Hi Michael, On 8/25/20 1:41 PM, Michael Kerrisk (man-pages) wrote: >> In C, successive multiplications are evaluated left to right (*), and >> therefore here is what happens: >> >> ``(sizeof(x) * INT_MAX * 2)``: >> >> 1) sizeof(x) * INT_MAX (the type is the largest of both, which is >> size_t (unsigned long; uint64_t)). >> 2) ANS * 2 (the type is again the largest: size_t) >> >> ``(INT_MAX * 2 * sizeof(x))``: >> >> 1) INT_MAX * 2 (the type is the largest of both, which is >> int as both are int (int; int32_t), so the >> result is already truncated as it doesn't fit >> an int; at this point, the intermediate result >> will be 2^32 - 2 (``INT_MAX - 1``) (if I did >> the math right)). >> 2) ANS * 2 (the type is again the largest of both: size_t; >> however, ANS was already incorrect, so the >> result will be an incorrect size_t value) >> >>> sizeof((sizeof(x) * INT_MAX * 2)) == 8 >> >> Here you were overflowing a uint64_t (if x were a char, it would not >> overflow, and the result would be close to UINT64_MAX). I also goofed :) I didn't see the sizeof applying to everything. And I was wrong: you aren't overflowing 64 bits, which would need x to be enormous. But the idea is there; the explanation above was right. >> >>> sizeof(INT_MAX * 2 * sizeof(x)) == 8 >> >> Here you were overflowing int32_t, and it would overflow regardless of >> sizeof(x). > > Thanks for the lesson in C basics. I clearly did need a refresher :-}. You're welcome! It happens me from time to time too. :) > > You and Jakub are of course correct. > > If you send the patches in the order (as I numbered in my previous reply): > > (1) > (3) > (2) as multiple pieces > > that would be best, since the first two patches should obviously be > applied, and then we can discuss the last patch(es) case by case. Ok. > > Thanks, > > Michael > Cheers, Alex.
