Arrow operator (->) usage in C

Question

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!

I'd just add to the answers the "why?".

. is standard member access operator that has a higher precedence than * pointer operator.

When you are trying to access a struct's internals and you wrote it as *foo.bar then the compiler would think to want a 'bar' element of 'foo' (which is an address in memory) and obviously that mere address does not have any members.

Thus you need to ask the compiler to first dereference whith (*foo) and then access the member element: (*foo).bar, which is a bit clumsy to write so the good folks have come up with a shorthand version: foo->bar which is sort of member access by pointer operator.

a->b is just short for (*a).b in every way (same for functions: a->b() is short for (*a).b()).

foo->bar is only shorthand for (*foo).bar. That's all there is to it.

struct Node {
    int i;
    int j;
};
struct Node a, *p = &a;

Here the to access the values of i and j we can use the variable a and the pointer p as follows: a.i, (*p).i and p->i are all the same.

Here . is a "Direct Selector" and -> is an "Indirect Selector".