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C++ reference question

Posted on 2014-12-17
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Last Modified: 2014-12-18
http://www.tutorialspoint.com/cplusplus/cpp_references.htm
says:

Once a reference is initialized to an object, it cannot be changed to refer to another object. Pointers can be pointed to another object at any time.

int main()
{
    int n = 56;
    int m=34;
     int &ref_n = n;
     ref_n=m;

    cout<<ref_n<<endl;
    return 0;
}

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But this code ouputs 34, why not 56?
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Comment
Question by:Nusrat Nuriyev
6 Comments
 
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Expert Comment

by:Rgonzo1971
ID: 40506474
HI,

Shouldn't it be

int& ref_n = n;

Regards
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Accepted Solution

by:
Tej Pratap Shukla ~Dexter earned 167 total points
ID: 40506499
hello..
 
 &ref_n means setting ref_n's ADDRESS to n's ADDRESS means ref_n is referring to n.

 int &ref_n = n;

means value of n is changed with the value of ref_n.

and ref_n = m

m= 34

That's why output is 34 .
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LVL 33

Assisted Solution

by:sarabande
sarabande earned 167 total points
ID: 40506677
to see that Dexter's answer is correct, check the value of n after the assignment. it is also 34. cause ref_n is only an alias of n.

Sara
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LVL 68

Expert Comment

by:Qlemo
ID: 40506694
The reference is constant, not the value it refers to.
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Expert Comment

by:sarabande
ID: 40506712
Shouldn't it be  int& ref_n = n;
the position of '&' is arbitrary at any place between type and name. same applies for the asterisk * for a pointer variable.

putting it to the variable is more a c like attitude. putting it to the type is c++-like and shows better that you are well aware of the differences between a reference or pointer type to a normal type. I personally prefer the middle way:

int & ref_n = n;

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Sara
0
 
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Assisted Solution

by:HooKooDooKu
HooKooDooKu earned 166 total points
ID: 40506930
It might help to think of it this way...

'Under the hood', a reference is really just a pointer.  But the language automatically dereferences the pointer for you so that you treat it like a normal variable.

You have to also understand the difference in ...
  int &ref_n = n
... which sets what ref_n points to and ...
  ref_n = m
... which is now just treating ref_n as a normal variable.  It's just that ref_n and n are THE same variable.

Again, it might help to see the difference like this...
  &ref_n = n  //<-- Sets a pointer value
... which is setting what ref_n points to and ...
  ref_n = m  //<-- Set the value of what the pointer points to
... which is setting the value of ref_n references.

So ...
  &ref_n = n    
and
  ref_n = m
... are sort of analogus to
  p = &n;
and
  *p = n;
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