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From my understanding of Rust, you will still pay the runtime cost of destruction and deallocation of objects with the object lifetime system ? It's like refere
by antpls 8y ago
From my understanding of Rust, you will still pay the runtime cost of destruction and deallocation of objects with the object lifetime system ? It's like reference counting, but precomputed by the compiler. Please correct me if I'm wrong.
- DaiPlusPlus 8y agoRight. It’s just easier to reason about the process with Rust’s ownership system compared to a reference-counting or GC system.
- unrealhoang 8y agoIt’s like manual allocate/deallocate. Only that the compiler will insert free (or drop, in case of Rust) for your variables when it reaches end of scope.
- steveklabnik 8y agoThe way the "it's like reference counting" mental model goes wrong is that things don't live arbitrarily wrong. Reference counting is used to extend the life of things, Rust will not do that. But yes, it's malloc and free.