CODEWITHSUNDEEP
c++arrayspointersdynamic-arrays
Basma
Basma

Reputation: 33

creating dynamic multidimensional arrays

I am a totally beginner in programming and I have a question in the following code it's about creating a dynamic multidimensional array I don't understand the reason of p what's the function of it here 

#include <iostream>
using namespace std;

void print2DArr(int** arp,int *ss,int nrows)
{
    for(int i=0;i<nrows;i++){
        for(int j=0;j<ss[i];j++){
            cout << arp[i][j] <<" ";
        }
        cout << endl;
    }
}

void main()
{
    int count;
    cout << "How many arrays you have?\n";
    cin >> count;
    int **arrs = new int*[count];
    int *sizes = new int[count];
    //int *arrs[3];//static array of pointers
    for(int i=0;i<count;i++)
    {
        int size;
        cout << "Enter size of array " << (i+1) << endl;
        cin >> size;
        sizes[i]=size;
        int *p = new int[size];
        arrs[i] = p;
        //arrs[i] = new int[size];
        cout << "Enter " << size << " values\n";
        for(int j=0;j<size;j++)
            //cin >> p[j];
            cin >> arrs[i][j];
    }

    print2DArr(arrs,sizes,count);

    //delete (dynamic de-allocation)
    for(int i=0;i<count;i++)
        delete[] arrs[i];
    delete[] arrs;


}

Upvotes: 0

Views: 111

Answers (2)

markus
markus

Reputation: 1651

you are creating a dynamically sized int array in your for loop. p is pointing to that newly created int array. You can then assign p (the adress of your new int array) to the i-th position of your array of pointers to int array (that's what the next statement does: arrs[i]=p).

Therefore your structure looks like:

arrs (array of pointers to int arrays):
- arrs0: pointing to int array 1 at adress 4711
- arrs1: pointing to int array 2 at adress 9876

sizes (array of int's) - holding the sizes of your int arrays (not of your array of pointers to int arrays!)
- 0: 2
- 1: 3

int array 1 (adr 4711)
- 0: 17
- 1: 23

int array 2 (adr 9876)

Upvotes: 0

R Sahu
R Sahu

Reputation: 206567

It's variable that does not do a whole lot. You can replace the lines

    int *p = new int[size];
    arrs[i] = p;

with

    arrs[i] = new int[size];

without any problem.

Upvotes: 3

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