x86 rte_memcpy_aligned possible optimization

[Morten Brørup]

Hi Bruce,

I think one of the loops in rte_memcpy_aligned() takes one too many rounds in the case where the catch-up could replace the last round.

Consider e.g. n = 128:

The 64 bytes block copy will take two rounds, and the catch-up will copy the last 64 bytes once again.

I think that the 64 bytes block copy could take only one round and let the catch-up copy the last 64 bytes.

I'm not sure if my suggested method is generally faster than the current method, so I'm passing the ball.

PS: It looks like something similar can be done for the other block copy loops in this file. I haven't dug into the details.


static __rte_always_inline void *
rte_memcpy_aligned(void *dst, const void *src, size_t n)
{
	void *ret = dst;

	/* Copy size < 16 bytes */
	if (n < 16) {
		return rte_mov15_or_less(dst, src, n);
	}

	/* Copy 16 <= size <= 32 bytes */
	if (n <= 32) {
		rte_mov16((uint8_t *)dst, (const uint8_t *)src);
		rte_mov16((uint8_t *)dst - 16 + n,
				(const uint8_t *)src - 16 + n);

		return ret;
	}

	/* Copy 32 < size <= 64 bytes */
	if (n <= 64) {
		rte_mov32((uint8_t *)dst, (const uint8_t *)src);
		rte_mov32((uint8_t *)dst - 32 + n,
				(const uint8_t *)src - 32 + n);

		return ret;
	}

	/* Copy 64 bytes blocks */
-	for (; n >= 64; n -= 64) {
+	for (; n > 64; n -= 64) {
		rte_mov64((uint8_t *)dst, (const uint8_t *)src);
		dst = (uint8_t *)dst + 64;
		src = (const uint8_t *)src + 64;
	}

	/* Copy whatever left */
	rte_mov64((uint8_t *)dst - 64 + n,
			(const uint8_t *)src - 64 + n);

	return ret;
}


Med venlig hilsen / Kind regards,
-Morten Brørup