3 ms·
On a memory unsafe language if you find your invariants violated all bets are off and it is possible that the integrity of the runtime is compromised. The only
by gpderetta 4y ago
On a memory unsafe language if you find your invariants violated all bets are off and it is possible that the integrity of the runtime is compromised. The only safe option is to bail out to the nearest protection boundary (i.e. the process).
On a memory safe language, the blast radius is potentially much smaller (in principle all objects transitively reachable from the failure point, but this is very pessimistic), so it is realistic to be able to isolate the failure in process.
- preseinger 4y agoWell, memory safety is not a boolean, it's a property that's defined by the memory model of the language. That model also defines the impact of a violation.