warning: passing argument 1 of 'display' from incompatible pointer type [-Wincompatible-pointer-types]

Question

#include<stdio.h>
/* create a 2D array by taking input from the user and write a display function to display the contents
   of this array*/
void display(int *ptr_Arr, int x, int y);

int main(){
    int arr[2][2], i, j;
    for(i=0; i<=1; i++){
        for(j=0; j<=1; j++){
            printf("Enter the value of element %dX%d: ", i, j);
            scanf("%d", &arr[i][j]);
        }    
    }
    display(arr, 2, 2);
    return 0;
}

void display(int *ptr_Arr, int a, int b){
    int x, y;
    for(x=0; x<a; x++){
        for(y=0; y<b; y++){
            printf("%d ", *ptr_Arr);
            ptr_Arr++;
        }
        printf("\n");
    }
}

I get below output for the above code and I am not able to understand and resolve the problem as I am a beginner. Please make me understand what the problem here is.

Output:

Answer 1

In this call

display(arr, 2, 2);

the two-dimensional array arr declared like

int arr[2][2], i, j;

is implicitly converted to a pointer to its first element of the type int ( * )[2] . However the corresponding function parameter has the type int * and there is no implicit conversion between these pointer types.

You need at least declare the first function parameter like

int ptr_Arr[][2]

or

int ( *ptr_Arr )[2]

Using pointers in the nested for loops within the function the function can be declared and defined for example the following way

void display( int ptr_Arr[][2], size_t a )
{
    for( int ( *p )[2] = ptr_Arr; p != ptr_Arr + a; ++p )
    {
        for ( int *q = *p; q != *p + sizeof( *p ) / sizeof( **p ); ++q )
        {
            printf("%d ", *q);
        }
        printf("\n");
    }
}

and called like

display( arr, sizeof( arr ) / sizeof( *arr ) );

If you want to write the function for two-dimensional arrays with various sizes then the function can be declared and defined the following way (provided that the compiler supports variable length arrays)

void display( size_t a, size_t b, int ptr_Arr[][b] )
{
    for( int ( *p )[b] = ptr_Arr; p != ptr_Arr + a; ++p )
    {
        for ( int *q = *p; q != *p + b; ++q )
        {
            printf("%d ", *q);
        }
        printf("\n");
    }
}

and called like for example

display( 2, 2, arr );

And at last you could reinterpret the two-dimensional array as a one-dimensional array within the function. In this case you will need implicitly to cast pointers. For example

void display( int *ptr_Arr, size_t a, size_t b )
{
    for ( int *p = ptr_Arr; p != ptr_Arr + a * b; p += b )
    {
        for ( int *q = *p; q != p + b; ++q )
        {
            printf("%d ", *q);
        }
        printf("\n");
    }
}

And the function is called like

display( ( int * )arr, 2, 2 );