Earlier quoted context omitted.
How does the rust compiler assure that when compiling to machine code? Machine code is less safe than C after all.
Machine code is generally much safer than C - e.g. it usually lacks undefined behaviour. If you're unsure about how a given piece of machine code behaves, it's usually sufficient to test it empirically.
But anyway this is kind of off-topic. I think OutOfHere was imagining that this somehow skips the type checking and borrow checking steps which of course it doesn't.