Move constructor and initialization list

Question

I want to implement move constructors (no copy constructor) for a certain type that needs to be a value type in a boost::unordered_map. Let's call this type Composite.

Composite has the following signature:

struct Base { Base(..stuff, no default ctor) : initialization list {} Base(Base&& other) : initialization list {} } struct Composite { Base member; Composite(..stuff, no default ctor) : member(...) {} Composite(Composite&& other) : member(other.member) {} // <---- I want to make sure this invokes the move ctor of Base }

I want to write this so boost::unordered_map< Key , Composite > does not require the copy constructor, and just uses the move constructor. If possible, I don't want to use the copy constructor of Base in the initialization list of move constructor of Composite.

Is this possible?

Kerrek SB · Accepted Answer · 2012-12-10 00:21:17Z

17

Say member(std::move(other.member)).

As a golden rule, whenever you take something by rvalue reference, you need to use it inside std::move, and whenever you take something by universal reference (i.e. deduced templated type with &&), you need to use it inside std::forward.

answered Dec 10, 2012 at 0:21

Kerrek SB

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3 Comments

mavam Over a year ago

+1 for a good use of the relatively new term universal reference.

Kerrek SB Over a year ago

@MatthiasVallentin: That term was contrived on the spot by Scott Meyers. The standards committee prefers the term "forwarding reference" nowadays.

mavam Over a year ago

We're almost two years in now, glad to see that the community has converged on a stable term.

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