secretflow/yacl

Trivial case in compile_time_utils.h

maths644311798 opened this issue · 3 comments

maths644311798 commented

compile_time_utils.h中有代码

template <>
constexpr uint32_t crc32<static_cast<size_t>(-1)>(
    [[maybe_unused]] const char* str) {
  return 0xFFFFFFFF;
}

这个特殊情况对应零多项式，应该返回0吧。

Jamie-Cui commented

可以参考一下这个逻辑，和 yacl 里面的 crc32 recursion 逻辑正好相反
https://web.mit.edu/freebsd/head/sys/libkern/crc32.c

/*
 * A function that calculates the CRC-32 based on the table above is
 * given below for documentation purposes. An equivalent implementation
 * of this function that's actually used in the kernel can be found
 * in sys/libkern.h, where it can be inlined.
 *
 *	uint32_t
 *	crc32(const void *buf, size_t size)
 *	{
 *		const uint8_t *p = buf;
 *		uint32_t crc;
 *
 *		crc = ~0U;  /* 初始化为 0xFFFFFFFF */
 *		while (size--)
 *			crc = crc32_tab[(crc ^ *p++) & 0xFF] ^ (crc >> 8);
 *		return crc ^ ~0U;
 *	}
 */
maths644311798 commented

我承认Yacl的实现和 https://web.mit.edu/freebsd/head/sys/libkern/crc32.c 有相同功能。但是

char c[2] = { 0x0,0x00 };
cout << hex << CT_CRC32(c) << '\n';

这段代码会输出 d202ef8d，如何解释？对于零多项式，按照CRC32算法预期应该输出0。

maths644311798 commented

明白了，我看了IEEE 802.3标准，确实首尾有取反操作。

  • 👍1