Arrow operator (->) usage in C

Question

I am currently learning C by reading a good beginner's book called "Teach Yourself C in 21 Days" (I have already learned Java and C# so I am moving at a much faster pace). I was reading the chapter on pointers and the -> (arrow) operator came up without explanation. I think that it is used to call members and functions (like the equivalent of the . (dot) operator, but for pointers instead of members). But I am not entirely sure. Could I please get an explanation and a code sample?

Answer 1

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.

Answer 2

#include<stdio.h>
struct examp{
int number;
};
struct examp a,*b=&a;`enter code here`
main()
{
a.number=5;
/* a.number,b->number,(*b).number produces same output. b->number is mostly used in linked list*/
   printf("%d \n %d \n %d",a.number,b->number,(*b).number);
}

output is 5 5 5

Answer 3

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

Answer 4

Dot is a dereference operator and used to connect the structure variable for a particular record of structure. Eg :

struct student
    {
      int s.no;
      Char name [];
      int age;
    } s1,s2;

main()
    {
      s1.name;
      s2.name;
    }

In such way we can use a dot operator to access the structure variable

Answer 5

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".

Answer 6

I had to make a small change to Jack's program to get it to run. After declaring the struct pointer pvar, point it to the address of var. I found this solution on page 242 of Stephen Kochan's Programming in C.

#include <stdio.h>

int main()
{
  struct foo
  {
    int x;
    float y;
  };

  struct foo var;
  struct foo* pvar;
  pvar = &var;

  var.x = 5;
  (&var)->y = 14.3;
  printf("%i - %.02f\n", var.x, (&var)->y);
  pvar->x = 6;
  pvar->y = 22.4;
  printf("%i - %.02f\n", pvar->x, pvar->y);
  return 0;
}

Run this in vim with the following command:

:!gcc -o var var.c && ./var

Will output:

5 - 14.30
6 - 22.40