Re: Returning a reference to a local variable

From:
pauldepstein@att.net
Newsgroups:
comp.lang.c++
Date:
Sat, 29 Dec 2007 21:31:36 -0800 (PST)
Message-ID:
<456330aa-9f72-4e21-ba35-848af99f31c8@b40g2000prf.googlegroups.com>
On Dec 30, 1:00 pm, Salt_Peter <pj_h...@yahoo.com> wrote:

On Dec 29, 10:40 pm, pauldepst...@att.net wrote:

#include <iostream>
using namespace std;

double & GetWeeklyHours()
{
    double h = 46.50;
    double &hours = h;
    return hours;}

//----------------------------------------------------------------------=

---=AD--

int main()
{
    double hours = GetWeeklyHours();

    cout << "Weekly Hours: " << hours << endl;

    return 0;

}

According to a (hopefully reliable) website, the above is correct
code.

Why is the above _not_ an example of the sin of "returning a reference
to a local variable"? What is the difference between the return-
reference-to-local problem and the above code?


It is an example of undefined behaviour. A compiler is not required to
generate a diagnostic either.
Is it accepteable? Not in a long shot.
Here, try the following and pay attention to the output and sequence
of events.

#include <iostream>

class Hours
{
  double m_d;
public:
  Hours() : m_d(0.0) { std::cout << "Hours()\n"; }
  Hours(double d) : m_d(d) { std::cout << "Hours(double)\n"; }
  ~Hours() { std::cout << "~Hours()\n"; }
  Hours(const Hours& copy)
  {
    std::cout << "Hours(const Hours& copy)\n";
    m_d = copy.m_d;
  }
  double get() const { return m_d; }

};

Hours& GetWeeklyHours()
{
  Hours h = 46.50;
  std::cout << "local initialized\n";
  Hours& hours = h;
  std::cout << "reference set\n";
  return hours;

}

//------------------------------------------------------------------------=

-=AD--

int main()
{
  Hours hours = GetWeeklyHours(); // is a copy (1)

  std::cout << "Weekly Hours: " << hours.get() << std::endl;

}

/*
Hours(double)
local initialized
reference set
~Hours() // local destroyed here
Hours(const Hours& copy) // is a copy (1) of a reference
Weekly Hours: 6.95329e-310
~Hours()
*/

The basic rule of thumb is: if a local invokes a default or
parametized ctor, not a copy, it only lives in that scope.
It doesn't matter whether you use a reference, a 'reference to const'
or a 'pointer to const', the residual garbage left over from the
destruction of the local variable can't be guarenteed. Anything can
happen.

this is fine, btw:
Hours GetWeeklyHours()
{
  return Hours(46.5);

}

but this is not:
Hours const& GetWeeklyHours()
{
  return Hours(46.5);

}- Hide quoted text -

- Show quoted text -- Hide quoted text -

- Show quoted text -


Thanks, Peter. You seem to have put a lot of time and effort towards
helping me and I appreciate that. However, I'm still a bit confused.
I understand why your example code is bugged. The mystery (to me) is
why the code I posted originally _does not_ appear to suffer from the
reference-to-local bug. It works on my compiler, and was copied from
what seemed like a decent website. Are you saying that my original
code suffers from the same bug but that sometimes the local is
destroyed at a late enough time so that the original code still gives
correct results? My question is: Why does the original code give
reliable results even though it appears to return a reference to a
local?

Paul Epstein

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