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I'm trying to deepen my understanding on syntax relevant to pointer in C, and I noticed that If I created an array of int first, int(*)[] is a way of giving a pointer to this array, which seemingly could also be done by simply using int *. So I'm confused what's the difference between pointer b and pointer c in Code below? If they're different, when should I use one way instead of choosing the other one? 

#include <stdio.h>

int main(void)
{
    int a[3];    
    int* b = a;
    int (*c)[3] = &a;

    printf("%p\n", a);
    printf("%p\n", b); 
    printf("%p\n", c); //these 3 outputs are same.
}

I just tried to add lines of code to further understand this:

    a[2] = 3;

    printf("%i\n", a[2]); // = 3
    printf("%i\n", b[2]); // = 3

It seems that b and c are different because printf("%i\n", c[2]) will generate an error. How to explain and understand all above?

Jabberwocky
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jg10
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    Duplicate: https://stackoverflow.com/questions/62209942/what-int-ptr4-really-means-and-how-is-it-different-than-ptr. For some reason, a flavour of this question keeps popping up every day lately. – Lundin Jun 10 '20 at 08:00

3 Answers3

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So I'm confused what's the difference between pointer b and pointer c in Code below?

Their types. b is of type int *, and c is of type int (*)[3].

The address of the array is the same as the address of the first element in the array, so the memory location they point to will be same here, but as per their types, operations on them, for example pointer arithmetic will produce completely different results.

Sourav Ghosh
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Both b and c point to the same memory but the difference is big:

  1. These are two different types (pointer to int and pointer to an array of 3 ints)
  2. b is dereferenced to a single integer pointed by b while c is dereferenced to an array of 3 ints which means you can assign an integer to *b (*b = 10;) but you can't do the same for c (but only with specified index (*c)[0] = 10)
  3. Pointer arithmetic is also different b + 1 will increase the value of b by sizeof(int) while c + 1 will increase the value of c by 3*sizeof(int)

If they're different, when should I use one way instead of choosing the other one?

As any other type in C, it should be used based on your needs and application. Most commonly used is the int *b; option since it gives you more flexibility. With such pointer you can handle array of any size, it is more commonly used and more readable. While pointer to an array binds you to an array of pre-defined size, its syntax is less readable (in my opinion) and thus will be harder to review/debug the code. In my experience I've never seen a production code which uses pointer to an array but this doesn't mean you cannot use it if you find any advantage of it in your application.

Alex Lop.
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  • But if **b** is a pointer to int, why I can do **printf("%i\n", b[2])** which is equal to a[2] – jg10 Jun 10 '20 at 06:10
  • @jg10 This is C syntax. `b[2]` is the same as *(b + 2). You see more examples here https://godbolt.org/z/p7axEK – Alex Lop. Jun 10 '20 at 06:15
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Let's start with a visual representation of what's going on:

           int [3]    int *     int (*)[3]
           -------    -----     ----------
   +---+
   |   |   a[0]       b[0]      c[0]
   + - +
   |   |   a[1]       b[1]
   + - +
   | 3 |   a[2]       b[2]
   +---+
   |   |                        c[1]
   + - +
   |   |
   + - +
   |   |
   +---+
   |   |                        c[2]
   + - +
   |   |
   + - +
   |   |
   +---+

On the far left is a view of memory as a sequence of 3-element arrays of int The remaining columns show how each of a[i], b[i], and c[i] are interpreted.

Since a is declared as int [3] and b is declared as int *, each a[i] and b[i] have type int (a[i] == *(a + i)).

But c is different - since c is declared as "pointer to 3-element array of int" (int (*)[3]), then each c[i] has type "3-element array of int" (int [3]). So instead of being the third element of a, c[2] is the first element of the third 3-element array following a.

John Bode
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