Conversation

I guess I don't understand the context. It seems to be about C, and I don't see how you can resolve that problem for C without coming up with a model to enforce a form of memory safety. What is the scope of UB that should be avoided? You mean, for a language like Rust or Swift?
2
The question of whether memory unsafety implies UB is sort of at the heart of the disconnect between the C spec and C practitioners. As a practitioner (and compiler guy) I view memory unsafety as a separate thing - after all a “bad” store still stores to a well defined place.
2
2
There is nothing well defined about what an out-of-bounds access or use-after-free will access. The compiler, linker and even runtime environment are assuming that is never going to happen and there's nothing defined about what the consequences are going to be from the C code.
3
1
Memory safety is certainly UB and certainly heavily impacted by optimizations. They would need to trap on memory corruption / type confusion bugs in order to get rid of undefined behavior while also still being able to heavily optimize without changing runtime behavior.
2
2
Tons of programs have latent memory corruption bugs and are depending on the specific way that the compiler, malloc implementation, etc. chose to lay things out in memory, what happens to be zeroed based on what they chose to do, etc. It's certainly UB with similar impacts.
2
C pointers can trivially be treated as addressed by the compiler and llvm will happily do that for you if you use int math rather than gep. The compiler is not optimizing based on the assumption that writing past the end of an array can’t happen unless you explicitly tell it to.
1
That's not true. Not marking GEPs as inbounds means that it's permitted to index outside of the bounds of the object. It absolutely does NOT mean that it's okay to dereference the pointer outside the bounds of the object. That's still incorrect and completely undefined.
2
Show replies