Array Pointer in C++

Question

int n;
int *array[8]
cout<<"Enter Number Between 0-9 Only"<<endl;
for(int i = 0; i< 9; i++){
    cout << "Enter Number " << (i + 1) << endl;
    cin >> n;
    if((n >= 0) && (n <= 9))
        array[i] = &n;
    else {
        cout << "Numbers from 0-9 only\n" << endl;
        i--;
    }

}
cout << *array[0] << endl;
}

I'm trying to store 9 entered numbers in a pointer array but its not working why?? Can you explain why to me and how to solve it or improve it. I'm just a beginner and its not homework im testing what i had read.

Why are you trying to mess around with pointers now if you are just a beginner? They belong much farther down the road if you are still confused on arrays. — Marlon
– Marlon, Commented Jul 18, 2012 at 20:37
this one seems pretty close to stackoverflow.com/questions/11548571 — Philipp
– Philipp, Commented Jul 18, 2012 at 21:42

Ed Heal · Accepted Answer · 2012-07-18 20:44:35Z

4

The line

array[i] = &n;

will store the same address in every entry in your array. This is just pointing to n so will always point to that value.

Either define the array as an array of integers

i.e. int array[9];

and then place the value into that array

i.e. array[i] = n;

OR

Allocate some memory off the heap

i.e.

int *array[9];
...
array[i] = new int;
*array[i] = n;

But you will then have to free this memory with delete to avoid a memory leak.

edited Jul 18, 2012 at 20:44

answered Jul 18, 2012 at 20:32

Ed Heal

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2 Comments

mmmmmm Over a year ago

The last bit will give rise to memory leaks unless you are careful - wether the answer is correct does depend on what exactly the OP is trying to do - but I suspect they do not need that complication.

Ed Heal Over a year ago

@mark - I know. I was just about to add that but the phone rang.

mmmmmm · Accepted Answer · 2012-07-18 20:40:55Z

2

There are several issues here.

You have nowhere to store the values. You have an array of 8 pointers which are all set to point to the same variable n, which is on the stack and so goes out of scope.
The array has 8 elements so the loop goes one past the end
This is C++ so really best not to use C arrays unless you have a justifiable reason to.

I would have something more like *NB not compiled and run)

{
...
std::vector<int> array;

cout<<"Enter Number Between 0-9 Only"<<endl;
for(int i = 0; i< 8; i++){
    int n;
    cout << "Enter Number " << (i + 1) << endl;
    cin >> n;
    if((n >= 0) && (n <= 9))
        array.push_back(n);
    else {
        cout << "Numbers from 0-9 only\n" << endl;
        i--;
    }

}

cout << array[0] << endl;
}

edited Jul 18, 2012 at 20:40

answered Jul 18, 2012 at 20:34

mmmmmm

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Comments

timos · Accepted Answer · 2012-07-18 20:29:30Z

1

You are saving a pointer to n in the array, but you constantly change the value of n.

answered Jul 18, 2012 at 20:29

timos

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Comments

ryanbwork · Accepted Answer · 2012-07-18 20:32:19Z

0

You don't really need to mess with pointers here. Change your array definition, how you populate it, and how you display and you should have better luck.

int array[9];
...
array[i] = n;
...
cout << array[0] << endl;

answered Jul 18, 2012 at 20:32

ryanbwork

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