Comment by tsimionescu

Comment by tsimionescu 18 hours ago

24 replies

It is absolutely knowable statically if ownership will be taken. It's not necessarily very easy to do so, but the decision is 100% up to the compiler, as part of overload resolution and optimization choices (like the NRVO analysis that the article mentions). Since ownership is an inherently static concept, it doesn't even make sense to think about "runtime ownership".

adrianN 18 hours ago

My function can choose to move or not to move from an object based on io input.

  • tsimionescu 18 hours ago

    Can you show an example of what you mean?

    My claim is that, if I call `foo(std::move(myObj))`, it is statically knowable if `foo` receives a copy of `myObj` or whether it is moved to it. Of course, `foo` can choose to further copy or move the data it receives, but it can't choose later on if it's copied or not.

    Now, if I give `foo` a pointer to myObj, it could of course choose to copy or move from it later and based on runtime info - but this is not the discussion we are having, and `std::move` is not involved from my side at all.

    • ben-schaaf 14 hours ago

      No, it is not statically knowable if it is actually moved.

          void foo(Obj && arg) {}
      
      Does not move `arg`. It's fairly easy to write code that assumes `std::move` moves the value, but that can lead to bugs. For example:

          void some_function(std::vector<int> &&);
          void some_function2(std::vector<int> &&);
      
          void main() {
              std::vector<int> a = { 1 };
              some_function(std::move(a));
      
              a.push_back(2);
              some_other_function(std::move(a));
          }
      
      The expectation is that `some_other_function` is always called with `{ 2 }`, but this will only happen if `some_function` actually moves `a`.
      • adrianN 10 hours ago

        Is pushing to a moved-from vector even legal? I thought in general the only guarantee you have after a move is that is save to destruct the object.

        • ben-schaaf an hour ago

          The state of a moved-from value is valid but unspecified (note, not undefined). IIRC the spec says vector must be `empty()` after a move. So all implementations do the obvious thing and revert back to an empty vector.

    • masklinn 17 hours ago

      > Can you show an example of what you mean?

          void foo(std::unique_ptr<int, Deleter>&& p) {
              std::random_device rdev {};
              auto dist = std::uniform_int_distribution<>(0, 1);
              if (dist(rdev)) {
                  auto pp = std::move(p);
              }
          }
      • tsimionescu 16 hours ago

        This is exactly what I meant as irrelevant.

        If I call `foo(std::move(my_unique_ptr))`, I know for sure, statically, that my_unique_ptr was moved from, as part of the function call process, and I can no longer access it. Whether `foo` chooses to further move from it is irrelevant.

charcircuit 16 hours ago

I don't understand the downvoted here. Either the compiler emits the code to call a move constructor or it doesn't.

  • Maxatar 15 hours ago

    Static analysis is about proving whether the code emitted by a compiler is actually called at runtime. It's not simply about the presence of that code.

    Code can be emitted but never executed.

    • charcircuit 7 hours ago

      >Static analysis is about proving whether the code emitted by a compiler is actually called at runtime.

      That is but one thing that can static analysis can prove. It can also prove whether source code will call a move contractor or a copy constructor. Static analysis is about analyzing a program without actually running it. Analysizing what code is emitted is one way a program can be analyzed.

      • knorker 5 hours ago

        The call to a move cons/move assign does not happen at call time. When a function taking rvalue reference is called, it can still have two code paths, one that copies the argument, and one that moves it.

        All the && does is prevent lvalues from being passed as arguments. It's still just a reference, not a move. Indeed, in the callee it's an lvalue reference.

        But yeah, you can statically check if there exists a code path that calls the copy cons/copy assign. But you'll need to check if the callee calls ANY type's copy cons/assign, because it may not be the same type as the passed in obj.

        At that point, what even is a move? char*p = smartptr.release() in the callee is a valid move into a raw pointer, satisfying the interface in callee. That's a move.[1] how could you detect that?

        [1] if this definition of move offends you, then instead remember that shared_ptr has a constructor that takes an rvalue unique_ptr. The move only happens inside the move constructor.

        How do you detect all cases of e.g. return cons(ptr.release()) ? It may even be the same binary code as return cons(std::move(ptr))

        Probably in the end shared pointer constructor probably calls .release() on the unique ptr. That's the move.

        Yup. That's what https://en.cppreference.com/w/cpp/memory/shared_ptr/shared_p... says.

        (Sorry, on phone so not full code. Hopefully I stopped autocorrect all times it interfered)

        • charcircuit 2 hours ago

          What the callee does it out of scope. We are talking about a single assignment or construction of a variable. This has nothing to do with tracing execution. It happens at one place, and you can look at the place to see if it is using a copy or move contructor.