我用了一种不同的方法来解决这个问题。我决定在我的项目中不使用这段代码，但我想我会把它放在相关的地方，以防它对某人有用。

Doesn't use bit-shifting or two's complement coercion. You choose the number of bits that comes out (it checks for valid values of '8', '16', '32', but I suppose you could change that) You choose whether to treat it as a signed or unsigned integer. It will check for range issues given the combination of signed/unsigned and number of bits, though you'll want to improve the error handling. It also has the "reverse" version of the function which converts the bits back to the int. You'll need that since there's probably nothing else that will interpret this output :D

function intToBitString(input, size, unsigned) { if ([8, 16, 32].indexOf(size) == -1) { throw "invalid params"; } var min = unsigned ? 0 : - (2 ** size / 2); var limit = unsigned ? 2 ** size : 2 ** size / 2; if (!Number.isInteger(input) || input < min || input >= limit) { throw "out of range or not an int"; } if (!unsigned) { input += limit; } var binary = input.toString(2).replace(/^-/, ''); return binary.padStart(size, '0'); } function bitStringToInt(input, size, unsigned) { if ([8, 16, 32].indexOf(size) == -1) { throw "invalid params"; } input = parseInt(input, 2); if (!unsigned) { input -= 2 ** size / 2; } return input; } // EXAMPLES var res; console.log("(uint8)10"); res = intToBitString(10, 8, true); console.log("intToBitString(res, 8, true)"); console.log(res); console.log("reverse:", bitStringToInt(res, 8, true)); console.log("---"); console.log("(uint8)127"); res = intToBitString(127, 8, true); console.log("intToBitString(res, 8, true)"); console.log(res); console.log("reverse:", bitStringToInt(res, 8, true)); console.log("---"); console.log("(int8)127"); res = intToBitString(127, 8, false); console.log("intToBitString(res, 8, false)"); console.log(res); console.log("reverse:", bitStringToInt(res, 8, false)); console.log("---"); console.log("(int8)-128"); res = intToBitString(-128, 8, false); console.log("intToBitString(res, 8, true)"); console.log(res); console.log("reverse:", bitStringToInt(res, 8, true)); console.log("---"); console.log("(uint16)5000"); res = intToBitString(5000, 16, true); console.log("intToBitString(res, 16, true)"); console.log(res); console.log("reverse:", bitStringToInt(res, 16, true)); console.log("---"); console.log("(uint32)5000"); res = intToBitString(5000, 32, true); console.log("intToBitString(res, 32, true)"); console.log(res); console.log("reverse:", bitStringToInt(res, 32, true)); console.log("---");

这就是解。事实上，这很简单

function binaries(num1){ 
        var str = num1.toString(2)
        return(console.log('The binary form of ' + num1 + ' is: ' + str))
     }
     binaries(3

)

        /*
         According to MDN, Number.prototype.toString() overrides 
         Object.prototype.toString() with the useful distinction that you can 
         pass in a single integer argument. This argument is an optional radix, 
         numbers 2 to 36 allowed.So in the example above, we’re passing in 2 to 
         get a string representation of the binary for the base 10 number 100, 
         i.e. 1100100.
        */

函数 dec2bin（dec） { return （dec >>> 0）.toString（2）; } console.log（dec2bin（1））;1 console.log（dec2bin（-1））;11111111111111111111111111111111 控制台.log（dec2bin（256））;100000000 console.log（dec2bin（-256））;11111111111111111111111100000000

您可以使用Number.toString(2)函数，但它在表示负数时存在一些问题。例如，(-1). tostring(2)输出为“-1”。

要解决这个问题，可以使用无符号右移位操作符(>>>)将数字强制转换为无符号整数。

如果你运行(-1 >>> 0). tostring(2)，你将把你的数字向右移动0位，这不会改变数字本身，但它将表示为一个无符号整数。上面的代码将正确地输出“111111111111111111111111111111111111111111111111”。

这个问题有进一步的解释。

-3 >>> 0(右逻辑移位)将其参数强制为无符号整数，这就是为什么你得到了-3的32位2的补数表示。

我们还可以计算正数或负数的二进制，如下所示:

函数toBinary (n) { Let binary = ""; 如果(n < 0) { N = N >>> 0; ｝ while(Math.ceil(n/2) > 0){ 二进制= n%2 +二进制; n = Math.floor(n/2); ｝ 返回二进制; ｝ console.log (toBinary (7)); console.log (toBinary (7));

逻辑可以被任何编程语言实现的实际解决方案:

如果你确定它只是积极的:

var a = 0;
var n = 12; // your input
var m = 1;
while(n) {
    a = a + n%2*m;
    n = Math.floor(n/2);
    m = m*10;
}

console.log(n, ':', a) // 12 : 1100

若能负或正——

(n >>> 0).toString(2)

我是这样处理的:

const decbin = nbr => {
  if(nbr < 0){
     nbr = 0xFFFFFFFF + nbr + 1
  }
  return parseInt(nbr, 10).toString(2)
};

从这个链接获得:https://locutus.io/php/math/decbin/