CODEWITHSUNDEEP
c++variablesreference
kumar_m_kiran
kumar_m_kiran

Reputation: 4022

C++ reference variable clarification

I think most would be surprised about the topic again, However I am referring to a book "C++ Common Knowledge: Essential Intermediate Programming" written by "Stephen C. Dewhurst".

In the book, he quotes a particular sentence (in section under Item 5. References Are Aliases, Not Pointers), which is as below

A reference is an alias for an object that already exists prior to the initialization of the reference. Once a reference is initialized to refer to a particular object, it cannot later be made to refer to a different object; a reference is bound to its initializer for its whole lifetime

Can anyone please explain the context of "cannot later be made to refer to a different object"

Below code works for me,

#include <iostream>
using namespace std;

int main(int argc, char *argv[])
{
 int i = 100;
 int& ref = i;
 cout<<ref<<endl;

 int k = 2000;
 ref = k;
 cout<<ref<<endl;

 return 0;
}

Here I am referring the variable ref to both i and j variable. And the code works perfectly fine.

Am I missing something? I have used SUSE10 64bit linux for testing my sample program.

Thanks for your input in advance.

Upvotes: 1

Views: 272

Answers (3)

fredoverflow
fredoverflow

Reputation: 263118

#include <iostream>

int main()
{
    int i = 100;
    int& ref = i;
    int k = 2000;
    ref = k;

    if (&ref == &i) std::cout << "ref is bound to i\n";
    if (&ref == &k) std::cout << "ref is bound to k\n";
}

Upvotes: 2

Tom
Tom

Reputation: 21108

The line ref = k is not changing what ref is referring to, it is assigning the value of k to the value of ref. If you subsequently print i you will find that i is now 2000 as well.

Upvotes: 5

Mark Rushakoff
Mark Rushakoff

Reputation: 258148

You aren't binding ref to another object: when you do ref = k, it is as though you did i = k, which simply reassigned the value of i. In other words, you're calling i's operator=.

Here's something that better demonstrates the idea:

#include <iostream>

int main()
{
    std::ostream& stream = std::cout;
    stream = std::cerr; // fails to compile!  Cannot access ostream::operator=
    return 0;
}

Upvotes: 11

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