This is an widely-held misconception, but C++ signed integers are _nothing_ like Z/nZ. Unsigned integers are (except for division), and machine arithmetic often is, but in the C++ abstract machine, signed overflow is undefined behavior.

That soundbite also repeats/paraphrases what thread author Joe Groff said "LLVM considers infinite recursion to be UB,"

But... the key is infinite loops without side effects can be optimized away. Endless loops even without a provable termination (The Halting Problem) are not UB.

But in this case, the value 'x' has a side effect of determining what bool of true/false value to return. That while() loop does not look like it's free of side effects.

That leaves only the loop itself, which can run infinitely long (which would be UB, and the compiler is free to assume that won't happen), or it won't, in which case it will always return true. Thus, like them or not, the rules of C++ allow this particular outcome.

I can't say I like this kind of optimisation. Sure, you can construct neat circus tricks with it like this, but other than that it seems pretty pointless for real-world software, and something that could easily turn a simple mistake into a debugging disaster.