Function returning a pointer to an int array

Question

I am learning C++ from Primer 5th edition and I am at Returning a Pointer to an Array. The declaration of this function is:

 int (*func(int i))[10];

and it's expected to return a pointer to an array.

I wrote code that does this:

#include <iostream>
#include <string>
using namespace std;

int *func(){
  static int a[]={1,2,3};
  return a;
}
int main (){
  int *p=func();
  for(int i=0;i!=3;++i){
    cout<<*(p+i);
  }
}

And it is working. But I want to know the difference between what I made here and

  int (*func(int i))[10];

How I can make this function call work, because in the book, there isn't any concrete example.

Yes because i don t understand this type of function and i made this code, and working but i want to work int (*func(int i))[10]; — alin, Jul 26 '13 at 13:10
I think you are confuse between [array name = `a` , `int*`and address of array `&a`, `int(*)[5]`](http://stackoverflow.com/questions/15177420/what-does-sizeofarr-return/15177499#15177499) read lined answer may help you. — Grijesh Chauhan, Jul 26 '13 at 13:11
@jrok: This question seems better than the other, I don't know why the votes to close, as it is well formed (well, on the average of the quality of the site I'd say) — David Rodríguez - dribeas, Jul 26 '13 at 13:11
you are returning address of first element. Actually type of `a` is `int[3]` that decays into `int*`. Important is you stores address into `int* p` and can assess elements of array as `p[i]`. Whereas if your function would be `int (*func(int i))[3]` then you return `&a` and assign to `int(*p)[3]` and can access `(*p)[i]` — Grijesh Chauhan, Jul 26 '13 at 13:14
@DavidRodríguez-dribeas Yeah, I feel a bit bad now :) voting to reopen. — jrok, Jul 26 '13 at 13:16
@GrijeshChauhan thank you very much, but for 20 minutes i reading and rereading your comment and i can t understand. Can you make like a code who working, because I can t understand where comes every declaration . — alin, Jul 26 '13 at 13:41
@alin let me know does my answer helps, ask your doubt to my answer.. feel free to ask .. — Grijesh Chauhan, Jul 26 '13 at 14:02
@alin In you previous question [you got the answer](http://stackoverflow.com/a/17871039/1673391) that no buddy notice, and voted. — Grijesh Chauhan, Jul 26 '13 at 14:39
@GrijeshChauhan Thank you very much for your answer, it helped me and i understan in finally, but i have one more question, how i escape by static int a[]{}, so, I want to return a array not from fuction, by exemple from int main(), or anywere — alin, Jul 26 '13 at 14:47
@alin (1) comment to answer instead to your question (else it will confuse other). (2) I couldn't understand your question --but Understand if you declares your array simply `int a[] = {1, 3, 3}` instead of using `static` keyword then is becomes buggy code. because life & scope of local array is till function not returns. — Grijesh Chauhan, Jul 26 '13 at 14:51
yes i know that if i don t use static it is wrong, it says this in book and i understand this, but i want to know if is any possibilities to return other array. — alin, Jul 26 '13 at 14:55
let us [continue this discussion in chat](http://chat.stackoverflow.com/rooms/34242/discussion-between-alin-and-grijesh-chauhan) — alin, Jul 26 '13 at 14:57

score 3 · Accepted Answer · edited May 23 '17 at 12:21

Read: What does sizeof(&array) return? to understand diffrence between array name and address of array.

Q1 I want to know the difference between:

In your code:

  int *func(){
    static int a[]={1,2,3};
    return a;
  }

you are returning address of first element. Actually type of a is int[3] that decays into int*. Important is
You stores address into int* p and can assess elements of array as p[i].

Whereas if your function would be int int (*func())[3] then you return &a, and assign to int(*p)[3] and can access (*p)[i].
Note: type of &a is int(*)[3].

Q2 How i can make this function call work, because in the book, there isn't any concrete example.

like:

int (*func())[3]{
    static int a[]={1,2,3};
    return &a;
}

And main():

int main(){ 
 int i=0;    
 int(*p)[3] = func();
 for(i=0; i<3; i++)
   printf(" %d\n", (*p)[i]);
 return 0;
}

You can check second version of code working id Ideone

Q1 I want to know the difference between:

As you are interested to know diffrence between two so now compare two different declarations of p in two versions of code:

1) : int* p; and we access array elements as p[i] that is equals to *(p + i).

2) : int (*p)[i] and we access array elements as (*p)[i] that is equals to *((*p) + i) or just = *(*p + i). ( I added () around *p to access array element because precedence of [] operator is higher then * So simple *p[i] means defense to the array elements).

Edit:

An addition information other then return type:

In both kind of functions we returns address that is of a static variable (array), and a static object life is till program not terminates. So access the array outsize func() is not a problem.

Consider if you returns address of simple array (or variable) that is not static (and dynamically allocated) then it introduce as Undefined behavior in your code that can crash.

Thank you very much for your answer, it helped me and i understan in finally, but i have one more question, how i escape by static int a[]{}, so, I want to return a array not from fuction, by exemple from int main(), or anywere — alin, Jul 26 '13 at 14:48
@alin see `main()` is quit different function then other its declaration is predefined you can't make a `main()` that returns pointer to array. Its return type has to be int only. — Grijesh Chauhan, Jul 26 '13 at 14:53
Sorry @ Grijesh chauhan, but when I "discover" this function (accept an answer) I give accept at all, and first was your answer, and when i accept any answer your answer was unaccepted. Sorry , i accept your answer. — alin, Jul 26 '13 at 20:25
@alin Ok no problem, You can vote all, but can accept only one that the reason ?I give a like to you. — Grijesh Chauhan, Jul 26 '13 at 20:30
@GrijeshChauhan i was reading your answer then i wanted to try your second and third code but i could not compile it (it tells me that Returning 'int[3] *' from a function/method/block returning 'int[3]': Array is not assignabel ) — Mohammad Abdollahzadeh, Jul 17 '17 at 10:30

score 2 · Answer 2 · answered Jul 26 '13 at 13:50

int(*)[10] is a pointer to an array of 10 ints. int* is a pointer to int. These are different types.

However, an array decays to a pointer to its first element, so you can do:

int a[10];
int(*p)[10] = &a;
int* q = a; // decay of int[10] to int*

But not:

q = p;
p = q;

Function returning a pointer to an int array

2 Answers2

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