C++ Programming - References - Discussion

Correct option is D:

It is perfectly valid & fine to return a reference to the global variable.

#include <iostream>
int s=9;
int& foo()
{
return s;
}
int main()
{
std::cout<<"value of s = "<<s<<'\n';
foo()=3;
std::cout<<"value of s = "<<s<<'\n';
}

It is syntactically valid to return a reference to a local variable but it doesn't make any sense to return a reference a local variable. Because the local variable is destroyed when execution of function finishes. So the reference to such variable is invalid & results in undefined behavior in C++.

Consider following program:

#include <iostream>
int& foo()
{
int s(9);
return s;
}
int main()
{
std::cout<<foo()<<'\n';
}

Compiler shows only warning when compiling above program.
[Warning] reference to local variable 's' returned [-Wreturn-local-addr]

Option C is correct.

We can return a global variable by reference.
If you workout a simple program you will get the answer !

And,
We cannot return a local variable by reference.
If you do so, you will get a warning.. which is a no right?

#include<iostream>
using namespace std;

int& a(){
int b=10;
return b;
}

int main(){
a();
}

bix.cpp [Warning] reference to local variable 'b' returned [-Wreturn-local-addr]

We can return global variable by reference.

But can not return a local variable by reference.

The code written by @Pravee is working fine because after returning from the function, the value goes out of scope but that memory is not allocated by some other variable. So we are able to print proper value. But if that memory will be allocated by some other variable, then we will not able to get the correct value.

We can return a global variable by reference and also we can return local variable by reference (it is not a language error), the user/ programmer has to take care not to return local variable by reference as it might not be there on the stack when he later uses it.

So, if we simply take what is allowed, at least by the compiler, the answer would be A.

Only A should be correct.

Otherwise they should specify what they mean by "can't", because you can actually return a local variable by reference, it's a potential bug, but you "can" do it you won't get any crash or compilation error by doing it.

And returning a reference to a global variable is totally legal.

Here local value is returned by reference so global and local both can be returned by reference.

int main()
{
float grade();
float gpa = grade();
cout<<gpa;
return 0;
}

float grade()
{
float gpa = 3.5;
float &a=gpa;
return a;
}

The output is 3.5.

int myGlobal = 1978;
int& returnGlobal()
{
return myGlobal;
}
int &myr = returnGlobal();
cout << "Global reference value=" << myr << endl;

It prints 1978 on the screen.
So returning global reference is possible.

You cannot return reference of local variable, by return statement function ends and memory attached to the local variable will be deleted or freed by the compiler. So the reference does not have anything to refer local variable is already deleted.

Why is 1st wrong??


int& ret ()
{
int &d=g;
d=100;cout<<g;
return(d);
}
works fine where g is a global variable

nd in main i can do
ret()+=10;
cout<<"global value"<<g<<endl;

its 110;

But if you see. ith following code works perfectly fine. It displays the value 9.

int& myfunc()
{
int i=9;
return i;
}
int main() {
int g=myfunc();
cout<<g<<endl;
return 0;
}