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fix typo in man page

make use of __uint128_t in scan_ulongn and provide long long based
implementation for 32-bit platforms
master
Felix von Leitner 7 years ago
parent
commit
4ab5705eec
  1. 2
      scan/scan_ulong.3
  2. 58
      scan/scan_ulongn.c

2
scan/scan_ulong.3

@ -35,6 +35,6 @@ scan_ulong("023",&i) -> i=23, return 3
scan_ulong("0x23",&i) -> i=0, return 1
scan_ulong("4294967296",&i" -> i=429496729, return 9 // 32-bit system
scan_ulong("4294967296",&i") -> i=429496729, return 9 // 32-bit system
.SH "SEE ALSO"
scan_xlong(3), scan_8long(3), fmt_ulong(3)

58
scan/scan_ulongn.c

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