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use fast version of the int reading code for the high-order digits too

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this increases code size slightly, but it's considerably faster,
especially for power-of-2 bases.
This commit is contained in:
Rich Felker 2012-04-13 04:38:56 -04:00
commit 11f3e33f9b
1 changed files with 13 additions and 3 deletions
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16
src/internal/intparse.c
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@ -83,9 +83,19 @@ int __intparse(struct intparse *v, const void *buf, size_t n)
v->state++; v->state++;
v->val = v->small; v->val = v->small;
case 5: case 5:
llim = UINTMAX_MAX/b; if (b==10) {
for (; n && (d=digits[*s])<b && v->val<=llim && d<=UINTMAX_MAX-v->val*b; n--, s++) for (; n && *s-'0'<10U && v->val<=UINTMAX_MAX/10 && (*s-'0')<=UINTMAX_MAX-10*v->val; n--, s++)
v->val = v->val * b + d; v->val = v->val * 10 + (*s-'0');
} else if ((b&-b) == b) {
int bs = "\0\1\2\4\7\3\6\5"[(0x17*b)>>5&7];
llim = UINTMAX_MAX>>bs;
for (; n && (d=digits[*s])<b && v->val<=llim; n--, s++)
v->val = (v->val<<bs) + d;
} else {
llim = UINTMAX_MAX/b;
for (; n && (d=digits[*s])<b && v->val<=llim && d<=UINTMAX_MAX-b*v->val; n--, s++)
v->val = v->val * b + d;
}
if (!n) return 1; if (!n) return 1;
if (d >= b) goto finished; if (d >= b) goto finished;
v->state++; v->state++;