Is this a correct way to define array of pointers to array?

Is this a correct way to define array of pointers to array in C programming language?

int* ptr[2];

int n1[5] = { 2,3,4,5,6 };
int n2[5] = { 2,3,4,5,6 };
ptr[0] = &n1;
ptr[1] = &n2;

I am getting errors like:

epi_2.c:20:12: warning: incompatible pointer types assigning to 'int *' from 'int (*)[5]' [-Wincompatible-pointer-types]
    ptr[1] = &n2;

You have to write

ptr[0] = n1;
ptr[1] = n2;

Array designators used in expressions with rare exceptions are converted to pointers to their first elements.

That is in the above statements expressions n1 and n2 have the type int * - the type of the left side expressions.

As for these statements

ptr[0] = &n1;
ptr[1] = &n2;

then the right side expressions have the type int ( * )[5] that is not compatible with the type int * of the left side expressions. So the compiler issues messages.

Otherwise you need to declare the array of pointers like

int ( * ptr[2] )[5];
//...
ptr[0] = &n1;
ptr[1] = &n2;

Here is a demonstration program.

#include <stdio.h>

int main( void ) 
{
    int* ptr[2];

    int n1[5] = { 2,3,4,5,6 };
    int n2[5] = { 2,3,4,5,6 };

    ptr[0] = n1;
    ptr[1] = n2;

    for ( size_t i = 0; i < 2; ++i )
    {
        for ( size_t j = 0; j < 5; j++ )
        {
            printf( "%d ", ptr[i][j] );
        }
        putchar( '\n' );
    }
}

The program output is

2 3 4 5 6 
2 3 4 5 6 

And here is another demonstration program.

#include <stdio.h>

int main( void ) 
{
    int ( * ptr[2] )[5];

    int n1[5] = { 2,3,4,5,6 };
    int n2[5] = { 2,3,4,5,6 };

    ptr[0] = &n1;
    ptr[1] = &n2;

    for ( size_t i = 0; i < 2; ++i )
    {
        for ( size_t j = 0; j < 5; j++ )
        {
            printf( "%d ", ( *ptr[i] )[j] );
        }
        putchar( '\n' );
    }
}

The program output is the same as shown above

2 3 4 5 6 
2 3 4 5 6