I'm confused about when a move constructor gets called vs a copy constructor. I've read the following sources:
Move constructor is not getting called in C++0x
Move semantics and rvalue references in C++11
All of these sources are either overcomplicated(I just want a simple example) or only show how to write a move constructor, but not how to call it. Ive written a simple problem to be more specific:
const class noConstruct{}NoConstruct;
class a
{
private:
int *Array;
public:
a();
a(noConstruct);
a(const a&);
a& operator=(const a&);
a(a&&);
a& operator=(a&&);
~a();
};
a::a()
{
Array=new int[5]{1,2,3,4,5};
}
a::a(noConstruct Parameter)
{
Array=nullptr;
}
a::a(const a& Old): Array(Old.Array)
{
}
a& a::operator=(const a&Old)
{
delete[] Array;
Array=new int[5];
for (int i=0;i!=5;i++)
{
Array[i]=Old.Array[i];
}
return *this;
}
a::a(a&&Old)
{
Array=Old.Array;
Old.Array=nullptr;
}
a& a::operator=(a&&Old)
{
Array=Old.Array;
Old.Array=nullptr;
return *this;
}
a::~a()
{
delete[] Array;
}
int main()
{
a A(NoConstruct),B(NoConstruct),C;
A=C;
B=C;
}
currently A,B,and C all have different pointer values. I would like A to have a new pointer, B to have C's old pointer, and C to have a null pointer.
somewhat off topic, but If one could suggest a documentation where i could learn about these new features in detail i would be grateful and would probably not need to ask many more questions.
A move constructor is called:
std::move(something)
std::forward<T>(something)
and T
is not an lvalue reference type (useful in template programming for "perfect forwarding")This is not a complete list. Note that an "object initializer" can be a function argument, if the parameter has a class type (not reference).
a RetByValue() {
a obj;
return obj; // Might call move ctor, or no ctor.
}
void TakeByValue(a);
int main() {
a a1;
a a2 = a1; // copy ctor
a a3 = std::move(a1); // move ctor
TakeByValue(std::move(a2)); // Might call move ctor, or no ctor.
a a4 = RetByValue(); // Might call move ctor, or no ctor.
a1 = RetByValue(); // Calls move assignment, a::operator=(a&&)
}