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>
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@ -383,9 +383,6 @@ static noinline_for_stack
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char *number(char *buf, char *end, unsigned long long num,
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struct printf_spec spec)
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{
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/* we are called with base 8, 10 or 16, only, thus don't need "G..." */
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static const char digits[16] = "0123456789ABCDEF"; /* "GHIJKLMNOPQRSTUVWXYZ"; */
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char tmp[3 * sizeof(num)];
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char sign;
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char locase;
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@ -422,7 +419,7 @@ char *number(char *buf, char *end, unsigned long long num,
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/* generate full string in tmp[], in reverse order */
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i = 0;
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if (num < spec.base)
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tmp[i++] = digits[num] | locase;
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tmp[i++] = hex_asc_upper[num] | locase;
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else if (spec.base != 10) { /* 8 or 16 */
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int mask = spec.base - 1;
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int shift = 3;
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@ -430,7 +427,7 @@ char *number(char *buf, char *end, unsigned long long num,
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if (spec.base == 16)
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shift = 4;
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do {
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tmp[i++] = (digits[((unsigned char)num) & mask] | locase);
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tmp[i++] = (hex_asc_upper[((unsigned char)num) & mask] | locase);
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num >>= shift;
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} while (num);
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} else { /* base 10 */
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