C Pointers Past Paper -

i'm looking @ past paper course i'm doing @ university, , there question c pointers.

i reckon have reasonable grasp of how work, question confusing me:

consider running c program fragment:   int x[4] = {0,2,4,6}; int *y; y = &x[2]; *(x + 2) = y[1] + 1;  value of expression *y afterwards? (a) 2 (b) 4 (c) 5 (d) 7 (e) 8 

now, in answers said question, says answer d.

i'm super confused, seeing as:

1. the value of x not declared, i'd have thought impossible evaluate x+2
2. y isn't array, how can y[1] evaluated?

why 7 correct answer here?

lets break down:

int x[4] = {0,2,4,6};

x [0] = 0 x [1] = 2 x [2] = 4 x [3] = 6 

int *y; pointer integer y point location in x

x [0] = 0       // <-- y ? x [1] = 2       // <-- y ? x [2] = 4       // <-- y ? x [3] = 6       // <-- y ? 

y = &x[2]; have specified y points x[2]

x [0] = 0 x [1] = 2 x [2] = 4       // <-- y (or y[0]) x [3] = 6 

*(x + 2) same x[2] so: x[2] = y[1] + 1;

x [0] = 0 x [1] = 2 x [2] = 4       // <-- x[2]   x [3] = 6       // <-- y[1]  

y[1] 6, y[1] + 1 = 7

note y[1] same *(y + 1). take address y points too, add size of 1 integer , obtain contents of points to.