我希望看到二进制形式的正整数或负整数。

很像这个问题,但是是针对JavaScript的。


当前回答

您可以编写自己的函数来返回一个比特数组。 示例如何将数字转换为比特

除数|被除数|位/余数

2 | | 1

2 4 | | 0

2 | 2 | 0

~ | | ~ 1

上面一行的例子:2 * 4 = 8,余数为1 9 = 1 0 0 1

function numToBit(num){
    var number = num
    var result = []
    while(number >= 1 ){
        result.unshift(Math.floor(number%2))
        number = number/2
    }
    return result
}

从下往上读余数。数字1从中间到上面。

其他回答

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)

你可以使用递归解决方案:

函数intToBinary(number, res = "") { If (number < 1) If (res === "")返回"0" 其他的 返回res 否则返回intToBinary(数学。楼层(数目/ 2),数目% 2 + res) } console.log (intToBinary (12)) console.log (intToBinary (546)) console.log (intToBinary (0)) console.log (intToBinary (125)) 只对正数有效。

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("---");

这是我的代码:

var x = prompt("enter number", "7");
var i = 0;
var binaryvar = " ";

function add(n) {
    if (n == 0) {
        binaryvar = "0" + binaryvar; 
    }
    else {
        binaryvar = "1" + binaryvar;
    }
}

function binary() {
    while (i < 1) {
        if (x == 1) {
            add(1);
            document.write(binaryvar);
            break;
        }
        else {
            if (x % 2 == 0) {
                x = x / 2;
                add(0);
            }
            else {
                x = (x - 1) / 2;
                add(1);
            }
        }
    }
}

binary();

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));