Reputation: 17272

When is pointer to array useful?

I can declare:

int (*ap)[N];

So ap is pointer to int array of size N. Why is this ever useful? If I pass it to function, what useful thing it can do with it that it could not do with just a normal pointer to the array's contents?

C FAQ say:

2.12: How do I declare a pointer to an array?

Usually, you don't want to.

Upvotes: 11

Answers (9)

Mayank Jaiswal

Reputation: 13076

Following code is a part of my article : Pointers and Arrays in C C++

You can check it out @ http://pointersandarrays.blogspot.com/

Pointers and 2D Arrays

Following code snippet illustrates how to declare and access a 2D array. Underneath 2D Array lies a single dimensional array. You will get be sure of this after playing with the following piece of code.

Code Snippet #4


 #include<iostream&rt;  
    using namespace std;  

    int main()  
    {  
     cout<< "Understanding Pointers and 2 D Arrays"<<endl;  
     cout<< "----------------------\n"<<endl;  

    //Declaration of a 2D Array.  
    int tab[3][5];  
    //Total space required : 3*5 * sizeof(int) = 15 * sizeof(int)   
    //Funda : Since the amount of required space is known by compiler, contiguous 15 memory cells are allocated here.     
    //Hence the case is similar to a 1 D Array of size 15. Lets try this out!  

    //Array initialization using array name   
    for(int i=0; i<3;i++)  
     for(int j=0; j<5;j++)  
       tab[i][j]=i+2*j;  

    //Print array using array name   
    cout << "\nPrint array using array name ..."<<endl;    
    for(int i=0; i<3;i++)  
     {  
      for(int j=0; j<5;j++)  
       printf("%2d ",tab[i][j] );  
      printf("\n");  
     }  

    //Print array using a pointer. Proof of 1 D array being allocated  
    cout << "\nPrint array using a pointer. Proof of 1 D array being allocated ..." << endl;       
    int *tptr;  
    tptr = &tab[0][0]; // pointer tptr points at first element of the array.   

    for(int i=0; i<15;i++)  
     printf("%d ",*(tptr+i) );  
    tptr = &tab[0][0];  
    cout << "\nNotice that array is printed row by row in a linear fashion."<<endl;  
     return 0;  
    }

Output #4:



Understanding Pointers and 2D Arrays  

Print array using array name ...  
 0  2  4  6  8   
 1  3  5  7  9   
 2  4  6  8 10   

Print array using a pointer. Proof of 1 D array being allocated ...  
0 2 4 6 8 1 3 5 7 9 2 4 6 8 10   
Notice that array is printed row by row in a linear fashion.

Upvotes: -1

zaharpopov

Reputation: 17272

I think the conclusion from this whole discussion is never. Nobody here really demonstrated a use for this construct where something else wouldn't work in a more comprehensible way.

Upvotes: -5

John Bode

Reputation: 123448

Generally, the only time you'll see a pointer to an array (T (*a)[N]) is as a function parameter, where a is meant to be a 2d array:

void foo(int (*a)[N], size_t count)
{
  size_t i;
  for (i = 0; i < count; i++)
      a[i][j] = ...;
  ...
}

void bar(void)
{
  int arr[M][N];
  foo(arr, M);
}

Note that for a function parameter declaration, int a[][N] is equivalent to int (*a)[N], but this is only true for function parameter declarations:

void foo (int a[][N], size_t count) {...}

Pointers to arrays are generally not as useful as pointers to the base type, since you need to know the array size to properly declare a pointer to it (a pointer to a 10-element array of int is a different type from a pointer to a 20-element array of int). Personally, I haven't found much use for them in 20-some-odd years of programming.

Remember that in most contexts, an array expression (such as arr above) will have its type implicitly converted from "N-element array of T" to "pointer to T" (except when the array expression is an operand of sizeof or &, or the array is a string literal being used as an initializer in a declaration). In this case, the type of arr in the call to foo is implicitly converted from "M-element array of N-element array of int" to "pointer to N-element array of int".

Given the declaration T a[M][N], all of the following expressions will evaluate to the same location (the address of the first element in the array), but the types will be different as shown below:

Expression            Type                Implicitly converted to
----------            ----                -----------------------
         a            T [M][N]            T (*)[N]
      a[0]            T [N]               T *
        &a            T (*)[M][N]        
     &a[0]            T (*)[N]
  &a[0][0]            T *

Upvotes: 3

Mark Tolonen

Reputation: 177610

A pointer to an array can be used to dynamically allocate a multi-dimensional array N, where N-1 dimensions are known. Below creates a Nx3 array.

int (*ap)[3];
ap = malloc(N * sizeof(*ap));
/* can now access ap[0][0] - ap[N-1][2] */

@Adam E/Cruachan, This is not the same thing as a pointer to a pointer. ap is a single pointer to a block of memory containing three consecutive integers. ap++ will advance the pointer address to the next block of three integers. for int **pp;, pp points to an integer pointer, each of which can point to an integer anywhere in memory.

         +-----+                +------+    +-----+
 ap ---> | int |   vs.  pp ---> | int* | -> | int |
         | int |                +------+    +-----+
         | int |        pp+1 -> | int* | -\
         +-----+                +------+   \   +-----+
 ap+1 -> | int |                  :  :      -> | int |
         | int |                               +-----+
         | int |
         +-----+
           : :

Upvotes: 14

scienty

Reputation: 314

There are situations where you want to pass the memory location across programs. e.g a windows API might expect you to pass a pointer to data structure or an array where as you are programming in some other language, say c#. Windows API does not care how the target language handles array's, for windows API it is just a stream of bytes in memory and it will fill it, send it back to you. to avoid cross language type missmatch in some cases we use pointer to array rather than implicit array name as pointer. More over it is not guaranteed that the implicit array name is a long pointer, some compilers might optimize it to be a relative value withing segment. a pointer to an array guarantees that it is of the order of machine register size and you can point to a location anywhere in the available RAM.

Upvotes: 2

chaosTechnician

Reputation: 1700

It seems pretty useless to me to do a pointer to an array. In C, an array is already a pointer to a block of that data type.

int (*ap)[N];
int **ipp;

are both the same data type (a pointer to a pointer to an Integer). The only difference there is that there is space for N integers allotted for ap.

There's no need to pass an array by a pointer to a function, like, for the purpose of changing the contents of the array within that function because it's already a pointer. As a general rule, I'd say it's unnecessary and just creates an additional need to dereference the pointer to get at the data in the array. But, I'm sure there's a program or algorithm somewhere that could find a legitimate use for it.

Upvotes: -1

bmargulies

Reputation: 100013

This will probably careen off into subjective/argumentative, but...

In my opinion, pointers to arrays are in the language because they fell into the language. There are pointers to every other declarable data type, so these are here too. I've never seen anyone get really useful work out of them. Hypothetically, they allow a prototype that demands an array and not a pointer to the element, but ...

Upvotes: 0

Steve Emmerson

Reputation: 7832

If you increment the pointer, it will then point to the start of the next group of N elements.

This is not a big deal and it's use is up to the developer.

Upvotes: 6

Adam Woś

Reputation: 2473

It's not useful, really. But sometimes pointers to arrays are used e.g. in Microsoft Windows API - I've seen a lot of this stuff there.

Upvotes: 2

When is pointer to array useful?

Answers (9)

Related Questions