What does "unspecified" mean in this context? If you mean "the value of a signed integer is unspecified if it overflows", then I can still construct cases in which you lose the optimization that widens a loop induction variable from 32-bit to 64-bit. For example, take this code:
for (int i = 0; i != n; i++)
sum += a[i];
If n is -2, will this function include the element at a[(uint64_t)INT_MAX + 1] in the sum? If signed overflow is UB, then the answer is "maybe", and thus i and n can legally be promoted to 64-bit. If signed overflow is unspecified, but the result must still fit in a 32-bit integer, then the answer is "no", and i and n can't legally be promoted.
"unspecified" in the standards sense of "unspecified behavior", i.e. any chosen behavior is permissible. The compiler doesn't have to document the behavior and the behavior isn't constrained, it just isn't undefined. That's still better than where we are today.
I looked this up after you mentioned it upthread, I thought you meant implementation-defined, but I learned that yeah, this exists too. The difference is that implementation-defined requires documentation of the choice made, whereas unspecified does not. However,
> the behavior isn't constrained
C++23, at least, does say
> The range of possible behaviors is usually delineated by this document.
