Reputation: 55247

Arrow operator (->) usage in C

I am reading a 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?

Upvotes: 348

Answers (11)

JgWangdu

Reputation: 317

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

Here 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".

Upvotes: 12

Jack

Reputation: 133619

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!

Upvotes: 162

71GA

Reputation: 1399

Well I have to add something as well. Structure is a bit different than array because array is a pointer and structure is not. So be careful!

Lets say I write this useless piece of code:

#include <stdio.h>

typedef struct{
        int km;
        int kph;
        int kg;
    } car;

int main(void){

    car audi = {12000, 230, 760};
    car *ptr = &audi;

}

Here pointer ptr points to the address (!) of the structure variable audi but beside address structure also has a chunk of data (!)! The first member of the chunk of data has the same address than structure itself and you can get it's data by only dereferencing a pointer like this *ptr (no braces).

But If you want to acess any other member than the first one, you have to add a designator like .km, .kph, .kg which are nothing more than offsets to the base address of the chunk of data...

But because of the preceedence you can't write *ptr.kg as access operator . is evaluated before dereference operator * and you would get *(ptr.kg) which is not possible as pointer has no members! And compiler knows this and will therefore issue an error e.g.:

error: ‘ptr’ is a pointer; did you mean to use ‘->’?
  printf("%d\n", *ptr.km);

Instead you use this (*ptr).kg and you force compiler to 1st dereference the pointer and enable acess to the chunk of data and 2nd you add an offset (designator) to choose the member.

Check this image I made:

But if you would have nested members this syntax would become unreadable and therefore -> was introduced. I think readability is the only justifiable reason for using it as this ptr->kg is much easier to write than (*ptr).kg.

Now let us write this differently so that you see the connection more clearly. (*ptr).kg ⟹ (*&audi).kg ⟹ audi.kg. Here I first used the fact that ptr is an "address of audi" i.e. &audi and fact that "reference" & and "dereference" * operators cancel eachother out.

Upvotes: 17

prashanth

Reputation: 11

#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

Upvotes: 1

Lukasz Matysiak

Reputation: 921

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.

Upvotes: 49

smwikipedia

Reputation: 64391

The -> operator makes the code more readable than the * operator in some situations.

Such as: (quoted from the EDK II project)

typedef
EFI_STATUS
(EFIAPI *EFI_BLOCK_READ)(
  IN EFI_BLOCK_IO_PROTOCOL          *This,
  IN UINT32                         MediaId,
  IN EFI_LBA                        Lba,
  IN UINTN                          BufferSize,
  OUT VOID                          *Buffer
  );


struct _EFI_BLOCK_IO_PROTOCOL {
  ///
  /// The revision to which the block IO interface adheres. All future
  /// revisions must be backwards compatible. If a future version is not
  /// back wards compatible, it is not the same GUID.
  ///
  UINT64              Revision;
  ///
  /// Pointer to the EFI_BLOCK_IO_MEDIA data for this device.
  ///
  EFI_BLOCK_IO_MEDIA  *Media;

  EFI_BLOCK_RESET     Reset;
  EFI_BLOCK_READ      ReadBlocks;
  EFI_BLOCK_WRITE     WriteBlocks;
  EFI_BLOCK_FLUSH     FlushBlocks;

};

The _EFI_BLOCK_IO_PROTOCOL struct contains 4 function pointer members.

Suppose you have a variable struct _EFI_BLOCK_IO_PROTOCOL * pStruct, and you want to use the good old * operator to call it's member function pointer. You will end up with code like this:

(*pStruct).ReadBlocks(...arguments...)

But with the -> operator, you can write like this:

pStruct->ReadBlocks(...arguments...).

Which looks better?

Upvotes: 1

Gopal Rao

Reputation: 9

#include<stdio.h>

int main()
{
    struct foo
    {
        int x;
        float y;
    } var1;
    struct foo var;
    struct foo* pvar;

    pvar = &var1;
    /* if pvar = &var; it directly 
       takes values stored in var, and if give  
       new > values like pvar->x = 6; pvar->y = 22.4; 
       it modifies the values of var  
       object..so better to give new reference. */
    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;
}

Upvotes: 0

Rich Vogt

Reputation: 93

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;
}

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