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Re: IP in range

by smahesh (Pilgrim)
on Jan 07, 2011 at 03:27 UTC ( #880996=note: print w/replies, xml ) Need Help??

in reply to IP in range


I solved this problem in my previous company in Java. I will present the idea here but leave the implementation to you.

A IPv4 address is just a collection of four octets and can be represented by a 32 bit integer. Just convert the start and end address of the ip range to a 32 bit integer representation. Convert the Ipv4 address to be checked also into a 32 bit number and if the number falls between the start and end numbers, it is in the range. If the address check is called lots of times for a particular range, a integer comparison will be faster than a string comparison.

For example,
Start IP = = 0x01020304 = 1057540 (decimal)
End IP = = 0x0C06FFFF = 201785343 (decimal)
Target IP = = 0x80010003 = 2147549187 (decimal)
2147549187 is not between 1057540 and 201785343 - so target ip is not in range.


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shmem is going to look for problems with C++. Urgh.
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[stevieb]: shmem I had to do that yesterday and earlier today (C++ issues). It was only fun after I figured it all out.
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