[-Wincompatible-pointer-types]. dont know what the problem is

Question

I am not able to find where it went wrong. When I run the program, it shows "access denied".

#include<stdio.h>
int main()
{
    char arr[4][40] =
    { "array of c string",
        "is fun to use",
        "make sure to properly",
        "tell the array size"
    };
char *p = arr;  /*Char-4-eg-Find-the-output.c:10:11: warning: initialization of 'char *' from 
incompatible pointer type 'char (*)[40]' [-Wincompatible-pointer-types]*/
    for (int i = 0; i < 100; i++)
    {
        printf("%c",*p );
        p++;
   }
    return 0;
}

Answer 1

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

So in this declaration

char *p = arr;

the array designator arr is converted to a pointer to its first element. Elements of the array arr have the type char[40]. So a pointer to an object of this type has the type char( * )[40].

However the initialized pointer has the type char *. And there is no implicit conversion from the type char ( * )[40] to the type char *. So the compiler issues a message.

Either you need to use casting like

char *p = ( char * )arr;

or to write

char *p = arr[0];

or

char *p = &arr[0][0];

If you want to output the array of strings as one sentence you could write for example

#include <stdio.h>

int main(void) 
{
    enum { N = 40 };
    
    char arr[][N] =
    { 
        "array of c string",
        "is fun to use",
        "make sure to properly",
        "tell the array size"
    };
    

    for ( char ( *p )[N] = arr; p != arr + sizeof( arr ) / sizeof( *arr ); ++p )
    {
        if ( p != arr ) putchar( ' ' );
        printf( "%s", *p );
    }
    
    putchar( '\n' );
    
    return 0;
}

The program output is

array of c string is fun to use make sure to properly tell the array size

Answer 2

Change char *p = arr; to char *p = &arr[0][0];.