sha1_to_hex() usage cleanup

Somebody on the #git channel complained that the sha1_to_hex() thing uses
a static buffer which caused an error message to show the same hex output
twice instead of showing two different ones.

That's pretty easily rectified by making it uses a simple LRU of a few
buffers, which also allows some other users (that were aware of the buffer
re-use) to be written in a more straightforward manner.
Signed-off-by: NLinus Torvalds <torvalds@osdl.org>
Signed-off-by: NJunio C Hamano <junkio@cox.net>

diff.c

@@ -1018,14 +1018,12 @@ static void run_diff(struct diff_filepair *p, struct diff_options *o)
 	}
 
 	if (memcmp(one->sha1, two->sha1, 20)) {
-		char one_sha1[41];
 		int abbrev = o->full_index ? 40 : DEFAULT_ABBREV;
-		memcpy(one_sha1, sha1_to_hex(one->sha1), 41);
 
 		len += snprintf(msg + len, sizeof(msg) - len,
 				"index %.*s..%.*s",
-				abbrev, one_sha1, abbrev,
-				sha1_to_hex(two->sha1));
+				abbrev, sha1_to_hex(one->sha1),
+				abbrev, sha1_to_hex(two->sha1));
 		if (one->mode == two->mode)
 			len += snprintf(msg + len, sizeof(msg) - len,
 					" %06o", one->mode);

merge-tree.c

@@ -24,16 +24,14 @@ static const char *sha1_to_hex_zero(const unsigned char *sha1)
 
 static void resolve(const char *base, struct name_entry *branch1, struct name_entry *result)
 {
-	char branch1_sha1[50];
-
 	/* If it's already branch1, don't bother showing it */
 	if (!branch1)
 		return;
-	memcpy(branch1_sha1, sha1_to_hex_zero(branch1->sha1), 41);
 
 	printf("0 %06o->%06o %s->%s %s%s\n",
 		branch1->mode, result->mode,
-		branch1_sha1, sha1_to_hex_zero(result->sha1),
+		sha1_to_hex_zero(branch1->sha1),
+		sha1_to_hex_zero(result->sha1),
 		base, result->path);
 }
 

sha1_file.c

@@ -108,9 +108,10 @@ int safe_create_leading_directories(char *path)
 
 char * sha1_to_hex(const unsigned char *sha1)
 {
-	static char buffer[50];
+	static int bufno;
+	static char hexbuffer[4][50];
 	static const char hex[] = "0123456789abcdef";
-	char *buf = buffer;
+	char *buffer = hexbuffer[3 & ++bufno], *buf = buffer;
 	int i;
 
 	for (i = 0; i < 20; i++) {