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

diff --git a/src/liblzma/common/memcmplen.h b/src/liblzma/common/memcmplen.h
index d8c42040d36874bcb369769af214015b044af263..394a4856dd6a0b2d3d42ba3ca20bc97c15e3f43e 100644 (file)
--- a/src/liblzma/common/memcmplen.h
+++ b/src/liblzma/common/memcmplen.h
@@ -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
        // other 64-bit processors too. On big endian one should use xor
        // 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
        while (len < limit) {
                const uint64_t x = read64ne(buf1 + len) - read64ne(buf2 + len);