C++ Programming - References - Discussion

I think the option 1 is correct.

A is the correct answer.

#include<iostream>
using namespace std;
int main()
{
int x = 10;
int& ref = x;
ref = 20;
cout<<ref;
int y=40;
ref=y;
cout<<ref;
return 0;
}

Here reference is referred to another variable 'y' after referring 'x'. How 1st option is correct?

When we are creating a reference to a variable we are actually giving it a different name so logically now ref and x are the same so whenever you are modifying ref you are modifying x and not changing what is ref referring to so the line.

ref=20 sets x also to 20 and line.
ref=y sets x equal to the value of y,
ref=y does not mean that ref is now a reference to y,
you can try printing x it will give you 40.

Therefore option 1 is incorrect.

#include<iostream>
using namespace std;
int main()
{
int x = 10;
const int const *ptr=&x;
int& ref = x;
ref = 20;
cout<<"before change ref = "<<ref<<endl;
int y=40;
ref=y;
cout<<"after change ref = "<<ref<<endl;
cout<<"but it actually change x and not change only ref i.e.x = "<<x<<endl;
cout<<"add of x= "<<&x<<endl;
cout<<"add of ref = "<<&ref<<endl;
cout<<"add of ptr = "<<&ptr<<endl;
cout<<"ptr = "<<ptr<<endl;


getchar();
return 0;
}

As you can see,

A reference is not a constant pointer.
as pointer has its own address and ref does not.

@Anu.

Option 4 is correct because reference cannot have its own address. That's why its is not a constant pointer.

One time it hold one variable address only.