CODEWITHSUNDEEP
c++
dkuznietsov
dkuznietsov

Reputation: 298

Pointers to structures in C++

In order to complete my homework I had to implement a list in C++, thus I defined a structure:

struct Node {
    int value;
    Node * next;
    Node * operator [] (int index)//to get the indexed node like in an array
    {
        Node *current = this;
        for (int i = 0; i<index; i++)
        {
            if (current==NULL) return NULL;
            current = current->next;
        }
        return current;
    }
};

When I used it with actual structures, it worked fine: 

Node v1, v2, v3;
v1.next = &v2;
v2.next = &v3;
v3.value = 4;
v3.next = NULL;
cout<<v1[2]->value<<endl;//4
cout<<v2[1]->value<<endl;//4
cout<<v3[0]->value<<endl;//4; works just as planned
cout<<v3[1]->value<<endl;//Segmentation fault

But when i tried to use it with pointers, the things got messed up:

Node *v4, *v5, *v6;
v4 = new Node;
v5 = new Node;
v6 = new Node;
v4->next = v5;
v4->value = 44;
v5->next = v6;
v5->value = 45;
v6->next = NULL;
v6->value = 4646;
//cout cout<<v4[0]->value<<endl; compiler says it's not a pointer
cout<<v4[0].value<<endl;//44
cout<<v4[1].value<<endl;//1851014134
cout<<v4[2].value<<endl;//45
cout<<v4[3].value<<endl;//1851014134
cout<<v4[4].value<<endl;//4646
cout<<v4[5].value<<endl;//1985297391;no segmentation fault
cout<<v6[1].value<<endl;//1985297391;no segmentation fault even though the next was NULL
delete v4;
delete v5;
delete v6;

Though it is possible to work with function, I've got some questions:

  1. Why the returned value in pointers example was a structure but not a pointer?
  2. Why elements now have doubled index and what are the elements between them?
  3. Why there was no segmentation fault?

I'd be very thankful if someone explained me these moments or gave me the sources I could learn from

Upvotes: 4

Views: 123

Answers (2)

ALEXANDER KONSTANTINOV
ALEXANDER KONSTANTINOV

Reputation: 913

By doing this 

v4 = new Node;
cout<<v4[0].value<<endl;//44
cout<<v4[1].value<<endl;//1851014134
cout<<v4[2].value<<endl;//45
cout<<v4[3].value<<endl;//1851014134
cout<<v4[4].value<<endl;//4646
cout<<v4[5].value<<endl;//1985297391;no segmentation fault

You are not calling operator[] of struct Node, you are making pointer dereferencing, v4[1] is equal to ++v4; *v4; So this code is causing unpredicted behavior, because you are dereferencing some garbage. To make it work as you want, you need to change it to:

cout<<v4->operator[](0).value<<endl;
cout<<v4->operator[](1).value<<endl;
cout<<v4->operator[](2).value<<endl;
...

Upvotes: 1

Barry
Barry

Reputation: 302643

That's because v4[0] (and the rest) aren't actually calling your Node::operator[]. That's because v4 isn't a Node, it's a Node*, and pointers have a builtin meaning behind operator[]: v4[i] == *(v4 + i) (that is, we're just indexing into that "array"). So when you write something like v4[3], that isn't calling operator[](3)... it's instead giving you back a Node three Nodes after v4 in memory somewhere, which is basically just garbage.

To get what you want to happen, you'd have to dereference the pointers first:

(*v4)[0]
(*v6)[1]
// etc

Upvotes: 5

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