使用指向结构C ++的指针创建动态分配的数组

Question

So I currently have a simple struct (linkedlist) that I will be using in a HashMap:

struct Node {
    std::string key, value;
    Node* head;
}

I'm currently trying to dynamically allocate an array with pointers to each struct. This is what I have right now ...

Node* nodes = new Node[100]

I understand this allocates an array of 100 nodes into memory (which I will have to delete later on); however, upon iteration to try to transverse these nodes (which I an implementing as a linked list)...

for (int x = 0; x < 100; x++) {
    Node current = nodes[x]; // Problem is I wanted an array to node pointers. This is not a pointer.
    while (current != nullptr) { // this isn't even legal since current is not a pointer.
        // DO STUFF HERE
        current = current.next; // This is not a pointer access to a method. I'm looking to access next with current->next;
    }
}

Hopefully I was clear enough. Can someone how to allocate a dynamic array of pointers to structs? So far I'm able to dynamically allocate an array of structs, just not an array of pointers to structs.

Answer 1

There are two approaches. Either you allocate an array of structures and introduce one more pointer that will point to the element in the array that will play the role of the head.

For example

Node *head = nodes;

(in this case head points to nodes[0])

After the list will not be needed you have to delete it using operator

delete [] nodes;

Or you can indeed to allocate an array of pointers to the structure like this

Node **nodes = new Node *[100];

But in this case each element of the array in turn should be a pointer to a dynamically allocated object;

And to delete the list you at first have to delete each object pointed to by elements of the array for example in a loop

for ( int i = 0; i < 100; i++ ) delete nodes[i];

and then to delete the array itself

delete [] nodes;

It is a good idea to initialize each element of the array with zeroes when the array is allocated for example

Node **nodes = new Node *[100]();

Answer 2

I suggested you this structure:

class myList {

struct Node {
string value;
Node* next;
}

/*Public methods .. Add/Set/Get/Next/isEmpty.. etc ... */
Node* head, *tail;
};

in main: myList* lis = new myList[number]; then you have number of lists! and do all work in class by method's and operators, like if you want the next node just call lis[0].getNext(); if you want to skip current node do lis[0].Next(); ... etc ..

this how to work, what you try to do is looks like C program!