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Lets say that I have an array of 4 32-bit integers which I use to store the 128-bit number
How can I perform left and right shift on this 128-bit number?
Thanks!
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If you only need to store it then you can store it in a byte array like "char num128[16]". If you need to manipulate it you need to use big numbers library like GMP. Show activity on this post. It is not possible to store it in one primitive data type, so we have to be slightly more creative.
The 128-bit data type can handle up to 31 significant digits (compared to 17 handled by the 64-bit long double). However, while this data type can store numbers with more precision than the 64-bit data type, it does not store numbers of greater magnitude.
1 in binary is 0001 , then bitshifting it by 0 won't do anything, which aligns with what you observed. So any number x << 0 is equivalent to x * 2^0 , which is x * 1 , which is just x .
Working with uint128? If you can, use the x86 SSE instructions, which were designed for exactly that. (Then, when you've bitshifted your value, you're ready to do other 128-bit operations...)
SSE2 bit shifts take ~4 instructions on average, with one branch (a case statement). No issues with shifting more than 32 bits, either. The full code for doing this is, using gcc intrinsics rather than raw assembler, is in sseutil.c (github: "Unusual uses of SSE2") -- and it's a bit bigger than makes sense to paste here.
The hurdle for many people in using SSE2 is that shift ops take immediate (constant) shift counts. You can solve that with a bit of C preprocessor twiddling (wordpress: C preprocessor tricks). After that, you have op sequences like:
LeftShift(uint128 x, int n) = _mm_slli_epi64(_mm_slli_si128(x, n/8), n%8)
for n = 65..71, 73..79, … 121..127 ... doing the whole shift in two instructions.
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