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>C++ moves are not destructive, as they are in Rust. This means that types must arrange for a "moved-from" state Using an object that has been moved from is on

>C++ moves are not destructive, as they are in Rust. This means that types must arrange for a "moved-from" state Using an object that has been moved from is only required not to lead to a crash - the object needs to be in a "valid but unspecified state". If there's no sentinel value, you can leave the object in pretty much any state at all, and it's on the caller to not do anything silly before resetting it. There's also https://clang.llvm.org/extra/clang-tidy/checks/bugprone-use-after-move.html https://clang.llvm.org/extra/clang-tidy/checks/bugprone-use-..., which will detect any such silliness.

roca 7y ago
Herb Sutter's article shows why this is actually a big problem in practice. He has an example of a class containing a guaranteed-non-null owning pointer. Leaving an object of this class in a "valid but unspecified state" means when you move out of it, you'll have to replace the internal pointer with a pointer to some new allocation, the very thing move semantics is supposed to help avoid. He suggests that such a class should simply not be moveable. So the fact that moves are not destructive is indeed very limiting.
- AnimalMuppet 7y ago
  Not quite. Let's suppose that I've got a pointer to a rather large array, and that array can't be null. When I move, the moved-from then has to allocate a new array, to satisfy the can't-be-null constraint. So I have the overhead of the allocation. But I don't have the overhead of copying all the bytes of the array.
  - evanpw 7y ago
    That workaround is also problematic, because move constructors are supposed to be noexcept, but 'new' can throw std::bad_alloc. So you either have to lie to the compiler and tell it that your move constructor is noexcept (maybe reasonable in this case since a bad allocation will then call std::terminate), or else you have bad consequences like std::vector silently copying your object instead of moving it.
  - roca 7y ago
    Sure, but that's almost as bad.
  - unlinked_dll 7y ago
    >So I have the overhead of the allocation. But I don't have the overhead of copying all the bytes of the array. The overhead of allocation is a magnitude greater evil than copying the bytes of the data structure the vast majority of the time. Most data is small and short lived. As well, copying bytes in memory doesn't have side effects and is far more deterministic than allocation. The fact the spec says non trivially-copyable data must be in some "valid but unspecified" state means that move semantics in C++ are borderline useless, except in contrived cases like the one you present.
    - AnimalMuppet 7y ago
      Not quite. It's the "pointer cannot be null" requirement that makes it useless here. If you allow "pointer can be null for a moved-out-of object", then they're much more useful. I'd say that it's the "pointer must not be null, even for a moved-out-of object" that's the "contrived" part.

roca 7y ago

Herb Sutter's article shows why this is actually a big problem in practice. He has an example of a class containing a guaranteed-non-null owning pointer. Leaving an object of this class in a "valid but unspecified state" means when you move out of it, you'll have to replace the internal pointer with a pointer to some new allocation, the very thing move semantics is supposed to help avoid. He suggests that such a class should simply not be moveable. So the fact that moves are not destructive is indeed very limiting.

AnimalMuppet 7y ago
Not quite. Let's suppose that I've got a pointer to a rather large array, and that array can't be null. When I move, the moved-from then has to allocate a new array, to satisfy the can't-be-null constraint. So I have the overhead of the allocation. But I don't have the overhead of copying all the bytes of the array.
- evanpw 7y ago
  That workaround is also problematic, because move constructors are supposed to be noexcept, but 'new' can throw std::bad_alloc. So you either have to lie to the compiler and tell it that your move constructor is noexcept (maybe reasonable in this case since a bad allocation will then call std::terminate), or else you have bad consequences like std::vector silently copying your object instead of moving it.
- roca 7y ago
  Sure, but that's almost as bad.
- unlinked_dll 7y ago
  >So I have the overhead of the allocation. But I don't have the overhead of copying all the bytes of the array. The overhead of allocation is a magnitude greater evil than copying the bytes of the data structure the vast majority of the time. Most data is small and short lived. As well, copying bytes in memory doesn't have side effects and is far more deterministic than allocation. The fact the spec says non trivially-copyable data must be in some "valid but unspecified" state means that move semantics in C++ are borderline useless, except in contrived cases like the one you present.
  - AnimalMuppet 7y ago
    Not quite. It's the "pointer cannot be null" requirement that makes it useless here. If you allow "pointer can be null for a moved-out-of object", then they're much more useful. I'd say that it's the "pointer must not be null, even for a moved-out-of object" that's the "contrived" part.