Templates, copy constructor and operator<<

Hello,

I have a problem with using a copy constructor to convert an object of a
templated class to object of another templated class. Let me first
include the code (my question is below):

<code:templates.h>
#include <string>
#include <iostream>

using namespace std;

template<class T> class tempo;

template<class T>
class delo
{
private:
tempo<T> a;
tempo<T> b;
public:
delo () : a(0), b(0) {}
delo ( T c, T d ) : a(c), b(d) {}

void evaluate( tempo<T>& c );
};

template<class T> void printValue( tempo<T> );
template<class T> ostream& operator<< ( ostream&, const tempo<T>& );

template <class T>
class tempo
{
private:
T value;

public:
tempo () : value(0) {}
tempo ( T init ) : value(init) {}
tempo ( delo<T> d ) { d.evaluate(*this); }

friend ostream& operator<< <T>( ostream&, const tempo& );
void compute( tempo& a, tempo& b );
};

template class delo<int>;
template class tempo<int>;
</code:templates.h>

<code:templates.cpp>
#include "templates.h"
#include <iostream>

template <class T>
void delo<T>::evaluate( tempo<T>& c )
{
c.compute( a, b );
}

template <class T>
void tempo<T>::compute( tempo<T>& a, tempo<T>& b )
{
value = a.value + b.value;
}

template <class T>
ostream& operator<< ( ostream& o, const tempo<T>& t )
{
o << "current value = " << t.value << endl;
return o;
}

template ostream& operator<< ( ostream&, const tempo<int>& );
</code:templates.cpp>

The class "delo" is used to delay the computation (compute, in this
example an addition) until it needs to be executed, e.g. when a left-side
object is given or when the result is requested for instance using cout.
In this case, the "delo" object should be converted to a "tempo" object
using the copy constructor, which calls the evaluate function, in which
the compute function is actually called.

So far, so good ... now look at a test program:

<code:test.cpp>
#include "templates.h"
#include <iostream>

using namespace std;

int main ()
{
tempo<int> anInt( 123 );
cout << anInt << endl;

delo<int> aDelInt( 789, 211 );
cout << aDelInt << endl;
}
</code:test.cpp>

The first cout line compiles without any problem, but the second cout line
results in a long error message, which goes like:

<error>
g++ -c test.cpp -o test.o
test.cpp: In function `int main()':
test.cpp:19: no match for `std::ostream& << delo<int>&' operator
/usr/include/c++/3.2.2/bits/ostream.tcc:55: candidates are:
std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT,
_Traits>::operator<<(std::basic_ostream<_CharT,
_Traits>&(*)(std::basic_ostream<_CharT, _Traits>&)) [with
_CharT = char,
_Traits = std::char_traits<char>]
/usr/include/c++/3.2.2/bits/ostream.tcc:77:
std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT,
_Traits>::operator<<(std::basic_ios<_CharT,
_Traits>&(*)(std::basic_ios<_CharT, _Traits>&)) [with _CharT =
char, _Traits
= std::char_traits<char>]
/usr/include/c++/3.2.2/bits/ostream.tcc:99:
std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT,
_Traits>::operator<<(std::ios_base&(*)(std::ios_ba se&)) [with
_CharT = char,
_Traits = std::char_traits<char>]
/usr/include/c++/3.2.2/bits/ostream.tcc:171:
std::basic_ostream<_CharT, _Traits>& std::basic_ostream<_CharT,
_Traits>::operator<<(long int) [with _CharT = char, _Traits =
std::char_traits<char>]
[...]
/usr/include/c++/3.2.2/ostream:211:
std::basic_ostream<_CharT,
_Traits>& std::operator<<(std::basic_ostream<_CharT, _Traits>&,
char) [with
_CharT = char, _Traits = std::char_traits<char>]
make: *** [test.o] Error 1
</error>

This says it cannot output the "delo" object, while it could copy this
object to a "tempo" object using the copy constructor. But it seems like
it does not know about this copy constructor, or that it cannot use it.
Copying it explicitly (that is, using "cout << tempo<int>( aDelInt )" in
test.cpp does work.

If I use non-templated versions of these classes, it works fine and it
calls the copy constructor like it should.

So my questions are: why does this give an error, and why is it not using
the copy constructor? Are there any limitations? And if so, is there
another way to solve this (without having to do it myself explicitly)?

-- Thanks for any reply.

Franky Backeljauw
University of Antwerp
Department of Mathematics and Computer Science

re: Templates, copy constructor and operator<<

<franky.backeljauw@ua.ac.be> wrote in message
news:Pine.GSO.4.50.0307040128160.6859-100000@hmacs.ruca.ua.ac.be...[color=blue]
> Hello,
>
> I have a problem with using a copy constructor to convert an object of a
> templated class to object of another templated class. Let me first
> include the code (my question is below):
>[/color]

// Snip a bunch_a_code...
[color=blue]
>
> So far, so good ... now look at a test program:
>
> <code:test.cpp>
> #include "templates.h"
> #include <iostream>
>
> using namespace std;
>
> int main ()
> {
> tempo<int> anInt( 123 );
> cout << anInt << endl;
>
> delo<int> aDelInt( 789, 211 );
> // cout << aDelInt << endl;[/color]

cout << (tempo <int>) aTemInt << endl;
[color=blue]
> }
>[/color]
// Snip. a bunch_a_error_message
[color=blue]
> Copying it explicitly (that is, using "cout << tempo<int>( aDelInt )" in
> test.cpp does work.[/color]

cout << (tempo <int>) aTemInt << endl;

Yields...

current value = 123

current value = 1000

That what you wanted??

Personally, I wouldn't recommend this or any other cast here...
[color=blue]
> If I use non-templated versions of these classes, it works fine
> and it calls the copy constructor like it should.[/color]
[color=blue]
> So my questions are: why does this give an error, and why is it
> not using the copy constructor? Are there any limitations?
> And if so, is there another way to solve this (without having to
> do it myself explicitly)?[/color]

That is what I'd do...

const tempo <int> aTemInt (aDelInt);
cout << aTemInt << endl;
[color=blue]
>
> -- Thanks for any reply.
>[/color]

HTH
[color=blue]
> Franky Backeljauw
> University of Antwerp
> Department of Mathematics and Computer Science[/color]

Regards

Brian

re: Templates, copy constructor and operator<<

On Thu, 3 Jul 2003, Brian MacBride wrote:
[color=blue]
> cout << (tempo <int>) aTemInt << endl;
>
> Yields...
>
> current value = 123
>
> current value = 1000
>
> That what you wanted??
>
> Personally, I wouldn't recommend this or any other cast here...
>
> That is what I'd do...
>
> const tempo <int> aTemInt (aDelInt);
> cout << aTemInt << endl;[/color]

Actually, I was wondering why "cout << aDelInt << endl;" does not work. I
have a copy constructor that can copy from "delo" to "tempo", but it does
not get called automatically. This means I cannot output a "delo" object,
although it works when these classes are not templated.

So my guess is that it has something to do with templates again ... doh.

-- Regards,

Franky Backeljauw
University of Antwerp
Department of Mathematics and Computer Science

re: Templates, copy constructor and operator<<

franky.backeljauw@ua.ac.be wrote in message news:<Pine.GSO.4.50.0307040128160.6859-100000@hmacs.ruca.ua.ac.be>...[color=blue]
> Hello,
>
> I have a problem with using a copy constructor to convert an object of a
> templated class to object of another templated class. Let me first
> include the code (my question is below):[/color]
[color=blue]
> template<class T> class tempo;
>
> template<class T>
> class delo
> { /* snip */
> public:
> delo () : a(0), b(0) {}
> delo ( T c, T d ) : a(c), b(d) {}
>
> void evaluate( tempo<T>& c );
> };[/color]
[color=blue]
> template <class T>
> class tempo
> { /* snip */
> public:
> tempo () : value(0) {}
> tempo ( T init ) : value(init) {}
> tempo ( delo<T> d ) { d.evaluate(*this); }
> };
> /*snip/[/color]
[color=blue]
> The class "delo" is used to delay the computation (compute, in this
> example an addition) until it needs to be executed, e.g. when a left-side
> object is given or when the result is requested for instance using cout.[/color]

This all looks simple and reasonable. However, the compiler can't assume
things are this simple. There might be template specializations of tempo
(e.g. when T==float: tempo<float>) where there is an additional ctor
(e.g. tempo( delo<double> d ) ). Now if the compiler has a delo<double>
it would be ambiguous which tempo<T> to create. Worse, unless the compiler
instantiates delo<T> for all Ts, it doesn't actually know which delo<T>
instances have a delo<double>. And "all Ts" is an infinitely large set.

Therefore the compiler isn't required or even allowed to take implicit
template instantiations into account when doing implicit casts. As
pointed out, if you specify the type T, it will succeed.

Regards,
--
Michiel Salters

re: Templates, copy constructor and operator<<

<franky.backeljauw@ua.ac.be> wrote in message
news:Pine.GSO.4.50.0307040958360.2668-100000@hmacs.ruca.ua.ac.be...[color=blue]
> On Thu, 3 Jul 2003, Brian MacBride wrote:
>[color=green]
> > cout << (tempo <int>) aTemInt << endl;
> >
> > Yields...
> >
> > current value = 123
> >
> > current value = 1000
> >
> > That what you wanted??
> >
> > Personally, I wouldn't recommend this or any other cast here...
> >
> > That is what I'd do...
> >
> > const tempo <int> aTemInt (aDelInt);
> > cout << aTemInt << endl;[/color]
>
> Actually, I was wondering why "cout << aDelInt << endl;" does not work. I
> have a copy constructor that can copy from "delo" to "tempo", but it does
> not get called automatically. This means I cannot output a "delo" object,
> although it works when these classes are not templated.
>[/color]

Well, I think it would be a bit of a stretch for the compiler to know
exactly what you intended without some guidance. It could reasonably assume
that you wanted to convert delo <onething> to tempo <whatever> ;-).
[color=blue]
> So my guess is that it has something to do with templates again ... doh.
>
> -- Regards,
>
> Franky Backeljauw
> University of Antwerp
> Department of Mathematics and Computer Science[/color]

Regards,

Brian