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Comment by tsimionescu

17 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.

12 comments

tsimionescu

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masklinn  17 hours ago

The only thing that is statically known here is that you’re wrong. The function I posted only moves its parameter half the time, at random. You may want to treat it as moved-from either way, but factually that’s just not what is happening.

  • charcircuit  17 hours ago

    This is like trying to defend that you can't statically know the result of 1 + 2 because:

      void foo() {
    std::random_device rdev {};
    auto dist = std::uniform_int_distribution<>(0, 1);
    if (dist(rdev)) {
      int res = 1 + 2;
    }
  }

    I can tell you for sure that the result of 1 + 2 will be 3.

    • masklinn  16 hours ago

      > This is like trying to defend that you can't statically know the result of 1 + 2

      It is completely unlike that. tsimionescu is asserting that they can always know statically whether `foo` will move its parameter. The function I provided is a counter-example to that assertion.

      Of course the branch body always moves, that's what it's there for. That has no bearing on the argument.

      3 replies →

tialaramex  15 hours ago

Yeah no.

In Rust if you pass say a Box<Goose> (not a reference, the actual object) into a function foo, it's gone, function foo might do something with that boxed goose or it might not, but it's gone anyway. If a Rust function foo wanted to give you it back they'd have to return the Box<Goose>

But C++ doesn't work that way, after calling foo my_unique_ptr is guaranteed to still exist, although for an actual unique_ptr it'll now be "disengaged" if foo moved from it. It has to still exist because C++ 98 (when C++ didn't have move semantics) says my_unique_ptr always gets destroyed at the end of its scope, so newer C++ versions also destroy my_unique_ptr for consistency, and so it must still exist or that can't work.

Creating that "hollowed out" state during a "move" operation is one of the many small leaks that cost C++ performance compared to Rust.

  • knorker  6 hours ago

    You should not be downvoted, which you appear to be. Your comparison is both correct and interesting.

    Maybe you're being too verbose for your point, and it would help readers if you summarize and narrow the argument to:

    In Rust a function signature can force a move to happen at call time (by being non-reference and not Copy), but in C++ a function taking rvalue reference (&&) only signals the callee that it's safe to move if you want, as it's not an lvalue in the caller.

    It's an added bonus that Rust prevents reusing the named variable in the caller after the move-call, but it's not what people seem to be confused about.

vlovich123  14 hours ago

aside from what others wrote, it’s also non local - whether std::move even does anything is dependent on the signature of foo - if foo takes it by const& you may think you’ve transferred ownership when it hasn’t actually happened.

  • masklinn  12 hours ago

    That is static though, that `foo` takes its parameter by `const&` and will thus not move it is available to the compiler (or other tooling) at compile time.

    The point of contention is whether that is always the case, or whether there are situations where moving from the parameter is a runtime decision.

    • vlovich123  4 hours ago

      Others have already answered that for you. I specifically said it’s non local which means it’s difficult for a human to reason about at the call site.

knorker  16 hours ago

No: https://godbolt.org/z/d7f6MWcb5

Look, the act of calling std::move and and calling a function taking an rvalue reference in no way invokes a move constructor or move assignment. It does not "move".

It's still just a reference, albeit an rvalue reference. std::move and the function shape is about the type system, not moving.

(Edit: amusingly, inside the callee it's an lvalue reference, even though the function signature is that it can only take rvalue references. Which is why you need std::move again to turn the lvalue into rvalue if you want to give it to another function taking rvalue reference)

I didn't reply to this thread until now because I thought you may simply be disagreeing about what "move" means (I would say move constructor or move assignment called), but the comment I replied to makes a more straightforward factually incorrect claim, that can easily be shown in godbolt.

If you mean something else, please sketch something up in godbolt to illustrate your point. But it does sound like you're confusing "moving" with rvalue references.

Edit: for the move to happen, you have to actually move. E.g. https://godbolt.org/z/b8M495Exq