My function can choose to move or not to move from an object based on io input.
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.
void foo(Obj && arg) {}
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));
}
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);
}
}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.
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
I mistakenly treated in my mind `foo(obj&& x)` as `foo(obj x)`, which would indeed move or copy statically. But with functions that take an r-value reference, you're of course absolutely right, it becomes impossible to determine statically if they will move or not.
void foo() {
std::random_device rdev {};
auto dist = std::uniform_int_distribution<>(0, 1);
if (dist(rdev)) {
int res = 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.
>That has no bearing on the argument.
That is the whole argument. Let me quote the other person: "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."
It is saying that for "auto pp = std::move(p);" we will know if it uses the move assign constructor or the copy assign constructor.
No, it is not.
> Let me quote the other person: "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."
Yes. `foo`.
> It is saying that for "auto pp = std::move(p);" we will know if it uses the move assign constructor or the copy assign constructor.
`pp` is not `foo`. That `pp` uses a move constructor is not the subject of the debate.
You can literally take the function I posted, build a bit of scaffolding around it, and observe that whether the parameter is moved into `foo` or not is runtime behaviour: https://godbolt.org/z/jrPKhP35s
Nothing is constructed at call time. Check out this example, which compiles just fine, even though Foo is neither copy nor move constructible/assignable: https://godbolt.org/z/Wj57o773d
"&&" is just a type system feature, resolving function polymorphism matching rvalue reference and not lvalue reference. It's not a thing that causes a move.
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.
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.
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.