Inconsistency in using pointer to an array and address of an array directly

Question

This code sample prints the array correctly.

int b[2] = {1, 2};
int *c = &b;
int  i, j,k = 0;
for (i = 0;i < 2; i++) {
    printf(\"%d \", *(c+i));
}
         


        

Answer 1

int *c = &b;     

This is actually not valid. You need a cast.

int *c = (int *) &b; 

The two expressions:

 *(c+i)

 and 

 *(&b+i)

are not same. In the first expression i is added to a int * and in the second expression i is added to a int (*)[2]. With int *c = (int *) &b; you convert a int (*)[2] to an int *. c + i points to the i-th int element of c but &b+i points to the int [2] element of b moving the pointer value outside the actual array object.