Returning const reference to local variable from a function

aJ. picture aJ. · Sep 23, 2009 · Viewed 23.9k times · Source

I have some questions on returning a reference to a local variable from a function:

class A {
public:
    A(int xx)
    : x(xx)
    {
        printf("A::A()\n");
    }
};

const A& getA1()
{
    A a(5);
    return a;
}

A& getA2()
{
    A a(5);
    return a;
}

A getA3()
{
    A a(5);
    return a;
}

int main()
{
    const A& newA1 = getA1(); //1
    A& newA2 = getA2(); //2
    A& newA3 = getA3(); //3
}

My questions are =>

  1. Is the implementation of getA1() correct? I feel it is incorrect as it is returning the address of a local variable or temporary.

  2. Which of the statements in main (1,2,3) will lead to undefined behavior?

  3. In const A& newA1 = getA1(); does the standard guarantee that a temporary bound by a const reference will not be destroyed until the reference goes out of scope?

Answer

Richard Corden picture Richard Corden · Sep 23, 2009

1. Is getA1() implementation correct ? I feel it is incorrect as it is returning address of local variable or temporary.

The only version of getAx() that is correct in your program is getA3(). Both of the others have undefined behaviour no matter how you use them later.

2. Which of the statements in main ( 1,2,3) will lead to undefined behavior ?

In one sense none of them. For 1 and 2 the undefined behaviour is as a result of the bodies of the functions. For the last line, newA3 should be a compile error as you cannot bind a temporary to a non const reference.

3. In const A& newA1 = getA1(); does standard guarantees that temporary bound by a const reference will not be destroyed until the reference goes out of scope?

No. The following is an example of that:

A const & newConstA3 = getA3 ();

Here, getA3() returns a temporary and the lifetime of that temporary is now bound to the object newConstA3. In other words the temporary will exist until newConstA3 goes out of scope.