Comment by joosters
3 years ago
The compiler didn't find UB. What it saw was a pointer dereference, followed by some code later on that checked if the pointer was null.
Various optimisation phases in compilers try to establish the possible values (or ranges) of variables, and later phases can then use this to improve calculations and comparisons. It's very generic, and useful in many circumstances. For example, if the compiler can see that an integer variable 'i' can only take the values 0-5, it could optimise away a later check of 'i<10'.
In this specific case, the compiler reasoned that the pointer variable could not be zero, and so checks for it being zero were pointless.
Yes - and the original post here is the same:
The compiler now knows x's possible range is non-negative.
A non-negative multiplied and divided by positive numbers means that i's possible range is also non-negative (this is where the undefinedness of integer overflow comes in - x * 0x1ff can't have a negative result without overflow occurring).
The first conditional is trivially true now, because of our established bounds on i, so it can just be replaced with "true". This is what causes the code to behave contrary to the OP's expectations: with his execution environment in the overflow case we can end up with a negative value in i.
It is probably more precise to say “if the pointer is null, then it doesn’t matter what I do here, so I am permitted to eliminate this” than to say that it can’t be null here. (It can’t be both null and defined behavior.)
I'm not sure that's right. The compiler isn't tracking undefined behaviour, it is tracking possible values. It just happens that one specific input into determining these values is the fact "a valid program can't dereference a null pointer", so if the source code ever dereferences a pointer, the compiler is free to reason that the pointer cannot therefore be null.
In essence, the compiler is allowed to assume that your code is valid and will only do valid things.