Question

What would be a efficient way to generate all possible IP v4 addresses? other than iterating all bytes in a one giant nested for loop.

Answer 1

Edit: My previous answer would have gone from 128.0.0.0 to 255.255.255.255 to 0.0.0.0 to 127.255.255.255. Presumably you want to go from 0.0.0.0 to 255.255.255.255, so I’ve edited my solution to do that.

int i = -1;
do {
  i++;

  int b1 = (i >> 24) & 0xff;
  int b2 = (i >> 16) & 0xff;
  int b3 = (i >>  8) & 0xff;
  int b4 = (i      ) & 0xff;

  //Now the IP is b1.b2.b3.b4

} while(i != -1);

Note: if you’re confused how this loop will ever end (i.e. how adding 1 to -1 enough times makes it -1 again), read up on two’s complement. Basically, adding one to Integer.MAX_VALUE results in Integer.MIN_VALUE, and does not throw any kind of exception.

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Old answer. Still hits all IPs, but probably not in the order you desire:

for(long n = Integer.MIN_VALUE; n <= Integer.MAX_VALUE; n++)
{
  int i = (int)n;

  int b1 = (i >> 24) & 0xff;
  int b2 = (i >> 16) & 0xff;
  int b3 = (i >>  8) & 0xff;
  int b4 = (i      ) & 0xff;

  //Now the IP is b1.b2.b3.b4
}

Please note: If the loop control variable was an int instead of a long, this would be an infinite loop (since all ints are always <= Integer.MAX_VALUE).