memory management error in C++

Question

I have doubts about the following test snippet. I use a Node pointer to point the node instance returned by function test2. Why the delete failed ?

Any comments would be appreciated. thanks!

struct Node
{
    int a;
};
Node& test2(Node &n)
{
    Node *p = new Node;
    p->a = n.a + 5;
    return *p;
}
Node* test3(Node &n)
{
    Node *p = new Node;
    p->a = n.a + 5;
    return p;
}
int main()
{
    Node m;
    m.a = 12;
    Node n = test2(m);
    Node *x = test3(n);
    cout << m.a << ";" << n.a << ";" << x->a << endl;
    delete x;  //OK
    x = &n;
    delete x;  //Error
    return 0;
}

Answer 1

When you write

    Node n = test2(m);

n is not a reference to the Node created by test2 . It is a copy of it. The Node created by test2 is leaked. You need:

    Node& n = test2(m);

As an aside: I assume this is an exercise to understand references and pointers. For a real program, you would always want to write:

std::unique_ptr<Node> test4(const Node& n)
{
    std::unique_ptr<Node> result{new Node};
    result->a = n.a + 5;
    return result;
}

or even better:

std::unique_ptr<Node> test4(const Node& n)
{
    auto result = std::make_unique<Node>();
    result->a = n.a + 5;
    return result;
}

With either of these solutions, you don't need to bother about deleting the allocated Node - the compiler will do that for you when the unique_ptr in your calling code goes out of scope.

Answer 2

test2 returns a reference of a Node object that's copied (or moved) to n because n is not a reference. This object now resides on the stack, calling delete on objects on the stack is illegal.

Answer 3

Why the delete failed ?

Here

x = &n;
delete x;  //Error

you are assigning an address of a local variable

Node n = test2(m);

which can't (neither needs to) be handled by delete .