Accidently I forgot to return value from function but when I returned reference in function declaration it worked.Why?

Question

#include <iostream>
#include<stdlib.h>
using namespace std;

class test{
    public:
    test(int a):i(a){
    }
    int display();
    private:
        int i;
};

int test::display(){
    i;
}
int main() {
    test obj(10);
    cout<<obj.display();
    return 0;
}

In above case some random value is printed. But when I changed function declaration as :

int& display();

and definition as :

int& test::display(){
    i;
}

It displayed correct value i.e. 10 I don't know why?

It is undefined behaviour. Your code has a bug, it does something unexpected. — juanchopanza, Dec 17 '14 at 17:04
Related to [Why does this C++ snippet compile (non-void function does not return a value)](http://stackoverflow.com/q/20614282/1708801) — Shafik Yaghmour, Dec 17 '14 at 17:22

score 7 · Answer 1 · answered Dec 17 '14 at 17:09

This is undefined behavior, so anything is possible - including a possibility when your code "works" as expected. Your compiler should have issued a warning about this - it is important to treat such warnings as errors, and fix all reported problems before testing your code.

Compilers use stack or CPU registers to return values from functions. When a return is missing, no data is placed in the space of the return value. However, the data that you planned to return may already be in a register or on the stack in just the right place, so the calling code exhibits the behavior you expect. It remains undefined, though.

Accidently I forgot to return value from function but when I returned reference in function declaration it worked.Why?

1 Answers1