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lib_generic memcpy: copy one word at a time if possible

If source and destination are aligned, this copies ulong values
until possible, trailing part is copied by byte. Thanks for the details
to Wolfgang Denk, Mike Frysinger, Peter Tyser, Chris Moore.

Signed-off-by: Alessandro Rubini <rubini@unipv.it>
Acked-by: Andrea Gallo <andrea.gallo@stericsson.com>
Acked-by: Mike Frysinger <vapier@gentoo.org>
Alessandro Rubini vor 15 Jahren
Ursprung
Commit
ecd830b863
1 geänderte Dateien mit 15 neuen und 4 gelöschten Zeilen
  1. 15 4
      lib_generic/string.c

+ 15 - 4
lib_generic/string.c

@@ -446,12 +446,23 @@ char * bcopy(const char * src, char * dest, int count)
  * You should not use this function to access IO space, use memcpy_toio()
  * or memcpy_fromio() instead.
  */
-void * memcpy(void * dest,const void *src,size_t count)
+void * memcpy(void *dest, const void *src, size_t count)
 {
-	char *tmp = (char *) dest, *s = (char *) src;
-
+	unsigned long *dl = (unsigned long *)dest, *sl = (unsigned long *)src;
+	char *d8, *s8;
+
+	/* while all data is aligned (common case), copy a word at a time */
+	if ( (((ulong)dest | (ulong)src) & (sizeof(*dl) - 1)) == 0) {
+		while (count >= sizeof(*dl)) {
+			*dl++ = *sl++;
+			count -= sizeof(*dl);
+		}
+	}
+	/* copy the reset one byte at a time */
+	d8 = (char *)dl;
+	s8 = (char *)sl;
 	while (count--)
-		*tmp++ = *s++;
+		*d8++ = *s8++;
 
 	return dest;
 }