如何在JS中将十进制转换为二进制？

Question

可以通过以下方式将二进制转换为十进制：

var binary = "110";
var int = parseInt(binary, 2);
document.getElementById("results").innerHTML = int;

<div id="results"></div>

但是，我如何做相反的操作：将 int 转换为二进制？

Answer 1

let decimal = prompt('please insert decimal number');
console.log(Number(decimal).toString(2));

Answer 2

Dec 到 Bin：原始（按位）

/**
*   Dec to Bin 
*   with bitwise operations
*
*   Eudes Serpa M.
**/

const numberToConvert = 5;
const numberOfBits = 32; // 32-bits binary
const arrBitwise = [0]; // save the resulting bitwise

for (let i=0; i<numberOfBits; i++) {
    let mask = 1;

    const bit = numberToConvert & (mask << i); // And bitwise with left shift

    if(bit === 0) {
        arrBitwise[i] = 0;
    } else {
        arrBitwise[i] = 1;
    }
}

const binary = arrBitwise.reverse().join("");

console.log(`This is the resulting binary: ${binary}`)
console.log(`This is the verification ${parseInt(binary, 2)}`);

解释：

• 第 2 行：我们指定组成生成的二进制文件的位数。

• 第 3 行：我们定义一个数组，用于保存位级操作产生的位。最后，这将是我们生成的二进制文件（反转它）

• 用于：用于“创建”位的二进制。

• Mask：表示我们在位级别移位到的数字（1 进行 AND 运算并获得要转换的数字的 1 中的位）。

• bit：执行操作的结果位，例如：

  位数 = 3;

  掩码=1；

  for (i = 0 -> 31) { // 32 位

    // Explanation of the operation to obtain the bit in position i
  
    // ---- For i = 0;
    1. mask << 0 = ...0001 (a 1 in decimal), since it does not do any shifting.
  
    2. 3 & 1
        /* At the bit level we have to
             3 = ...0011
             1 = ...0001,
             so when doing the AND operation at the bit level, we have to:
         0011
        &0001
        ------
        0001 === 1 decimal
     */
  
    // bit then takes the value resulting from the previous operations. This is: 
    bit = 1;
  
    // The if is not meet, so it enters the else:
    arrBitwise[0] = 1;
  
    // ---- For i = 1;
    1. mask << 1 = ...0010 (a 2 in decimal)
  
    2. 3 & 2
        /* At the bit level we have to
             3 = ...0011
             2 = ...0010,
             so when doing the AND operation at the bit level, we have to:
        0011
       &0010
       -------
        0010 === 2 decimal
     */
  
    // bit then takes the value resulting from the previous operations. This is: bit = 2;
  
    // The if is not meet, so it enters the else:
    arrBitwise[1] = 1;
  
  
    // ----- For i = 2;
    1. mask << 2 = ...0100 (a 4 in decimal)
  
    2. 3. 4
        /* At the bit level we have to
             3 = ...0011
             4 = ...0100,
             so when doing the AND operation at the bit level, we have to:
        0011
       &0100
      -------
        0000 === 0 decimal
     */
  
    // bit then takes the value resulting from the previous operations. This is: 
    bit = 0;
  
   // The if meet, so:
    arrBitwise[2] = 0;
  

  }

因此，arrBitwise 将是： arrBitwise = [1, 1, 0, 0, ..., 0];

arrBitwise.reverse() // [0, ..., 0, 0, 1, 1]

与 .join()

"0...0011"

二进制表示3。

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Bitwise_AND

https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Left_shift

Answer 3

var x = 6;
console.log(x.toString(2));