fix typo in man page
make use of __uint128_t in scan_ulongn and provide long long based implementation for 32-bit platformsmaster
parent
9b6671efee
commit
74d676fc62
@ -1,23 +1,51 @@
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#include "scan.h"
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#include "scan.h"
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#include "haveuint128.h"
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size_t scan_ulongn(const char* src,size_t n,unsigned long int* dest) {
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size_t scan_ulongn(const char* src,size_t n,unsigned long int* dest) {
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register const char *tmp=src;
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register const char *tmp=src;
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register unsigned long int l=0;
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register unsigned long int l=0;
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register unsigned char c;
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register unsigned char c;
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while (n-->0 && (c=(unsigned char)(*tmp-'0'))<10) {
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/* Since the conditions can be computed at compile time, the compiler
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unsigned long int n;
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* should only emit code for one of the implementations, depending on
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/* we want to do: l=l*10+c
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* which architecture the code is compiled for. */
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* but we need to check for integer overflow.
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#ifdef HAVE_UINT128
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* to check whether l*10 overflows, we could do
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if (sizeof(unsigned long)==sizeof(unsigned long long) && sizeof(unsigned long)<sizeof(__uint128_t)) {
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* if ((l*10)/10 != l)
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/* implementation for 64-bit platforms with gcc */
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* however, multiplication and division are expensive.
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for (; n-->0 && (c=(unsigned char)(*tmp-'0'))<10; ++tmp) {
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* so instead of *10 we do (l<<3) (i.e. *8) + (l<<1) (i.e. *2)
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__uint128_t L=(__uint128_t)l*10+c;
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* and check for overflow on all the intermediate steps */
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if ((L >> ((sizeof(L)-sizeof(l))*8))) break;
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n=l<<3; if ((n>>3)!=l) break;
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l=(unsigned long)L;
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if (n+(l<<1)+c < n) break;
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}
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l=n+(l<<1)+c;
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*dest=l;
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++tmp;
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return (size_t)(tmp-src);
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} else
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#endif
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if (sizeof(unsigned long)<sizeof(unsigned long long)) {
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/* implementation for 32-bit platforms */
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for (; n-->0 && (c=(unsigned char)(*tmp-'0'))<10; ++tmp) {
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unsigned long long L=(unsigned long long)l*10+c;
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if ((unsigned long)L != L) break;
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l=(unsigned long)L;
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}
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*dest=l;
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return (size_t)(tmp-src);
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} else {
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/* implementation for 64-bit platforms without gcc */
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while (n-->0 && (c=(unsigned char)(*tmp-'0'))<10) {
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unsigned long int n;
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/* we want to do: l=l*10+c
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* but we need to check for integer overflow.
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* to check whether l*10 overflows, we could do
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* if ((l*10)/10 != l)
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* however, multiplication and division are expensive.
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* so instead of *10 we do (l<<3) (i.e. *8) + (l<<1) (i.e. *2)
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* and check for overflow on all the intermediate steps */
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n=l<<3; if ((n>>3)!=l) break;
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if (n+(l<<1)+c < n) break;
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l=n+(l<<1)+c;
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++tmp;
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}
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if (tmp-src) *dest=l;
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return (size_t)(tmp-src);
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}
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}
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if (tmp-src) *dest=l;
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return (size_t)(tmp-src);
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}
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}
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