2 ms·

Hmm, it could be that once UB is encountered the entire program becomes invalid, then. In practice, a lot of UB is quite subtle and may not necessarily result i

Hmm, it could be that once UB is encountered the entire program becomes invalid, then. In practice, a lot of UB is quite subtle and may not necessarily result in complete disaster, but of course once it's occurred you could end up in any number of completely invalid states and that would be the fault of the UB.

masklinn 2y ago
> Hmm, it could be that once UB is encountered the entire program becomes invalid, then. The UB doesn't actually need to be encountered, just guaranteed to be encountered eventually (in a non-statistical meaning), that is where the time travel comes from e.g. if you have if (condition) { printf("thing\n"); 1/0; } else { // other thing } the compiler can turn this into if (condition) { // crash } else { // other thing } as well as // other thing In the first case you have "time travel" because the crash occurs before the print, even though in a sequentially consistent world where division by zero was defined as a crash (e.g. in python) you should see the print first.
- LoganDark 2y ago
  > The UB doesn't actually need to be encountered, just guaranteed to be encountered eventually That's what I meant. Anything that leads to UB itself has UB.