Display 2D array using pointers and function in c -

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• how pass 2d array through pointer in c [duplicate] 2 answers

respected members of stackoverflow, complete armature c program,i want access element of matrix using pointers.like want print elements of matrix using point.i have attached code along incorrect output.please me .thank you

` #include<stdio.h> #include<conio.h>  void print(int **arr, int m, int n) {     int i, j;     (i = 0; < m; i++)     {       (j = 0; j < n; j++)       {     printf("%d ",*(arr+i*n+j));                   //print elements of matrix     //printf("%d ", *((arr+i*n) + j));       }       printf("\n");     } }   int main() {     int arr[20][20],m,n,c,d;     clrscr();     printf("row=\n");     scanf("%d",&m);     printf("col=\n");     scanf("%d",&n);     for(c=0;c<m;c++)     {      for(d=0;d<n;d++)       {        scanf("%d",&arr[c][d]);       }     }     for(c=0;c<m;c++)                      //print matrix without function calling     {      for(d=0;d<n;d++)      {      printf("%d ",arr[c][d]);      }      printf("\n");     }     print((int **)arr, m, n);            //print matrix using function calling     getch();     return 0; }`  

above code produce output shown below

row= 2 col= 3 //elements of matrix 2 3 4 5 6 7 //print without using function 2 3 4 5 6 7 //print using function"print(int **a,int m,int n)" 2 3 4 0 0 0 

when using function not getting exact matrix value. note:print(int **a,int m,int n) should used.

as @whozcraig told in coment, arr should int**

int** arr = malloc(sizeof(int*)*m); int a; for(a=0;a<m;a++)     arr[a]=malloc(sizeof(int)*n); for(c=0;c<m;c++) {  for(d=0;d<n;d++)   {    scanf("%d", &arr[c][d]); //scanf("%d",arr+c*n+d);   } } 

note: not tested
edit: malloc not casted