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
}挑战是编写三个函数来计算两个数字的按位或、位或和位异或。
发布于 2022-01-19 07:33:42
您的代码很好,唯一缺少的就是在返回值上将二进制转换为整数。
.的
return +x试一试
return parseInt(x, 2)https://stackoverflow.com/questions/70766038
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