I do not understand this challenge in C

Question

#include<stdio.h>
struct point { int x; int y; };
void foo(struct point*);
int main() {
struct point p1[] = {1, 2, 3, 4, 5};
foo(p1);
}
void foo(struct point p[]) {
printf("%d %d\n", p->x, (p + 2) ->y);
return 0;
}

what the p->x and (p + 2) -> y mean?
please help!
thank you

Answer 1

if p struct pointer then it's elements are accessed by p -> element;

So there p -> x refers to p[0] -> x
and p[0] is {1, 2};
since struct has { x, y} which values are stored on p1[] array sequentially..

p +2 points to p[2] =>{5} of {x, y} , y is not assinged any value so default is 0;

Here is clear code for the above program: hope it helps to understand clearly.

#include<stdio.h>
struct point { int x; int y; };
void foo(struct point*);
int main(){
struct point p1[] = { {1, 2}, {3, 4}, {5} };

foo(p1);
}
void foo(struct point *p){
printf("%d %d\n", (p+0)->x, (p + 2)->y);
}

Answer 2

Ana🌺Đurica🌼🌸 arrow operator used to dereferencing a pointer, so the syntax is the same as (*ptr). Dereferencing a pointer means to get the value of that address and as Jayakrishna🇮🇳 mentioned y had no assignment.

Answer 3

Jayakrishna🇮🇳
so it goes like this
struct point p1[] = { {x,y}, {x,y}, {x}}
?
and
p -> x => 1
p -> y => 2
(p + 1) -> x => 3
(p + 1) -> y => 4
(p + 2) -> x => 5
(p + 2) -> y => 0

Answer 4

Yes.
p1[] array is array of 'point ' structure.
and each structure holding 2 variables (x, y) .

Answer 5

The output is 1 0