Arrow operator (->) usage in C

foo->bar is equivalent to (*foo).bar, i.e. it gets the member called bar from the struct that foo points to.

Yes, that's it.

It's just the dot version when you want to access elements of a struct/class that is a pointer instead of a reference.

struct foo
{
  int x;
  float y;
};

struct foo var;
struct foo* pvar;
pvar = malloc(sizeof(struct foo));

var.x = 5;
(&var)->y = 14.3;
pvar->y = 22.4;
(*pvar).x = 6;

That's it!