Thread: dynamic arrays of structs...

this sounds silly, but some reason (which i can't spot) when i declrare an array of simple structs on the free store, in the following program i can't access them through (->) derefrerencing a pointer instead it only let's use the dot operator (.)...

Code:

#include<iostream>
using namespace std;

struct node
{
	int data;
};


int main()
{
	int size;

	cout << "Enter the size: ";
	cin >> size;

	node *ptr = new node[size];
	

	for(int i = 0; i < size; i++)
	{
		cin >> ptr[i]->data;
	}

	
	for(i = 0; i < size; i++)
	{
		cout << ptr[i]->data;
	}


	return 0;
}

i don't think i'm making a mistake when it comes time to declaring an array dynamically, however the problem pertains to the 2 (for) loops by trying to access (ptr[i]->data)....the only way to compile it is to use the (.) dot operator, but as my memory serves me correctly i'm supposed to use (->) when i try to access data through the pointer...weird....???

hmmm.... either i haven't done this stuff for a while or i just don't see it....

any help would be appreciated...


matheo917

let me see if i understand you correctly...

by typing: ptr[i] --- whatever (i) is in this case, what i'm impilictly doing is derefrencing the pointer, so in result i must use (.) dot operator to access its data members....

hmmm.....sounds reasonable, but i can't really picture this....
hmmm....

ptr....is a pointer to the first 'struct' in the array of the free store...


can you elaborate on this a bit more ....

thanx in advance


matheo917

node * p = new node;

p is a pointer to a single node. To reference the data component of the node you have to use the arrow operator or the derference opertor with the parentheisis combo.

node *p1 = new node[size];

p1 is a pointer to a collection of nodes, where the memory for the group is consecutive. In acutality p1 points to the first piece of memory for the section of memory reserved for the group of nodes. That means p1, for all intents and purposes, is the name of an array or is an array (the name of an array "can be considered" a constant pointer to the first element of the array) . Therefore use the dot operator. if you are using the [] preceding it and the -> if you are using just the name with offset, just like you would for an array declared with static memory/on the stack rather than on the free store/heap/dynamic memory.

thanx...that makes everything a lot clrearer....


matheo917

If you truly like dereferencing stuff so much:

Code:

(ptr+i)->member = something;

nifty....

good to know

thanx