I'm storing the IP address in sockaddr_in6
which supports an array of four 32-bit, addr[4]. Essentially a 128 bit number.
I'm trying to calculate number of IPs in a given IPv6 range (how many IPs between). So it's a matter of subtracting one from another using two arrays with a length of four.
The problem is since there's no 128bit data type, I can't convert into decimal.
Thanks a ton!
You could use some kind of big-int library (if you can tolerate LGPL, GMP is the choice). Fortunately, 128 bit subtraction is easy to simulate by hand if necessary. Here is a quick and dirty demonstration of computing the absolute value of (a-b), for 128 bit values:
#include <iostream>
#include <iomanip>
struct U128
{
unsigned long long hi;
unsigned long long lo;
};
bool subtract(U128& a, U128 b)
{
unsigned long long carry = b.lo > a.lo;
a.lo -= b.lo;
unsigned long long carry2 = b.hi > a.hi || a.hi == b.hi && carry;
a.hi -= carry;
a.hi -= b.hi;
return carry2 != 0;
}
int main()
{
U128 ipAddressA = { 45345, 345345 };
U128 ipAddressB = { 45345, 345346 };
bool carry = subtract(ipAddressA, ipAddressB);
// Carry being set means that we underflowed; that ipAddressB was > ipAddressA.
// Lets just compute 0 - ipAddressA as a means to calculate the negation
// (0-x) of our current value. This gives us the absolute value of the
// difference.
if (carry)
{
ipAddressB = ipAddressA;
ipAddressA = { 0, 0 };
subtract(ipAddressA, ipAddressB);
}
// Print gigantic hex string of the 128-bit value
std::cout.fill ('0');
std::cout << std::hex << std::setw(16) << ipAddressA.hi << std::setw(16) << ipAddressA.lo << std::endl;
}
This gives you the absolute value of the difference. If the range is not huge (64 bits or less), then ipAddressA.lo
can be your answer as a simple unsigned long long
.
If you have perf concerns, you can make use of compiler intrinsics for taking advantage of certain architectures, such as amd64 if you want it to be optimal on that processor. _subborrow_u64
is the amd64 intrinsic for the necessary subtraction work.
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