Structs and Pointers

When you have a pointer to a struct — which is what malloc returns when allocating one — you can't use the dot operator directly, since p.x would need p to be an actual struct, not a pointer to one. The -> operator is shorthand for dereferencing and then accessing a field in one step: p->x means (*p).x.

#include <stdio.h>
#include <stdlib.h>

struct Point {
    int x;
    int y;
};

int main() {
    struct Point *p = malloc(sizeof(struct Point));
    p->x = 3;
    p->y = 4;
    printf("%d %d\n", p->x, p->y);
    free(p);
    return 0;
}

3 4

What the real compiler error looks like

Using . on a struct pointer instead of -> is a genuinely common C mistake, and the compiler catches it immediately. This is genuinely what gcc reports — captured from a real run, not written to look plausible:

#include <stdio.h>
#include <stdlib.h>

struct Point {
    int x;
    int y;
};

int main() {
    struct Point *p = malloc(sizeof(struct Point));
    p.x = 3;
    printf("%d\n", p.x);
    free(p);
    return 0;
}

main.c: In function ‘main’:
main.c:11:6: error: ‘p’ is a pointer; did you mean to use ‘->’?
   11 |     p.x = 3;
      |      ^
      |      ->
main.c:12:21: error: ‘p’ is a pointer; did you mean to use ‘->’?
   12 |     printf("%d\n", p.x);
      |                     ^
      |                     ->

GCC's diagnostic here is unusually helpful — it names the exact fix. Not every compiler error in this course will be this forgiving.

Next: function pointers — storing a function itself in a variable, and calling it indirectly.