> Do you mean to say the compiler will insert locks?
I think you're right, it's the compiler that'll insert the locks according to the semantics of the language.
> For an out-of-order superscalar machine, mutability has nothing to do with it;
Yes it does, because it creates data dependencies. If a piece of code B loads the contents of memory where a piece of code A writes that's before B, then you can never execute B before or in parallel with A.
I think you're right, it's the compiler that'll insert the locks according to the semantics of the language.
> For an out-of-order superscalar machine, mutability has nothing to do with it;
Yes it does, because it creates data dependencies. If a piece of code B loads the contents of memory where a piece of code A writes that's before B, then you can never execute B before or in parallel with A.