Add a fast version of bcmp which compares longwords at a time.

Submitted by: Peter Jeremy <jeremyp@gsmx07.alcatel.com.au>

sys/libkern/bcmp.c

@@ -30,10 +30,16 @@
  * OUT OF THE USE OF THIS SOFTWARE, EVEN IF ADVISED OF THE POSSIBILITY OF
  * SUCH DAMAGE.
  *
- * $Id$
+ * $Id: bcmp.c,v 1.4 1997/02/22 09:39:51 peter Exp $
  */
 
 #include <string.h>
+#include <machine/endian.h>
+
+typedef	const void	*cvp;
+typedef	const unsigned char	*ustring;
+typedef unsigned long	ul;
+typedef const unsigned long	*culp;
 
 /*
  * bcmp -- vax cmpc3 instruction
@@ -43,6 +49,91 @@ bcmp(b1, b2, length)
 	const void *b1, *b2;
 	register size_t length;
 {
+#if BYTE_ORDER == LITTLE_ENDIAN
+	/*
+	 * The following code is endian specific.  Changing it from
+	 * little-endian to big-endian is fairly trivial, but making
+	 * it do both is more difficult.
+	 *
+	 * Note that this code will reference the entire longword which
+	 * includes the final byte to compare.  I don't believe this is
+	 * a problem since AFAIK, objects are not protected at smaller
+	 * than longword boundaries.
+	 */
+	int	shl, shr, len = length;
+	ustring	p1 = b1, p2 = b2;
+	ul	va, vb;
+
+	if (len == 0)
+		return (0);
+
+	/*
+	 * align p1 to a longword boundary
+	 */
+	while ((long)p1 & (sizeof(long) - 1)) {
+		if (*p1++ != *p2++)
+			return (1);
+		if (--len <= 0)
+			return (0);
+	}
+
+	/*
+	 * align p2 to longword boundary and calculate the shift required to
+	 * align p1 and p2
+	 */
+	shr = (long)p2 & (sizeof(long) - 1);
+	if (shr != 0) {
+		p2 -= shr;			/* p2 now longword aligned */
+		shr <<= 3;			/* offset in bits */
+		shl = (sizeof(long) << 3) - shr;
+
+		va = *(culp)p2;
+		p2 += sizeof(long);
+
+		while ((len -= sizeof(long)) >= 0) {
+			vb = *(culp)p2;
+			p2 += sizeof(long);
+			if (*(culp)p1 != (va >> shr | vb << shl))
+				return (1);
+			p1 += sizeof(long);
+			va = vb;
+		}
+		/*
+		 * At this point, len is between -sizeof(long) and -1,
+		 * representing 0 .. sizeof(long)-1 bytes remaining.
+		 */
+		if (!(len += sizeof(long)))
+			return (0);
+
+		len <<= 3;		/* remaining length in bits */
+		/*
+		 * The following is similar to the `if' condition
+		 * inside the above while loop.  The ?: is necessary
+		 * to avoid accessing the longword after the longword
+		 * containing the last byte to be compared.
+		 */
+		return ((((va >> shr | ((shl < len) ? *(culp)p2 << shl : 0)) ^
+		    *(culp)p1) & ((1L << len) - 1)) != 0);
+	} else {
+		/* p1 and p2 have common alignment so no shifting needed */
+		while ((len -= sizeof(long)) >= 0) {
+			if (*(culp)p1 != *(culp)p2)
+				return (1);
+			p1 += sizeof(long);
+			p2 += sizeof(long);
+		}
+
+		/*
+		 * At this point, len is between -sizeof(long) and -1,
+		 * representing 0 .. sizeof(long)-1 bytes remaining.
+		 */
+		if (!(len += sizeof(long)))
+			return (0);
+
+		return (((*(culp)p1 ^ *(culp)p2)
+			 & ((1L << (len << 3)) - 1)) != 0);
+	}
+#else
 	register char *p1, *p2;
 
 	if (length == 0)
@@ -54,4 +145,5 @@ bcmp(b1, b2, length)
 			break;
 	while (--length);
 	return(length);
+#endif
 }