问题
//code from https://skillsmatter.com/skillscasts/2188-move-semanticsperfect-forwarding-and-rvalue-references
class Widget {
public:
Widget(Widget&& rhs)
: pds(rhs.pds) // take source’s value
{
rhs.pds = nullptr; // why??
}
private:
struct DataStructure;
DataStructure *pds;
};
I can't understand the reason for setting rhd.pds
to nullptr
.
What will happen if we remove this line : rhs.pds = nullptr;
回答1:
Some details of the class have been removed. In particular, the constructor dynamically allocates the DataStructure
object and the destructor deallocates it. If, during a move, you just copied the pointer from one Widget
to another, both Widget
s would have pointers to the same allocated DataStructure
object. Then, when those objects are destroyed, they would both attempt to delete
it. This would give undefined behaviour. To avoid this, the Widget
that is being moved from has its internal pointer to set to nullptr
.
This a standard pattern when implementing a move constructor. You want to move ownership of some dynamically allocated objects from one object to another, so you need to make sure the original object no longer owns those allocated objects.
Diagrammatically, you start off with this situation, wanting to move ownership of the DataStructure
from one Widget
to the other:
┌────────┐ ┌────────┐
│ Widget │ │ Widget │
└───╂────┘ └────────┘
┃
▼
┌───────────────┐
│ DataStructure │
└───────────────┘
If you just copied the pointer, you'd have:
┌────────┐ ┌────────┐
│ Widget │ │ Widget │
└───╂────┘ └───╂────┘
┗━━━━━━━━┳━━━━━━━┛
▼
┌───────────────┐
│ DataStructure │
└───────────────┘
If you then set the original Widget
pointer to nullptr
, you have:
┌────────┐ ┌────────┐
│ Widget │ │ Widget │
└────────┘ └───╂────┘
┃
▼
┌───────────────┐
│ DataStructure │
└───────────────┘
Ownership has successfully been transferred, and when both Widget
s can be destroyed without causing undefined behaviour.
回答2:
The DataStructure
object is likely "owned" by the Widget
, and resetting the pointer prevents it from being accidentally deleted when the Widget
is destroyed.
Alternately, it's conventional to reset objects to an "empty" or "default" state when they are moved-from, and resetting the pointer is a harmless way to follow the convention.
回答3:
class Widget {
public:
Widget(Widget&& rhs)
: pds(rhs.pds) // take source’s value
{
rhs.pds = nullptr; // why??
}
~Widget() {delete pds}; // <== added this line
private:
struct DataStructure;
DataStructure *pds;
};
I added a destructor in the above class.
Widget make_widget() {
Widget a;
// Do some stuff with it
return std::move(a);
}
int main {
Widget b = make_widget;
return 0;
}
To illustrate what would happen if you remove the nullptr assignment, check the above methods. A widget a would be created in a helper function and assigned to widget b.
Since widget a goes out of scope its destructor its called, which deallocates memory, and you are left with widget b which is pointing to invalid memory address.
If you assign nullptr to rhs, a destructor is also called, but since delete nullptr does nothing all is good :)
来源:https://stackoverflow.com/questions/22114025/why-do-we-need-to-set-rvalue-reference-to-null-in-move-constructor