> The intermediate code for a checked array access, though, should be indicating that the value of the indexing expression is to be moved into a temporary register. The code which checks the value and performs the access refers to that temporary register. Only if the storage for the temporary registers (the storage to which they are translated by the back end) changes randomly would there be a problem.
You'd almost certainly pass it as a function parameter, prima facie in a register/on the stack, sure, and therefore in unoptimised code nothing weird would happen. But an optimising compiler might inline the function call, observe that the value doesn't escape, and then if registers are already full it might choose to access the same memory address twice (no reason to copy it onto the stack, and spilling other registers would cost more).
I don't know how likely this exact scenario is, but it's the kind of thing that can happen. Today's compilers stack dozens of optimisation passes, most of which don't know anything about what the others are doing, and all of which make basic assumptions like that the values at memory addresses aren't going to change under them (unless they're specifically marked as volatile). When one of those assumptions is broken, even compiler authors can't generally predict what the effects will be.
Makes sense. When a temporary is the result of a simple expression with no side effects that is expected to evaluate to the same value each time, the temporary can be taken back. An obvious example of this is constant folding. We set a temporary t27 to 42. Well, that can just be 42 everywhere, so we don't need the temporary. The trust "evaluate to same value each time" is based on assumptions, which, if they are wrong, things are screwed.
You'd almost certainly pass it as a function parameter, prima facie in a register/on the stack, sure, and therefore in unoptimised code nothing weird would happen. But an optimising compiler might inline the function call, observe that the value doesn't escape, and then if registers are already full it might choose to access the same memory address twice (no reason to copy it onto the stack, and spilling other registers would cost more).
I don't know how likely this exact scenario is, but it's the kind of thing that can happen. Today's compilers stack dozens of optimisation passes, most of which don't know anything about what the others are doing, and all of which make basic assumptions like that the values at memory addresses aren't going to change under them (unless they're specifically marked as volatile). When one of those assumptions is broken, even compiler authors can't generally predict what the effects will be.