10

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
Community
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auroranil
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    You should be aware that using bitwise operators will limit you to 32-bit signed integers. `((Math.pow(2,32)/2)-1)|0; // 2147483647` Remove the `-1` and you'll not get the desired result. `((Math.pow(2,32)/2))|0; // -2147483648` –  Jan 28 '12 at 23:57
  • Interesting. That's probably the reason that this function is slightly faster than the `Math.floor(x)` function. http://jsperf.com/floor-or-or – auroranil Jan 30 '12 at 03:01
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    it is not actually 'flooring', try with `-1.23` to see what happens – ajax333221 Jun 05 '12 at 03:53

3 Answers3

14

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).

Hamish
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5

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’)
georg
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    `s/Most (.*?) don't/Strongly typed \1 generally don't/` (This is, however, very language-dependent.) –  Jan 28 '12 at 23:59
2

When doing a floor, although it would be possible to convert the argument to an integer, this is not what most languages would do because the original type is a floating-point number.

A better way to do it while preserving the data type is to go to exponent digits into the mantissa and zero the remaining bits.

If you're interested you can take a look at the IEEE spec for floating point numbers.

ose
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