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liblzma: memcmplen.h: Add a comment why subtraction is used. 

(cherry picked from commit 0b99783d63)
This commit is contained in:
Lasse Collin 2024-03-22 17:46:30 +02:00
parent 94939a145f
commit b3a7561880
1 changed files with 13 additions and 0 deletions
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13
src/liblzma/common/memcmplen.h
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@ -67,6 +67,19 @@ lzma_memcmplen(const uint8_t *buf1, const uint8_t *buf2,
// This is only for x86-64 and ARM64 for now. This might be fine on // This is only for x86-64 and ARM64 for now. This might be fine on
// other 64-bit processors too. On big endian one should use xor // other 64-bit processors too. On big endian one should use xor
// instead of subtraction and switch to __builtin_clzll(). // instead of subtraction and switch to __builtin_clzll().
//
// Reasons to use subtraction instead of xor:
//
// - On some x86-64 processors (Intel Sandy Bridge to Tiger Lake),
// sub+jz and sub+jnz can be fused but xor+jz or xor+jnz cannot.
// Thus using subtraction has potential to be a tiny amount faster
// since the code checks if the quotient is non-zero.
//
// - Some processors (Intel Pentium 4) used to have more ALU
// resources for add/sub instructions than and/or/xor.
//
// The processor info is based on Agner Fog's microarchitecture.pdf
// version 2023-05-26. https://www.agner.org/optimize/
#define LZMA_MEMCMPLEN_EXTRA 8 #define LZMA_MEMCMPLEN_EXTRA 8
while (len < limit) { while (len < limit) {
const uint64_t x = read64ne(buf1 + len) - read64ne(buf2 + len); const uint64_t x = read64ne(buf1 + len) - read64ne(buf2 + len);