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ATTN: oot	by hadji	2014-02-08 11:11:44
...I think I understand now	by oot	2014-02-08 12:33:47
When I saw your method (which makes a lot of intuitive sense) I thought it meant my method was wrongly calculating every IP as the start of a network range, eg. the impossible 199.74.83.3 - 199.74.83.130. But it wasn't, because for the last octet I used 2 rather than 256. So it was actually about the number of numbers between 11000111010010100101001100000000 (199.74.83.0, the lowest possible subnet) through 11001111010011100101010110000000 (207.78.85.128, the highest possible subnet) that were multiples of 128 (eg. ended in 0000000) and therefore valid network addresses for a /25 subnet. So, putting those big binary numbers in decimal, (3478017408 - 3343536896)/128 = 1050629. Then add 1 to avoid "fencepost error" (the lowest subnet isn't just a boundary, it's a member of the set of valid subnets) and get 1050630, verifying kahuana's answer as correct.
[ Reply ]
What can I say, except.... e!	by kahuana	2014-02-08 16:01:56

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