Correct way to sum 128bit numbers represented by 4 32bit unsigned integers?

Asked Jan 17 '22 at 13:43

Active Jan 17 '22 at 13:47

Viewed 44 times

I'm creating a kernel in C++ / CUDA in which I need to sum two 128bit numbers represented as follows:

struct Int128 {
    unsigned int bit0_31;
    unsigned int bit32_63;
    unsigned int bit64_95;
    unsigned int bit96_127;
};

Int128 sum(Int128 a, Int128 b) {
    ...?
}

What is the most efficient C++ way to achieve this? Thanks.

P.S. Overflow to 129bit can be ignored.

edited Jan 17 '22 at 13:47

Daniel Langr

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asked Jan 17 '22 at 13:43

Frax

Which is the wider integral type allowed? – Jarod42 Jan 17 '22 at 13:45
3

Possibly relevant: https://stackoverflow.com/q/6162140/580083. – Daniel Langr Jan 17 '22 at 13:46
Since CUDA 11.5, it's possible to use __int128 in device code with compatible host compilers – Abator Abetor Jan 17 '22 at 13:51
I'd like to stick to 32bit math if possible. – Frax Jan 17 '22 at 13:52
Use `unsigned long long` as the accumulator. If you can't use that, add them in 16-bit chunks. – Eljay Jan 17 '22 at 13:55
@Eljay can... you give me an example using the ULL as the accumulator? – Frax Jan 17 '22 at 14:09
1

`uint64_t acc = a.bit0_31; acc += b.bit0_31; result.bit0_31 = static_cast(acc); acc >>= 32; acc += a.bit32_63;` and so forth. – Eljay Jan 17 '22 at 14:16
1

If you need to do a lot of bit twiddling code, I highly recommend Hacker's Delight (2nd ed) by Henry S. Warren Jr. It's a wonderful cookbook of bit twiddling routines, collected and collated over many years. – Eljay Jan 17 '22 at 14:26

Correct way to sum 128bit numbers represented by 4 32bit unsigned integers?

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