Why does this function in C++ work, but only if printed to stdout?

Question

I'm trying to understand how references/pointers/dereferencing works in c++. Please see below for some example code :

#include <iostream>
#include <cstdio>

int& plusThree(int num)
{
    int threeMore = num + 3;
    //std::cout << "threeMore is " << threeMore << "\n";
    //printf("threeMore is %d", threeMore);
    return threeMore;
}


int main()
{
    int three = plusThree(0);
    std::cout << "three is " << three << "\n";
    return 0;
}

The function plusThree() shouldn't work, and if you run the code, it doesn't. It'll return three is 0. However, if you uncomment either of the lines which prints threeMore, main will now print three is 3...

So my questions are as follows:

Why does the function plusThree() not work? is it because int& means it should return an integer reference, but return threeMore is returning an int?
Why does the function plusThree() now work, if one of std::cout or printf is uncommented?

Thanks!

Not all invalid code in c++ is required to crash or fail to c compile. There are many types of errors that simply lead to undefined behavior, which means that anything can happen. "Anything" includes "appearing to work as intended". — François Andrieux, May 10 '19 at 14:12
Related: https://stackoverflow.com/questions/4643713/c-returning-reference-to-local-variable — drescherjm, May 10 '19 at 14:13
Change `int& plusThree(int num)` to `int plusThree(int num)` — Jesper Juhl, May 10 '19 at 14:13
there is no guarantee to get a wrong result when you do something wrong. — 463035818_is_not_an_ai, May 10 '19 at 14:13
The duplicate answers question 1. Question 2 relies on the false assumption that uncommenting the output expressions makes the function work. It doesn't, it only changes how the Undefined Behavior is expressed. — François Andrieux, May 10 '19 at 14:14

score 0 · Answer 1 · edited May 10 '19 at 15:47

As @Retired Ninja mentioned, returning a reference to a local variable from a function has undefined behavior...

The int threeMore; should be defined outside the function, or it must be returned by value not a reference.

You must change plusThree's implementation, or the int threeMore; variable:

int threeMore;

Return by reference:

int threeMore = 0;

int& plusThree(int num)
{
    threeMore = num + 3;
    //std::cout << "threeMore is " << threeMore << "\n";
    //printf("threeMore is %d", threeMore);
    return threeMore;
}

Return by value:

int plusThree(int num)
{
    int threeMore = num + 3;
    //std::cout << "threeMore is " << threeMore << "\n";
    //printf("threeMore is %d", threeMore);
    return threeMore;
}

Why does this function in C++ work, but only if printed to stdout?

1 Answers1