C++ Programming - References - Discussion

It would be nice if s/o with access to modify the answer would correct it, because you can return a reference to a global variable and can not return a reference to a local variable.

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]

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;

#include<iostream.h>
int i=9;
int& myfunc()
{
return i;
}
int main() {
int g=myfunc();
cout<<g<<endl;
return 0;
}

Here, we can return global variable by reference.

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 not return a global variable by reference.

We can return a local variable by reference.

You CAN do both in the sense that it syntactically allowed, but returning a local variable by reference yields a program that has undefined runtime behavior. So the point is what is meant by 'CAN'.

@Pravee. Already proved that 2 is correct. Is there someone who knows why 1 is incorrect?

We can Return a global variable, but not a local as a local variable goes out of scope once a function call is over.

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;
}