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gliptic | 1 month ago

The borrow checker doesn't decide when things are dropped. It only checks reference uses and doesn't generate any code. This will work exactly the same as long as your program doesn't violate any borrowing rules.

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kace91|1 month ago

No, I get that from an architectural perspective they are separate processes. The point is, unlike in other languages, the compiler is developed assuming the input has been borrow checked, right? So it is surprising to me that it doesn’t blow up somewhere when that invariant doesn’t hold.

andrewaylett|1 month ago

In a correct program, the borrow checker has no effect.

Languages like C compile code with the understanding that if the compiler can't prove the code is incorrect, it'll assume it's correct. Rust compiles with the expectation that unless the compiler can prove the code correct (according to the language rules), it won't compile it. In C, all programs that only perform defined behaviour are valid, but many programs which exhibit undefined behaviour are also valid. In safe Rust, all programs which exhibit undefined behaviour are invalid. But as a trade-off, many programs which would actually execute perfectly well are also considered invalid.

In both cases, once you get past the layers that check stuff, you may normally assume that whatever you have has already been shown to be OK and you probably don't have enough information to re-check while compiling. It might blow up at runtime, it might not.

masklinn|1 month ago

> So it is surprising to me that it doesn’t blow up somewhere when that invariant doesn’t hold.

The final program may be broken in various manners because you don't respect the language's prescribed semantics, in about the same way they do in C and C++. From the compiler's perspective the borrow checker validates that rules it assumes are upheld are actually upheld.

mrustc already compiles rust code without having a borrow checker (well IIRC recent-ish versions of mrustc have some borrow checking bits, but for the most part it still assumes that somebody else has done all the borrow checking).

pornel|1 month ago

The compiler has deep assumptions about exclusive ownership and moves, which affects destructors and deallocation of objects.

It doesn't actually depend on the borrow checker. All lifetime labels are discarded after being checked. Code generation has no idea about borrow checking. Once the code is checked, it is compiled just like C or C++ would, just assuming the code is valid and doesn't use dangling pointers.

Borrow checker doesn't affect program behavior. It either stops compilation or does nothing at all. It's like an external static analysis tool.