Fun with uninitialized variables and compiler (GCC)

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I’m just curious to know why this sudden change in the behavior of uninitialized bool?

Disassemble the code and see what the compiler’s doing.

My guess: since the value is now only used locally, the compiler optimizes it away completely. Since the behaviour is undefined anyway, the compiler can safely just assume any value, e.g. false. This is a pretty obvious optimization since the value of b is constant as far as the compiler is concerned, and the whole logic of the switch is redundant. So why put it in the executable?

(The important point here is really that b is only ever used locally in the second code, and that in turn will trigger more optimizations even in unoptimized code. The first code has to be inlined before the compiler can do any such optimizations, or the code paths have to be traced which isn’t trivial).

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