Returning Address with Return Statement?

Code:

---

class array
{
 int arr[3];
 public:
 int &access(int n)
 {


  return arr[n];
 }     
      };
int main()
{
array a;
for(int i=0;i<3;i++)
cin>>a.access(i);
for(int i=0;i<3;i++)
cout<<a.access(i)<<endl;
cin.get();
}

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My question is "Is return arr[n] returning the address of arr[n]?",If so then why does the cout print the values rather than address?

Quote:

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Originally Posted by Waleed Mujeeb

Is return arr[n] returning the address of arr[n]?

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No, it returns a reference to arr[n] since the return type of the function is int&.

If that is the case then why does scanf("%d",a.access(i)); not work and scanf("%d",&a.access(i)); works ?

Quote:

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Originally Posted by Waleed Mujeeb

If that is the case then why does scanf("%d",a.access(i)); not work and scanf("%d",&a.access(i)); ?

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Because scanf expects an int* argument to correspond to the %d format specifier.

Are you aware of the differences between C++ references and pointers?

Thanks, I messed up addressess and References. Need to check em out.

Btw, avoid scanf in C++. Use std::cin.