apply the bitwise operator @ to Result(5) and Result(6). The result is a signed 32 bit integer.

8. Return Result(7).

Answer 2

Bitwise operators convert their arguments to integers (see http://es5.github.com/#x9.5). Most languages I know don't support this type of conversion:

    $ python -c "1.0|0"
    Traceback (most recent call last):
      File "", line 1, in 
    TypeError: unsupported operand type(s) for |: 'float' and 'int'

    $ ruby -e '1.0|0'
    -e:1:in `': undefined method `|' for 1.0:Float (NoMethodError)

    $ echo "int main(){1.0|0;}" | gcc -xc -
    : In function ‘main’:
    :1: error: invalid operands to binary | (have ‘double’ and ‘int’)