Take ownership of parameter by rvalue-reference

问题

I want to make clear that the constructor of my class A will take ownership of the passed Data parameter. The obvious thing to do is take a unique_ptr by value:

class A
{
public:
    A(std::unique_ptr<Data> data) : _data(std::move(data)) { }

    std::unique_ptr<Data> _data;
};

However, for my use-case, there is no reason why Data should be a pointer, since a value type would suffice. The only remaining option that I could think of to make really clear that Data will be owned by A is pass by rvalue-reference:

class A
{
public:
    A(Data&& data) : _data(std::move(data)) { }

    Data _data;
};

Is this a valid way to signal ownership or are there better options to do this without using unique_ptr?

回答1:

Yes, I think it is a valid way.

In the case of unique_ptr, it is non-copyable, so there is no danger of someone accidentally making a copy when they didn't intend to so both pass-by-value and pass-by-rvalue-reference signify taking ownership.

In the case of Data, pass-by-rvalue-reference documents that you are taking ownership and there is no danger of the caller accidentally making a copy when they didn't intend to.

回答2:

Yes, it's a valid way. You can also pass it by value:

class A {
    A(Data data) : _data(std::move(data)) { }
};
Data data;
A a(std::move(data));

来源：https://stackoverflow.com/questions/39721654/take-ownership-of-parameter-by-rvalue-reference