use fast version of the int reading code for the high-order digits too
authorRich Felker <dalias@aerifal.cx>
Fri, 13 Apr 2012 08:38:56 +0000 (04:38 -0400)
committerRich Felker <dalias@aerifal.cx>
Fri, 13 Apr 2012 08:38:56 +0000 (04:38 -0400)
this increases code size slightly, but it's considerably faster,
especially for power-of-2 bases.

src/internal/intparse.c

index e30a1a5805847d1a986f433c1e33a01313d0fde4..fba38c0ac6371370b407ca8eacd40519b52358ca 100644 (file)
@@ -83,9 +83,19 @@ int __intparse(struct intparse *v, const void *buf, size_t n)
                v->state++;
                v->val = v->small;
        case 5:
-               llim = UINTMAX_MAX/b;
-               for (; n && (d=digits[*s])<b && v->val<=llim && d<=UINTMAX_MAX-v->val*b; n--, s++)
-                       v->val = v->val * b + d;
+               if (b==10) {
+                       for (; n && *s-'0'<10U && v->val<=UINTMAX_MAX/10 && (*s-'0')<=UINTMAX_MAX-10*v->val; n--, s++)
+                               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 (d >= b) goto finished;
                v->state++;