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
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
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edited May 23, 2017 at 10:34
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asked Jan 28, 2012 at 23:53
auroranilauroranil
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3 Answers
Reset to default 18Because, according to the ECMAScript specifications, bitwise 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:
Evaluate
A
.Call
GetValue(Result(1))
.Evaluate
B
.Call
GetValue(Result(3))
.Call
ToInt32(Result(2)).
Call
ToInt32(Result(4)).
Apply the bitwise operator
@
toResult(5)
andResult(6)
. The result is a signed 32 bit integer.Return
Result(7)
.
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’)
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.
((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
– user1106925 Commented Jan 28, 2012 at 23:57Math.floor(x)
function. jsperf.com/floor-or-or – auroranil Commented Jan 30, 2012 at 3:01-1.23
to see what happens – ajax333221 Commented Jun 5, 2012 at 3:53