## The challenge

Take the following IPv4 address: 128.32.10.1 This address has 4 octets where each octet is a single byte (or 8 bits).

• 1st octet 128 has the binary representation: 10000000
• 2nd octet 32 has the binary representation: 00100000
• 3rd octet 10 has the binary representation: 00001010
• 4th octet 1 has the binary representation: 00000001

So 128.32.10.1 == 10000000.00100000.00001010.00000001

Because the above IP address has 32 bits, we can represent it as the 32 bit number: 2149583361.

Write a function `ipToInt32(ip)` that takes an IPv4 address and returns a 32 bit number.

 ``````1 `````` `````` ipToInt32("128.32.10.1") => 2149583361 ``````

## The solution in Javascript

Option 1:

 ``````1 2 3 `````` ``````function ipToInt32(ip){ return ip.split(".").reduce(function(int,v){ return int*256 + +v } ) } ``````

Option 2:

 ``````1 2 3 4 `````` ``````function ipToInt32(ip){ ip = ip.split('.'); return ((ip << 24) + (ip << 16) + (ip << 8) + (ip << 0))>>>0; } ``````

Option 3:

 ``````1 2 3 4 5 `````` ``````function ipToInt32(ip){ var array8 = new Uint8Array(ip.split('.').reverse().map(Number)) var array32 = new Uint32Array(array8.buffer); return array32; } ``````

## Test cases to validate our solution

 ``````1 2 3 4 5 6 7 `````` ``````describe("Tests", () => { it("test", () => { Test.assertEquals(ipToInt32("128.32.10.1"),2149583361, "wrong integer for ip: 128.32.10.1") Test.assertEquals(ipToInt32("128.114.17.104"),2154959208, "wrong integer for ip: 128.114.17.104") Test.assertEquals(ipToInt32("0.0.0.0"),0, "wrong integer for ip: 0.0.0.0") }); }); ``````