Pointer to the Address of Element of an Array

Question

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

Output Showing is

1234,56
1212,33
1434,80
1312,78

As we know *(&Variable) will print the Value of the Variable But when We implement the same concept in above program...

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

output is showing the Address of each element of array.

Why this is happening? Why Output is not equal to value of elements of Array??

Answer 1

Your array is bi-dimensional. So, s[i] is the address of line i (i.e. s[i] is of type int *), and &s[i] is an address of an address (i.e. of type int **). When you apply * on it, you get an address (i.e. int *).

It is true that the operator & means "address-of" and *(&x) refers to the value of x.