我的代码通过了我所有的测试，但有些 edabit 不认可它

Question

function bitwiseAND(n1, n2) {
  let a = n1.toString(2).padStart(8, "0");
  let b = n2.toString(2).padStart(8, "0");
  let x = "";
  for (let i = 0; i < 8; i++) {
     if (+a[i] === 1 && +a[i] === +b[i]) x = x + "1";
     else x = x + "0";
}
return +x
}

function bitwiseOR(n1, n2) {
  let a = n1.toString(2).padStart(8, "0");
  let b = n2.toString(2).padStart(8, "0");
  let x = "";
  for (let i = 0; i < 8; i++) {
    if (+a[i] === 1 || +b[i] === 1) x = x + "1";
    else x = x + "0";
}
return +x
}

function bitwiseXOR(n1, n2) {
  let a = n1.toString(2).padStart(8, "0");
  let b = n2.toString(2).padStart(8, "0");
  let x = "";
  for (let i = 0; i < 8; i++) {
    if ((+a[i] === 1 && +b[i] === 0) || (+a[i] === 0 && +b[i] === 1))
      x = x + "1";
    else x = x + "0";
}
return +x
}

This is link to the challenge

挑战是编写三个函数来计算两个数字的按位与、按位或和按位异或。

Answer 1

您的代码很好，唯一缺少的是将二进制转换为 return 值的整数。

代替

return +x

尝试

return parseInt(x, 2)

我的代码通过了我所有的测试，但有些 edabit 不认可它

My code passes all my tests but some how edabit is not approving it

javascript

bitwise-operators

bitwise-and

bitwise-or

bitwise-xor