Question

Well, here is a full sample that works but the console vanishes right after the last print and i cant make it stay. Also there are a few queries that I include in some lines

//bidimensional array dynamic memory allocation

#include <stdio.h>
#include <stdlib.h>

void main()

{
     int **p; // pointer to pointer
     int n,m,i,j,k; // n is rows, m is cols, i and j are the indexes of the array, k  is going to be like m, but used to print out

     do
     {
     printf("\n how many rows?");
     scanf ("\%d", &n);
     }
     while (n <= 0);

// booking memory for an array of n elements, each element is a pointer to an int (int *)

//Query: a pointer to an int? wouldnt it be a pointer to a pointer ? It uses **

     p = (int **) malloc (n * sizeof(int *)); // 
     if(p == NULL)
          {
          printf("Insuficient memory space");
          exit( -1);
          }

          for (i = 0; i < n; i++)  // now lets tell each row how many cols it is going to have
          {
              printf("\n\nNumber of cols of the row%d :", i+1); // for each row it can be different

              scanf("%d", &m); // tell how many cols
              p[i] = (int*)malloc(m * sizeof(int)); // we allocate a number of bytes equal to datatype times the number of cols per row

/Query: I cant grasp the p[i] because if p was a pointer to a pointer, what is that array notation, i mean the square brackets/

              if(p[i] == NULL)
              { printf("Insuficient memory space");
              exit(-1);
              }

              for (j=0;j<m;j++)
              {
                  printf("Element[%d][%d]:", i+1,j+1);
                  scanf("%d",&p[i][j]); // reading through array notation
              }

              printf("\n elements of row %d:\n", i+1);
              for (k = 0; k < m; k++)
              // printing out array elements through pointer notation
              printf("%d ", *(*(p+i)+k));
          }

              // freeing up memory assigned for each row
              for (i = 0; i < n; i++)
              free(p[i]);
              free(p);// freeing up memory for the pointers matrix

              getchar(); // it cannot stop the console from vanishing
              fflush(stdin); // neither does this
}

// ********thanks a lot******

Answer 1

it’s easy to understand pointers in context of arrays.
So if

int * p

is the one-dimensional array of int, then int ** p will be two -dimensional array of int. In other words it is an array that containt a pointers to one-dimensional array.

so

p = (int **) malloc (n * sizeof(int *)); // 

is a pointer to the pointer

and p[i] is current pointer to the int.