Difference between “pointer to int” and “pointer to array of ints”

Question

int main()
{
    int (*x)[5];                 //pointer to an array of integers
    int y[6] = {1,2,3,4,5,6};    //array of integers
    int *z;                              


        

Answer 1

One should understand the internal representation of (*x)[i]. Internally, it is represented as *((*x)+i), which is nothing but the ith element of the array to which x is pointing. This is also a way to have a pointer pointing to 2d array. The number of rows is irrelevant in a 2d array.

For example:

int arr[][2]={{1,2},{3,4}};
int (*x)(2);
x=arr; /* Now x is a pointer to the 2d array arr.*/

Here x is pointing to a 2d array having 2 integer values in all columns, and array elements are stored contiguously. So (*x)[0] will print arr[0][0] (which is 1), (*x)[1] will print the value of arr[0][1] (which is 2) and so on. (*x+1)[0] will print the value of arr[1][0] (3 in this case) (*x+1)[1] will print the value of arr[1][1] (4 in this case) and so on.

Now, a 1d array could be treated as nothing but a 2d array having only one row with as many columns.

int y[6] = {1,2,3,4,5,6};
int (*x)[6];
x =y;

This means x is a pointer to an array having 6 integers. So (*x)[i] which is equivalent to *((*x)+i) will print ith index value of y.