Move Constructor, move vector between two objects using std::move?

Question

I have an object something like the following and I'm trying to implement a move constructor for so you can have an insert for std::vector<Mesh>.

struct Mesh
{    
    std::vector<Vector3> vPoint;
    bool Valid;

    Mesh(Mesh&& other)
    {
        vPoint = std::move(other.vPoint);
        Valid = std::move(other.Valid);
    }
};

Is this the correct way? And if so what is the value of other.Valid after std::move operates on it?

Edit:

Also if I have an instance of this object do I need to use std::move in the following scenario?

std::vector<Mesh> DoSomething()
{
    Mesh mesh; //Imagine vPoint is filled here to

    std::vector<Mesh> meshes;
    meshes.push_back(std::move(mesh)); // Here is my question, std::move? or just pass mesh here?

    return meshes;
}

@us2012 I'm just curious what happens with std::move<bool> or other integral types. Does it copy the value to this.Valid and then set other.Valid to the default value of bool (aka false)? — NtscCobalt
– NtscCobalt, Commented Feb 9, 2013 at 1:45

Praetorian · Accepted Answer · 2013-02-09 01:57:49Z

15

You should write your move constructor as follows:

Mesh( Mesh&& other )
: vPoint( std::move( other.vPoint ) )
, Valid( std::move( other.Valid ) )
{}

The disadvantage of assignment within the constructor body as opposed to using the constructor initializer list is that in the former case the member objects of the Mesh object that you're moving to are default constructed and then assigned to within the body. In the latter case they're constructed directly from the result of the std::move call.

You shouldn't read from an object, be it an integral type, or a more complex object, after moving it. Such objects exist in an unspecified state.

answered Feb 9, 2013 at 1:57

Praetorian

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1 Comment

NtscCobalt Over a year ago

Ok thank you, but what about within the Move Assignment operator? Is vPoint = std::move(other.vPoint) appropriate?

us2012 · Accepted Answer · 2013-02-09 01:56:35Z

3

(partial answer - answering the bit about moveing bool)

cppreference.com has the following to say about std::move:

The library code is required to leave a valid value in the argument, but unless the type or function documents otherwise, there are no other constraints on the resulting argument value. This means that it's generally wisest to avoid using a moved from argument again.

So you cannot rely on a bool to be either true or false after moveing it.

answered Feb 9, 2013 at 1:56

us2012

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4 Comments

NtscCobalt Over a year ago

Ah ok thank you, from testing it appears that built-in types do not have a move constructor so bool b = std::move<bool>(other.b) is basically just an assignment.

David Schwartz Over a year ago

An assignment meets all the requirements for a move and there doesn't seem to be any way to improve on it.

NtscCobalt Over a year ago

@DavidSchwartz Yeah I was hoping for a compiler warning or etc stating that std::move() will be downgraded to assignment only when no move-constructor is available for the type.

us2012 Over a year ago

@NtscCobalt The hope for a warning is probably the result of the unfortunate name of std::move - it should really be called std::obtain_rvalue_reference. The compiler has no way of knowing (short of computing a full call graph and probably a lot of further semantic analysis) whether the obtained reference will be used to move stuff.

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