Anyone knows why javascript Number.toString
function does not represents negative numbers correctly?
//If you try
(-3).toString(2); //shows "-11"
// but if you fake a bit shift operation it works as expected
(-3 >>> 0).toString(2); // print "11111111111111111111111111111101"
I am really curious why it doesn't work properly or what is the reason it works this way?
I've searched it but didn't find anything that helps.
Short answer:
The toString()
function basically takes the decimal, converts it
to binary and adds a "-" sign.
A zero fill right shift converts it's operand to a 32 signed bit
integer in two complements format.
A more detailed answer:
Question 1:
//If you try
(-3).toString(2); //show "-11"
It's in the function .toString()
. When you output a number via .toString()
:
Syntax
numObj.toString([radix])
If the numObj is negative, the sign is preserved. This is the case
even if the radix is 2; the string returned is the positive binary
representation of the numObj preceded by a - sign, not the two's
complement of the numObj.
So basically it takes the decimal, converts it to binary and adds a "-" sign.
- Base 10 "3" converted to base 2 is "11"
- Add a sign gives us "-11"
Question 2:
// but if you fake a bit shift operation it works as expected
(-3 >>> 0).toString(2); // print "11111111111111111111111111111101"
A zero fill right shift converts it's operand to a 32 signed bit integer.
The operands of all bitwise operators are converted to signed 32-bit
integers in two's complement format.
-3 >>> 0 (right logical shift) coerces its arguments to unsigned integers, which is why you get the 32-bit two's complement representation of -3.
http://en.wikipedia.org/wiki/Two%27s_complement
http://en.wikipedia.org/wiki/Logical_shift
var binary = (-3 >>> 0).toString(2); // coerced to uint32
console.log(binary);
console.log(parseInt(binary, 2) >> 0); // to int32
on jsfiddle
output is
11111111111111111111111111111101
-3