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/*
 * Checksum routine for Internet Protocol family headers.
 *
 * This routine is very heavily used in the network
 * code and should be modified for each CPU to be as fast as possible.
 *
 * This implementation is 386 version.
 *
 *  $Id$
 */

#include <stdio.h>                /* for puts */

#undef	ADDCARRY
#define ADDCARRY(x)     if ((x) > 0xffff) (x) -= 0xffff
#define REDUCE          {sum = (sum & 0xffff) + (sum >> 16); ADDCARRY(sum);}

/*
 * Thanks to gcc we don't have to guess
 * which registers contain sum & w.
 */
#define ADD(n)	__asm__ volatile \
    ("addl " #n "(%2), %0" : "=r" (sum) : "0" (sum), "r" (w))
#define ADDC(n)	__asm__ volatile \
    ("adcl " #n "(%2), %0" : "=r" (sum) : "0" (sum), "r" (w))
#define LOAD(n)	__asm__ volatile \
     ("movb " #n "(%1), %0" : "=q" (junk) : "r" (w))
#define MOP	__asm__ volatile \
    ("adcl         $0, %0" : "=r" (sum) : "0" (sum))

int
in_cksum(
	struct mbuf *m,
	int len )
{
	register u_short *w;
	register unsigned sum = 0;
	register int mlen = 0;
	int byte_swapped = 0;
	union { char	c[2]; u_short	s; } su;

	for (;m && len; m = m->m_next) {
		if (m->m_len == 0)
			continue;
		w = mtod(m, u_short *);
		if (mlen == -1) {
			/*
			 * The first byte of this mbuf is the continuation
			 * of a word spanning between this mbuf and the
			 * last mbuf.
			 */

			/* su.c[0] is already saved when scanning previous
			 * mbuf.  sum was REDUCEd when we found mlen == -1
			 */
			su.c[1] = *(u_char *)w;
			sum += su.s;
			w = (u_short *)((char *)w + 1);
			mlen = m->m_len - 1;
			len--;
		} else
			mlen = m->m_len;
		if (len < mlen)
			mlen = len;
		len -= mlen;
		/*
		 * Force to long boundary so we do longword aligned
		 * memory operations
		 */
		if (3 & (int) w) {
			REDUCE;
			if ((1 & (int) w) && (mlen > 0)) {
				sum <<= 8;
				su.c[0] = *(char *)w;
				w = (u_short *)((char *)w + 1);
				mlen--;
				byte_swapped = 1;
			}
			if ((2 & (int) w) && (mlen >= 2)) {
				sum += *w++;
				mlen -= 2;
			}
		}
		/*
		 * Advance to a 486 cache line boundary.
		 */
		if (4 & (int) w && mlen >= 4) {
			ADD(0);
			MOP;
			w += 2;
			mlen -= 4;
		}
		if (8 & (int) w && mlen >= 8) {
			ADD(0);
			ADDC(4);
			MOP;
			w += 4;
			mlen -= 8;
		}
		/*
		 * Do as much of the checksum as possible 32 bits at at time.
		 * In fact, this loop is unrolled to make overhead from
		 * branches &c small.
		 */
		mlen -= 1;
		while ((mlen -= 32) >= 0) {
			u_char junk;
			/*
			 * Add with carry 16 words and fold in the last
			 * carry by adding a 0 with carry.
			 *
			 * The early ADD(16) and the LOAD(32) are to load
			 * the next 2 cache lines in advance on 486's.  The
			 * 486 has a penalty of 2 clock cycles for loading
			 * a cache line, plus whatever time the external
			 * memory takes to load the first word(s) addressed.
			 * These penalties are unavoidable.  Subsequent
			 * accesses to a cache line being loaded (and to
			 * other external memory?) are delayed until the
			 * whole load finishes.  These penalties are mostly
			 * avoided by not accessing external memory for
			 * 8 cycles after the ADD(16) and 12 cycles after
			 * the LOAD(32).  The loop terminates when mlen
			 * is initially 33 (not 32) to guaranteed that
			 * the LOAD(32) is within bounds.
			 */
			ADD(16);
			ADDC(0);
			ADDC(4);
			ADDC(8);
			ADDC(12);
			LOAD(32);
			ADDC(20);
			ADDC(24);
			ADDC(28);
			MOP;
			w += 16;
		}
		mlen += 32 + 1;
		if (mlen >= 32) {
			ADD(16);
			ADDC(0);
			ADDC(4);
			ADDC(8);
			ADDC(12);
			ADDC(20);
			ADDC(24);
			ADDC(28);
			MOP;
			w += 16;
			mlen -= 32;
		}
		if (mlen >= 16) {
			ADD(0);
			ADDC(4);
			ADDC(8);
			ADDC(12);
			MOP;
			w += 8;
			mlen -= 16;
		}
		if (mlen >= 8) {
			ADD(0);
			ADDC(4);
			MOP;
			w += 4;
			mlen -= 8;
		}
		if (mlen == 0 && byte_swapped == 0)
			continue;       /* worth 1% maybe ?? */
		REDUCE;
		while ((mlen -= 2) >= 0) {
			sum += *w++;
		}
		if (byte_swapped) {
			sum <<= 8;
			byte_swapped = 0;
			if (mlen == -1) {
				su.c[1] = *(char *)w;
				sum += su.s;
				mlen = 0;
			} else
				mlen = -1;
		} else if (mlen == -1)
			/*
			 * This mbuf has odd number of bytes.
			 * There could be a word split betwen
			 * this mbuf and the next mbuf.
			 * Save the last byte (to prepend to next mbuf).
			 */
			su.c[0] = *(char *)w;
	}

	if (len)
		puts("cksum: out of data");
	if (mlen == -1) {
		/* The last mbuf has odd # of bytes. Follow the
		   standard (the odd byte is shifted left by 8 bits) */
		su.c[1] = 0;
		sum += su.s;
	}
	REDUCE;
	return (~sum & 0xffff);
}