4 ms·
Trivially, NaN != NaN, so this function is not even properly reflexive (i.e. for a = b = NaN we find cmp(&a,&b) = cmp(&b,&a) = 1).
by scythe 3y ago
Trivially, NaN != NaN, so this function is not even properly reflexive (i.e. for a = b = NaN we find cmp(&a,&b) = cmp(&b,&a) = 1).
- jstimpfle 3y agoI mean, yeah. It's not like there is a lot of meaning to NaN. For most applications, trying to sort NaNs is probably a bug in itself. So I think this is likely a perfectly valid comparison function. To make a point, here is a sorting function I'm using. static int cmp_seqnos(const void *a, const void *b) { uint64_t x = (const uint64_t *) a; uint64_t y = (const uint64_t *) b; if (y - x < ((uint64_t) 1 << 63)) return -1; return x != y; } I think this should work if there are uint64_t a, b: b - a < ((uint64_t) 1 << 63) and all values x from the set to be sorted are in the range a..b. Note that a < b isn't a requirement here. Sorting such numbers can be useful when dealing with sliding windows.