So my book is explaining me pointers to an array using ts example

#include <stdio.h>

int main()
{
    int s[4][2] = {
        {1234,56},{1212,33},{1434,80},{1312,78}
    };
    int(*p)[2];
    int i, j, * pint;
    for (i = 0; i <= 3; i  )
    {
        p = &s[i];
        pint = (int*)p;
        printf("\n");
        for (j = 0; j<= 1; j  )
        {
            printf("%d ", *(pint   j));
        }

    }
    return 0;
}

The output is Given as

1234 56
1212 33
1434 80
1312 78

No issue I am getting the same output.
My question is what was the need of using another pointer pint ? Why can't we directly use P?

So When I tried to do it using P directly it didn't work

printf("%d ", *(p   j));

I got garbage values in output, Why is this happening?
I also tried printing p and pint they are the same.

CodePudding user response：

Although p and pint have the same value, p 1 and pint 1 do not. p 1 is the same as (char *)p sizeof *p, and pint 1 is the same as (char *)pint sizeof *pint. Since the size of the object pointed to is different, the arithmetic gives different results.

CodePudding user response：

The pointer p is declared like

int(*p)[2];

So dereferencing the pointer expression with the pointer in this call of printf

printf("%d ", *(p   j));

you will get the j-th "row" of the type int[2] of the two dimensional array that in turn will be implicitly converted to a pointer of the type int * that will point to the first element of the j-th "row".

So instead of outputting elements of each row you will output first elements of each row that moreover results in undefined behavior when i will be greater than 2.