Thread (22 messages) 22 messages, 6 authors, 2016-03-05

Re: [9/9] powerpc: optimise csum_partial() call when len is constant

From: Scott Wood <oss@buserror.net>
Date: 2016-03-05 05:29:14
Also in: linuxppc-dev, lkml

On Tue, Sep 22, 2015 at 04:34:36PM +0200, Christophe Leroy wrote:
+/*
+ * computes the checksum of a memory block at buff, length len,
+ * and adds in "sum" (32-bit)
+ *
+ * returns a 32-bit number suitable for feeding into itself
+ * or csum_tcpudp_magic
+ *
+ * this function must be called with even lengths, except
+ * for the last fragment, which may be odd
+ *
+ * it's best to have buff aligned on a 32-bit boundary
+ */
+__wsum __csum_partial(const void *buff, int len, __wsum sum);
+
+static inline __wsum csum_partial(const void *buff, int len, __wsum sum)
+{
+	if (__builtin_constant_p(len) && len == 0)
+		return sum;
+
+	if (__builtin_constant_p(len) && len <= 16 && (len & 1) == 0) {
+		__wsum sum1;
+
+		if (len == 2)
+			sum1 = (__force u32)*(u16 *)buff;
+		if (len >= 4)
+			sum1 = *(u32 *)buff;
+		if (len == 6)
+			sum1 = csum_add(sum1, (__force u32)*(u16 *)(buff + 4));
+		if (len >= 8)
+			sum1 = csum_add(sum1, *(u32 *)(buff + 4));
+		if (len == 10)
+			sum1 = csum_add(sum1, (__force u32)*(u16 *)(buff + 8));
+		if (len >= 12)
+			sum1 = csum_add(sum1, *(u32 *)(buff + 8));
+		if (len == 14)
+			sum1 = csum_add(sum1, (__force u32)*(u16 *)(buff + 12));
+		if (len >= 16)
+			sum1 = csum_add(sum1, *(u32 *)(buff + 12));
+
+		sum = csum_add(sum1, sum);
Why the final csum_add instead of s/sum1/sum/ and putting csum_add in the
"len == 2" and "len >= 4" cases?

The (__force u32) casts are unnecessary.  Or rather, it should be
(__force __wsum) -- on all of them, not just the 16-bit ones.

The pointer casts should be const.
+	} else if (__builtin_constant_p(len) && (len & 3) == 0) {
+		sum = csum_add(ip_fast_csum_nofold(buff, len >> 2), sum);
It may not make a functional difference, but based on the csum_add()
argument names and other csum_add() usage, sum should come first
and the new content second.

-Scott
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