the difference of pointers is a signed type ptrdiff_t; if it is only
32-bit, left-shifting it by 30 bits produces undefined behavior. cast
the difference to an appropriate unsigned type, uint32_t, before
shifting to avoid this.
the a64l function is specified to return a signed 32-bit result in
type long. as noted in the bug report by Ed Schouten, converting
implicitly from uint32_t only produces the desired result when long is
a 32-bit type. since the computation has to be done in unsigned
arithmetic to avoid overflow, simply cast the result to int32_t.
further, POSIX leaves the behavior on invalid input unspecified but
not undefined, so we should not take the difference between the
potentially-null result of strchr and the base pointer without first
checking the result. the simplest behavior is just returning the
partial conversion already performed in this case, so do that.
{
int e;
uint32_t x = 0;
- for (e=0; e<36 && *s; e+=6, s++)
- x |= (strchr(digits, *s)-digits)<<e;
- return x;
+ for (e=0; e<36 && *s; e+=6, s++) {
+ const char *d = strchr(digits, *s);
+ if (!d) break;
+ x |= (uint32_t)(d-digits)<<e;
+ }
+ return (int32_t)x;
}
char *l64a(long x0)