Returning a reference to a local variable

**Salt_Peter** · Dec 30 '07, 05:05 AM

Re: Returning a reference to a local variable

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

#include <iostream>
using namespace std;
>
double & GetWeeklyHours( )
{
double h = 46.50;
double &hours = h;
return hours;}
>
//---------------------------------------------------------------------------
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(cons t 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;
}

//---------------------------------------------------------------------------
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);
}

**pauldepstein@att.net** · Dec 30 '07, 05:35 AM

Re: Returning a reference to a local variable

On Dec 30, 1:00 pm, Salt_Peter <pj_h...@yahoo. comwrote:

On Dec 29, 10:40 pm, pauldepst...@at t.net wrote:
>
>
>
>
>

#include <iostream>
using namespace std;

>

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

>

//---------------------------------------------------------------------------
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(cons t 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;
>
}
>
//---------------------------------------------------------------------------
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

**Kira Yamato** · Dec 30 '07, 05:35 AM

Re: Returning a reference to a local variable

On 2007-12-29 22:40:31 -0500, pauldepstein@at t.net said:

#include <iostream>
using namespace std;
>
double & GetWeeklyHours( )
{
double h = 46.50;
double &hours = h;
return hours;
}
//---------------------------------------------------------------------------
int main()
{
double hours = GetWeeklyHours( );
>
cout << "Weekly Hours: " << hours << endl;
>
return 0;
}
>
>
According to a (hopefully reliable) website, the above is correct
code.

No. As far as I know, the standard makes no guarantee a behavior for
this code. This is no correct code.

BTW, please give the link to that website.

--

-kira

**pauldepstein@att.net** · Dec 30 '07, 06:05 AM

Re: Returning a reference to a local variable

On Dec 30, 1:31 pm, Kira Yamato <kira...@earthl ink.netwrote:

On 2007-12-29 22:40:31 -0500, pauldepst...@at t.net said:
>
>
>
>
>

#include <iostream>
using namespace std;

>

double & GetWeeklyHours( )
{
double h = 46.50;
double &hours = h;
return hours;
}
//---------------------------------------------------------------------------
int main()
{
double hours = GetWeeklyHours( );

>

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

>

return 0;
}

>

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

>
No. As far as I know, the standard makes no guarantee a behavior for
this code. This is no correct code.
>
BTW, please give the link to that website.
>
--
>
-kira- Hide quoted text -
>
- Show quoted text -

Thanks Kira and Peter

The link is http://www.functionx.com/cpp/example...nreference.htm
The reason I was reluctant to include the link at the beginning is
that I didn't want to be seen as unfairly accusing the people who
designed the website.

Paul Epstein

**Salt_Peter** · Dec 30 '07, 07:05 AM

Re: Returning a reference to a local variable

On Dec 30, 12:31 am, pauldepst...@at t.net wrote:

On Dec 30, 1:00 pm, Salt_Peter <pj_h...@yahoo. comwrote:
>
>
>

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

>

#include <iostream>
using namespace std;

>

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

>

//---------------------------------------------------------------------------
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(cons t 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;

>

}

>

//---------------------------------------------------------------------------
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

First off - you may very well find others to have a different point of
view than mine.
You might even find a compiler with some scheme that makes the above
work with primitives (stored in a register? who knows - i really don't
care).

The fact remains, once the scope ends:
a) the local variable is no more
b) that area of memory is therefore left unprotected (its writeable)

So if you have a busy program (multi-threaded, etc)... its only a
question of time.
Debugging an issue that only happens intermittently is near
impossible. Testing for UB is easy comparatively.

A completely different Issue...
The standard specifically says you can't bind a 'reference to non-
const' to a temporary. Many beleive this means that its ok to *return*
a local through a reference to const. Well, not quite. The standard,
as i understand it, says this:

void DisplayHours(co nst double& ref) // must be a ref to const
{
std::cout << ref << std::endl;
}

int main()
{
DisplayHours(do uble(99));
}

The above works because double(99), a temporary, has a lifetime that
lasts until the function ceases to exist.
That, by the way, is the better way to write your GetWeeklyHours
function, pass the variable by reference_to_co nst which safely
modifies the original. At least thats guarenteed 100% of the time.

Hopefully, a little light was thrown on this and if not, someone else
can give it a go.

**Daniel T.** · Dec 30 '07, 02:05 PM

Re: Returning a reference to a local variable

pauldepstein@at t.net wrote:

#include <iostream>
using namespace std;
>
double & GetWeeklyHours( )
{
double h = 46.50;
double &hours = h;
return hours;
}
//---------------------------------------------------------------------------
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?

As I understand it, the example code works because RVO constructs the
local variable at the site of main's 'hours' variable.

Without RVO, the above code would not work, and RVO cannot be guaranteed.

**Ron Natalie** · Dec 30 '07, 03:55 PM

Re: Returning a reference to a local variable

Daniel T. wrote:

>
As I understand it, the example code works because RVO constructs the
local variable at the site of main's 'hours' variable.
>
Without RVO, the above code would not work, and RVO cannot be guaranteed.

Well more likely, the RVO just causes the copy to the main:hours
variable before the GetWeeklyHours: h is obliterated.

One of those insidious forms of undefined behavior is things appear to
work normally, TODAY.

Returning a reference to a local variable

Returning a reference to a local variable

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