lib/vsprintf.c: eliminate duplicate hex string array
gcc doesn't merge or overlap const char[] objects with identical contents
(probably language lawyers would also insist that these things have
different addresses), but there's no reason to have the string
"0123456789ABCDEF" occur in multiple places. hex_asc_upper is declared in
kernel.h and defined in lib/hexdump.c, which is unconditionally compiled
in.
Signed-off-by: Rasmus Villemoes <linux@rasmusvillemoes.dk>
Cc: Peter Zijlstra <a.p.zijlstra@chello.nl>
Cc: Tejun Heo <tj@kernel.org>
Signed-off-by: Andrew Morton <akpm@linux-foundation.org>
Signed-off-by: Linus Torvalds <torvalds@linux-foundation.org>
diff --git a/lib/vsprintf.c b/lib/vsprintf.c
index 98f1ce9..7a299d4 100644
--- a/lib/vsprintf.c
+++ b/lib/vsprintf.c
@@ -383,9 +383,6 @@
char *number(char *buf, char *end, unsigned long long num,
struct printf_spec spec)
{
- /* we are called with base 8, 10 or 16, only, thus don't need "G..." */
- static const char digits[16] = "0123456789ABCDEF"; /* "GHIJKLMNOPQRSTUVWXYZ"; */
-
char tmp[3 * sizeof(num)];
char sign;
char locase;
@@ -422,7 +419,7 @@
/* generate full string in tmp[], in reverse order */
i = 0;
if (num < spec.base)
- tmp[i++] = digits[num] | locase;
+ tmp[i++] = hex_asc_upper[num] | locase;
else if (spec.base != 10) { /* 8 or 16 */
int mask = spec.base - 1;
int shift = 3;
@@ -430,7 +427,7 @@
if (spec.base == 16)
shift = 4;
do {
- tmp[i++] = (digits[((unsigned char)num) & mask] | locase);
+ tmp[i++] = (hex_asc_upper[((unsigned char)num) & mask] | locase);
num >>= shift;
} while (num);
} else { /* base 10 */