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>
This commit is contained in:
Родитель
e26c12c777
Коммит
3ea8d440a8
|
@ -383,9 +383,6 @@ static noinline_for_stack
|
|||
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 @@ char *number(char *buf, char *end, unsigned long long num,
|
|||
/* 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 @@ char *number(char *buf, char *end, unsigned long long num,
|
|||
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 */
|
||||
|
|
Загрузка…
Ссылка в новой задаче