Binary and decimal conversion, bit operations
Under Record on codewar do a topic and harvest
128.32.10.1 == 10000000.00100000.00001010.00000001
Because the above IP address has 32 bits, we can represent it as the unsigned 32 bit number: 2149583361
Complete the function that takes an unsigned 32 bit number and returns a string representation of its IPv4 address.
Example : 2149583361 ==> "128.32.10.1"
Their problem-solving idea is to decimal numbers into binary (0 up less than 32), and then successively taken into eight decimal number, then .
the connection shall be the IP .
Record a few points inside it:
- Decimal to binary
numObj.toString([radix])
radix can specify decimal, the default is 10
let x = 2149583361;
x.toString(2) // "10000000001000000000101000000001"
- Binary to decimal
Number.parseInt(string[, radix])
radix can specify decimal, the default is 10
Number.parseInt("10000000001000000000101000000001",2) // 2149583361
- When less than 32 how to quickly fill
0
Array(len + 1).join('0')
let x = 998, //指定值
x_2 = x.toString(2),
len = 32 - x_2.length; // 需要补0的个数
x_2 += Array(len + 1).join('0');
Complete problem-solving as follows:
function int32ToIp(int32){
let int2 = int32.toString(2),
len = 32 - int2.length,
begins = [0,8,16,24],
ipArr = [];
if (len) {
int2 += Array(len + 1).join('0')
}
begins.forEach((begin) => {
ipArr.push(Number.parseInt(int2.slice(begin,begin + 8),2))
})
return ipArr.join('.');
}
int32ToIp(2149583361) // '128.32.10.1'
After submitting the bigwigs find other simple idea is to use the shift operator
let x = 2149583361; // 按位移动会先将操作数转换为大端字节序顺序(big-endian order)的32位整数
x >> 24 & 0xff // 128 //右移24位即可得到原来最左边8位,然后&运算取值
Similarly 16,8,0 right to get to the corresponding IP field.
Function is as follows:
function int32ToIp(int32){
return `${int32 >> 24 & 0xff}.${int32 >> 16 & 0xff}.${int32 >> 8 & 0xff}.${int32 >> 0 & 0xff}`
}
int32ToIp(2149583361) // '128.32.10.1'