How do I move multiple parameter via Move semantics? - c++

Question

Let us assume I have an Object MeasurementValues, which has n different pointers (this examples just show pointers to primitive types, but pointers to other complex objects mivght occur as well).
class MesaurementValues {
private:
int *measurement_1;
double *measurement_2;
long long *measurement_3;
//..
float *measurement_n;
int noPointer;
}
I know, this example might be a little bit contrived, anyway. I try to fullfill the Rule of Five in my code.
Do I have to
this.measurement_x = old.measurement_x ;// for all x = {1,..,n} ?

Answer 1

First off just be aware that move-semantics give no advantage over copy-semantics for plain-old-data type (POD) members, but it sure does when your class contains other class objects and/or arrays. When you implement move-semantics, it means you have a "move constructor" and/or a "move assignment operator" which would look like this in your class:
class MesaurementValues { private:
int *measurement_1;
double *measurement_2;
long long *measurement_3;
//..
float *measurement_n;
int noPointer;
//a couple different objects
someObject* pObj1;
differentObject* pObj2;
public:
MeasurementValues( MeasurementValues&& move ); //move-constructor
MeasurementValues& operator= (MeasurementValues&& move); //move-assignment
}
-Assuming your class has POD data and class objects, and
-assuming there are alot of variables to move over:
MeasurementValues::MeasurementValues( MeasurementValues&& old) {
//copy plain-old-data over
measurement_1 = old.measurement_1;
measurement_2 = old.measurement_2;
//copy over values of the pointers
pObj1 = old.pObj1;
pObj2 = old.pObj2;
}
Keep in mind that move-semantics, as other posters have said, only have an advantage if your data members are other moveable objects, or dynamically-allocated memory.
EDIT:
pointers in must become "invalid", as they've been moved. Therefore I would set them to null to prevent unexpected behavior:
MeasurementValues::MeasurementValues( MeasurementValues&& old)
: measurement_1() //null...
//,...
{
//Swapping null into old...
std::swap(measurement_1, old.measurement_1);
//...
}

Answer 2

Do I have to
this.measurement_x = old.measurement_x ;// for all x = {1,..,n} ?
I would rely on the pimpl idiom for this, and use something like a unique pointer. Below is a detailed example. Note you only have to work with Impl, and rely on its defaults (as it contains no pointers).
#include <iostream>
#include <memory>
struct Moveable
{
public:
Moveable();
~Moveable();
Moveable(const Moveable& m);
Moveable& operator=(const Moveable& m);
Moveable(Moveable&& m);
Moveable& operator=(Moveable&& m);
int foo() const;
private:
struct Impl;
std::unique_ptr<Impl> pimpl_;
};
//Moveable.cpp
struct Moveable::Impl
{
Impl(): a(1), b(2), c(3), buffer(){}
int a, b, c;
char buffer[10000]; //Make it worth our while...
};
int Moveable::foo() const{ return pimpl_->a+pimpl_->b+pimpl_->c;}
Moveable::Moveable()
: pimpl_(new Impl)
{
std::cout << "Default " << (void*)this << std::endl;
}
Moveable::~Moveable()
{
std::cout << "Destruct " << (void*)this << std::endl;
//automagically...
}
Moveable::Moveable(const Moveable&m)
: pimpl_(new Impl(*m.pimpl_))
{
std::cout << "Copying " << &m << " to " << (void*)this << std::endl;
}
Moveable& Moveable::operator=(const Moveable& m)
{
std::cout << "Copy assign " << (void*)&m << " to " << (void*)this << std::endl;
*pimpl_ = *m.pimpl_; //Relying on their defaults...
}
Moveable::Moveable(Moveable&& m)
: pimpl_(move(m.pimpl_))
{
std::cout << "Moving " << (void*)&m << " to " << (void*)this << std::endl;
}
Moveable& Moveable::operator=(Moveable&& m)
{
pimpl_ = move(m.pimpl_);
std::cout << "Move assigning " << &m << " to " << (void*)this << std::endl;
}
int main()
{
Moveable x;
Moveable y(x); //Copying...
y = x; //Copying assign
y = Moveable(); //Default construct, then move assignment, destruct temp
Moveable z(std::move(y));
std::cout << "Calling foo " << z.foo() << std::endl;
return 0;
}