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.

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  ipToInt32("128.32.10.1") => 2149583361

The solution in Javascript

Option 1:

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

Option 2:

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

Option 3:

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function ipToInt32(ip){
  var array8 = new Uint8Array(ip.split('.').reverse().map(Number))
  var array32 = new Uint32Array(array8.buffer);
  return array32[0];
}

Test cases to validate our solution

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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")
  });
});