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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.

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  • 1
    I think you are confuse between array name = a , int*and address of array &a, int(*)[5] read lined answer may help you. Commented Jul 26, 2013 at 13:11
  • 4
    @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) Commented Jul 26, 2013 at 13:11
  • 1
    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] Commented Jul 26, 2013 at 13:14
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    @alin let me know does my answer helps, ask your doubt to my answer.. feel free to ask .. Commented Jul 26, 2013 at 14:02
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    @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. Commented Jul 26, 2013 at 14:51

2 Answers 2

Reset to default
3

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.

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5 Comments

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 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.
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 Ok no problem, You can vote all, but can accept only one that the reason ?I give a like to you.
@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 )
2

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;
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