Anyone knows why javascript Number.toString function does not represents negative numbers correctly?
//If you try
(-3).toString(2); //shows \"-11\"
// but if
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
.toString()
is designed to return the sign of the number in the string representation. See EcmaScript 2015, section 7.1.12.1:
- If m is less than zero, return the String concatenation of the String "-" and ToString(−m).
This rule is no different for when a radix is passed as argument, as can be concluded from section 20.1.3.6:
- Return the String representation of this Number value using the radix specified by radixNumber. [...] the algorithm should be a generalization of that specified in 7.1.12.1.
Once that is understood, the surprising thing is more as to why it does not do the same with -3 >>> 0
.
But that behaviour has actually nothing to do with .toString(2)
, as the value is already different before calling it:
console.log (-3 >>> 0); // 4294967293
It is the consequence of how the >>>
operator behaves.
It does not help either that (at the time of writing) the information on mdn is not entirely correct. It says:
The operands of all bitwise operators are converted to signed 32-bit integers in two's complement format.
But this is not true for all bitwise operators. The >>>
operator is an exception to the rule. This is clear from the evaluation process specified in EcmaScript 2015, section 12.5.8.1:
- Let lnum be ToUint32(lval).
The ToUint32 operation has a step where the operand is mapped into the unsigned 32 bit range:
- Let int32bit be int modulo 232.
When you apply the above mentioned modulo operation (not to be confused with JavaScript's %
operator) to the example value of -3, you get indeed 4294967293.
As -3 and 4294967293 are evidently not the same number, it is no surprise that (-3).toString(2)
is not the same as (4294967293).toString(2)
.
-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
Just to summarize a few points here, if the other answers are a little confusing:
(-3 >>> 0).toString(2)
, let's call it A, does the job; but we want to know why and how it worksvar num = -3; num.toString(-3)
we would have gotten -11
, which is simply the unsigned binary representation of the number 3 with a negative sign in front, which is not what we want1) (-3 >>> 0)
The >>>
operation takes the left operand (-3), which is a signed integer, and simply shifts the bits 0 positions to the left (so the bits are unchanged), and the unsigned number corresponding to these unchanged bits.
The bit sequence of the signed number -3 is the same bit sequence as the unsigned number 4294967293, which is what node gives us if we simply type -3 >>> 0
into the REPL.
2) (-3 >>> 0).toString
Now, if we call toString
on this unsigned number, we will just get the string representation of the bits of the number, which is the same sequence of bits as -3.
What we effectively did was say "hey toString, you have normal behavior when I tell you to print out the bits of an unsigned integer, so since I want to print out a signed integer, I'll just convert it to an unsigned integer, and you print the bits out for me."
The toString()
function takes the decimal, converts it
to binary and adds a "-" sign.
A zero fill right shift converts it's operands to signed 32-bit integers in two complements format.
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.
It takes the decimal, converts it to binary and adds a "-" sign.
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 operands to signed 32-bit integers. The result of that operation is always an unsigned 32-bit integer.
The operands of all bitwise operators are converted to signed 32-bit integers in two's complement format.