Difference between "->" and "." in C [duplicate]

Question

Possible Duplicate:
Arrow operator (->) usage in C

I am trying to figure out the difference between the "." and "->" styles of data access in C language. Eg.

struct div{
    int quo;
    int rem;
};

1) Using "->"

struct div *d = malloc(sizeof(struct div));
d->quo = 8/3;
d->rem = 8%3;
printf("Answer:\n\t Quotient = %d\n\t Remainder = %d\n-----------\n",d->quo,d->rem);

2) Using "."

struct div d;
d.quo = 8/3;
d.rem = 8%3;
printf("Answer:\n\t Quotient = %d\n\t Remainder = %d\n-----------\n",d.quo,d.rem);

I am getting the same output for both the cases.

Answer: Quotient = 2 Remainder = 2

How these two approaches are working internally? And at what time which one should be used? I tried searching it on the internet but of not much help. Any relevant link is also appreciated.

Also is there any difference between their storage in memory?

Answer 1

Basic difference is this:

. is the member of a structure
-> is the member of a POINTED TO structure

So the . is used when there is a direct access to a variable in the structure. But the -> is used when you are accessing a variable of a structure through a pointer to that structure.

Say you have struct a:

struct a{
    int b;
}

However say c is a pointer to a then you know that: c = (*a). As a result -> is being used as a alternative for a certain case of the .. So instead of doing (*a).b, you can do c->b, which are the exact same.