Negative numbers to binary string in JavaScript

早过忘川 提交于 2019-12-17 15:23:15

问题


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.


回答1:


Short answer:

  1. The toString() function takes the decimal, converts it to binary and adds a "-" sign.

  2. A zero fill right shift converts it's operands to signed 32-bit integers 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.

It takes the decimal, converts it to binary and adds a "-" sign.

  1. Base 10 "3" converted to base 2 is "11"
  2. 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 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.




回答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




回答3:


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 



回答4:


.toString() is designed to return the sign of the number in the string representation. See EcmaScript 2015, section 7.1.12.1:

  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:

  1. 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:

  1. Let lnum be ToUint32(lval).

The ToUint32 operation has a step where the operand is mapped into the unsigned 32 bit range:

  1. 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).



来源:https://stackoverflow.com/questions/16155592/negative-numbers-to-binary-string-in-javascript

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