Edward Hu Edward Hu - 1 year ago 102
C++ Question

C++ return by reference stack allocation details

can someone walk through exactly what happens with the memory in this operator overload function? I am confused on how exactly the object created inside the operator function gets deallocated in the main.

Object& operator+(const Object& other) {
Object o(*this); //create instance of o that deep copies first argument
//copy contents of other and add onto o
return o;
int main() {
Object b;
Object c;
Object a = b + c;

Edit: to be more specific, isn't it bad practice to create a local object in a function and then return it by reference? Wouldn't that cause a memory leak?

Edit 2: I am referencing my textbook Data abstraction & problem solving with c++ carrano which suggests an operator + overload for LinkedLists in this format:
LinkedList<ItemType>& operator+(const LinkedList<ItemType>& rightHandSide) const;
. They implemented the method in the way I described.

Edit 2.5: the full method pseudocode given by the book:

LinkedList<ItemType>& operator+(const LinkedList<ItemType>& rightHandSide) const {
concatList = a new, empty instance of LinkedList
concatList.itemCount = itemCount + rightHandSide.itemCount
leftChain = a copy of the chain of nodes in this list
rightChain = a copy of the chain of nodes in the list rightHandSide
concatList.headPtr = leftChain.headPtr
return concatList

Edit 3: Asked my professor about this. Will get to the bottom of this by tomorrow.

Edit 4: The book is wrong.

Answer Source

Returning a reference to a local object

As everyone else correctly states, returning a reference to a local object results in undefined behaviour. You will end up with a handle to a destroyed function-scope object.

Returning references in arithmetic operators

If you think about it, a+b should give you a result, but it shouldn't change a nor b. C++ however leaves it up to you to define how operators work on your own types so it's possible to implement the behaviour you need. This is why the operator+ usually has to create a new object and can't return a reference.

Compound assignments (+=, -=, etc) on the other hand do change the object itself so a += b is changing a. This is why it's usually being implemented by returning a reference (not to a local object, but to the instance itself):

Object& Object::operator+=(const Object& rhs)
    // do internal arithmetics to add 'rhs' to this instance
    return *this; // here we return the reference, but this isn't a local object!
Recommended from our users: Dynamic Network Monitoring from WhatsUp Gold from IPSwitch. Free Download