How to create an array of pointers to structs? C++

Question

I wrote this piece of code that I am not sure exactly how it works, but it works. This is the code:

  struct node
{
    string data;
    node *chain;
};

   int tablesize=10;

  node *ptr [tablesize];

  for (i=0; i<tablesize; i++)
{
    ptr[i]=new node;
    ptr[i]->data="Empty";
    ptr[i]->chain=NULL;
}

If I understand it correctly, first I create an array of 10 pointers, then I assign each pointer with a new node? Why does it work only when I dereference it twice though? ( ptr[i]->data="Empty";)

Because ptr[i] is a pointer. You are not de-referencing twice. BTW you are using variable length arrays (VLAs), which are not standard C++. — juanchopanza
– juanchopanza, Commented Oct 30, 2013 at 8:20

Udo Klein · Accepted Answer · 2013-10-30 08:22:02Z

1

Because ptr is declared as an array of pointers. Thus ptr[i] is a pointer to node. Hence you need to dereference it in order to access the pointed to entitiy.

answered Oct 30, 2013 at 8:22

Udo Klein

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3 Comments

Reboot_87 Over a year ago

So is it right to say that node *ptr [tablesize]; is actually pointer to pointer array?

juanchopanza Over a year ago

@Reboot_87 no, it is an array of pointers to node. Just like int a[10] is an array of ints.

Reboot_87 Over a year ago

I think it is because: node *ptr [tablesize]; is the same as writing: node **ptr=new node *[tablesize];

RvdK · Accepted Answer · 2013-10-30 08:21:31Z

0

ptr[i] is a node* therefore you have to use ->

(you are not de-referencing)

answered Oct 30, 2013 at 8:21

RvdK

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1 Comment

Lightness Races in Orbit Over a year ago

No? What would you call it, then?

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