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basic question on c++ linking

 
 
Peskov Dmitry
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      06-29-2008
It is a very basic question.Surely i got something wrong in my basic
understanding.

//Contents of file1.cpp

using namespace std;
#include <iostream>

template <typename T>
class my_stack;

int main(){
my_stack<int> st1;

int top_element;
top_element = st1.pop();
cout << "Hello World !! Top = " << top_element << endl;
return 0;
}

The class my_stack is defined in another file. say (template.cpp)

file.o template.o -> project

why does it give linker error, since i provided the class definition
while linking too !!!

../src/file1.cpp: In function ‘int main()’:
../src/file1.cpp:8: error: ‘my_stack’ was not declared in this scope
 
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Rolf Magnus
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      06-29-2008
Peskov Dmitry wrote:

> It is a very basic question.Surely i got something wrong in my basic
> understanding.
>
> //Contents of file1.cpp
>
> using namespace std;
> #include <iostream>
>
> template <typename T>
> class my_stack;
>
> int main(){
> my_stack<int> st1;
>
> int top_element;
> top_element = st1.pop();
> cout << "Hello World !! Top = " << top_element << endl;
> return 0;
> }
>
> The class my_stack is defined in another file.


It's not a class. It's a class template.

> say (template.cpp)
>
> file.o template.o -> project
>
> why does it give linker error, since i provided the class definition
> while linking too !!!


A class template is not a class. It's a description for the compiler that
tells it how to build the class once it knows the template arguments. But
for that, the compiler must know the template's definition, not just a
declaration.

 
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Peskov Dmitry
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      06-29-2008
Ok ...

It is not working even for a class definition.
 
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Rolf Magnus
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      06-29-2008
Peskov Dmitry wrote:

> Ok ...
>
> It is not working even for a class definition.


What does not work, and how does it not work? Show some code and the error
you get.

 
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Erik Wikström
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      06-29-2008
On 2008-06-29 10:47, Peskov Dmitry wrote:
> Ok ...
>
> It is not working even for a class definition.


Please quote the text you are replying to.

You do not have a link-error but a compile-error because you are trying
to use an undeclared and undefined type (my_stack). To solve the problem
you need to include the header-file of my_stack (which should contain
the class definition) in file1.cpp.

Notice also that if you put the template definition in the header-file
and include it (i.e. rename template.cpp to template.h) you can use the
template class in file1.cpp.

--
Erik Wikström
 
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James Kanze
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      06-29-2008
On Jun 29, 10:26 am, Rolf Magnus <(E-Mail Removed)> wrote:
> Peskov Dmitry wrote:
> > It is a very basic question.Surely i got something wrong in my basic
> > understanding.


> > //Contents of file1.cpp


> > using namespace std;
> > #include <iostream>


> > template <typename T>
> > class my_stack;


> > int main(){
> > my_stack<int> st1;


> > int top_element;
> > top_element = st1.pop();
> > cout << "Hello World !! Top = " << top_element << endl;
> > return 0;
> > }


> > The class my_stack is defined in another file.


> It's not a class. It's a class template.


> > say (template.cpp)


> > file.o template.o -> project


> > why does it give linker error, since i provided the class
> > definition while linking too !!!


> A class template is not a class. It's a description for the
> compiler that tells it how to build the class once it knows
> the template arguments. But for that, the compiler must know
> the template's definition, not just a declaration.


But that doesn't really change anything here. The statement
template< typename T > class my_stack ;
is a declaration, not a definition. The same would be true
without the "template< typename T >" (although it would declare
a class, rather than a class template).

Certain uses require a definition; a simple declaration is not
sufficient. Instantiating a template is one, and defining a
variable with a given type is another. Since he does both in
main, a definition must be available, and he's not made one
available.

--
James Kanze (GABI Software) email:(E-Mail Removed)
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
 
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Peskov Dmitry
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Posts: n/a
 
      06-29-2008
On Jun 29, 3:04*pm, James Kanze <(E-Mail Removed)> wrote:
> On Jun 29, 10:26 am, Rolf Magnus <(E-Mail Removed)> wrote:
>
>
>
> > Peskov Dmitry wrote:
> > > It is a very basic question.Surely i got something wrong in my basic
> > > understanding.
> > > //Contents of file1.cpp
> > > using namespace std;
> > > #include <iostream>
> > > template <typename T>
> > > class my_stack;
> > > int main(){
> > > my_stack<int> st1;
> > > int top_element;
> > > top_element = st1.pop();
> > > cout << "Hello World *!! Top = *" << top_element << endl;
> > > return 0;
> > > }
> > > The class my_stack is defined in another file.

> > It's not a class. It's a class template.
> > > say (template.cpp)
> > > file.o template.o -> project
> > > why does it give linker error, since i provided the class
> > > definition while linking too !!!

> > A class template is not a class. It's a description for the
> > compiler that tells it how to build the class once it knows
> > the template arguments. But for that, the compiler must know
> > the template's definition, not just a declaration.

>
> But that doesn't really change anything here. *The statement
> * * template< typename T > class my_stack ;
> is a declaration, not a definition. *The same would be true
> without the "template< typename T >" (although it would declare
> a class, rather than a class template).
>
> Certain uses require a definition; a simple declaration is not
> sufficient. *Instantiating a template is one, and defining a
> variable with a given type is another. *Since he does both in
> main, a definition must be available, and he's not made one
> available.
>
> --
> James Kanze (GABI Software) * * * * * * email:(E-Mail Removed)
> Conseils en informatique orientée objet/
> * * * * * * * * * *Beratung in objektorientierter Datenverarbeitung
> 9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34


Let me put my question differently ? which one below

a)
class new_class;

or

b)
class new_class
{
int a;
public:
new_class();
};

is a class declaration ?

Because if i include (b) in my code as this one it works.

using namespace std;
#include <iostream>

class new_class
{
int a;
public:
new_class();

};

int main(){
new_class st1;
return 0;
}


I assume class definition includes the function that you are defining
something like
class new_class
{
int a;
public:
new_class();
};

new_class::new_class
{
a = 10;
}

is a class definition ?

Please let me know if my understanding is incorrect ?
 
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Erik Wikström
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Posts: n/a
 
      06-29-2008
On 2008-06-29 13:10, Peskov Dmitry wrote:
> On Jun 29, 3:04 pm, James Kanze <(E-Mail Removed)> wrote:
>> On Jun 29, 10:26 am, Rolf Magnus <(E-Mail Removed)> wrote:
>>
>>
>>
>> > Peskov Dmitry wrote:
>> > > It is a very basic question.Surely i got something wrong in my basic
>> > > understanding.
>> > > //Contents of file1.cpp
>> > > using namespace std;
>> > > #include <iostream>
>> > > template <typename T>
>> > > class my_stack;
>> > > int main(){
>> > > my_stack<int> st1;
>> > > int top_element;
>> > > top_element = st1.pop();
>> > > cout << "Hello World !! Top = " << top_element << endl;
>> > > return 0;
>> > > }
>> > > The class my_stack is defined in another file.
>> > It's not a class. It's a class template.
>> > > say (template.cpp)
>> > > file.o template.o -> project
>> > > why does it give linker error, since i provided the class
>> > > definition while linking too !!!
>> > A class template is not a class. It's a description for the
>> > compiler that tells it how to build the class once it knows
>> > the template arguments. But for that, the compiler must know
>> > the template's definition, not just a declaration.

>>
>> But that doesn't really change anything here. The statement
>> template< typename T > class my_stack ;
>> is a declaration, not a definition. The same would be true
>> without the "template< typename T >" (although it would declare
>> a class, rather than a class template).
>>
>> Certain uses require a definition; a simple declaration is not
>> sufficient. Instantiating a template is one, and defining a
>> variable with a given type is another. Since he does both in
>> main, a definition must be available, and he's not made one
>> available.


Please do not quota signatures.

> Let me put my question differently ? which one below
>
> a)
> class new_class;


This is a declaration, it tells the compiler that there exists a class
called new_class, but nothing else. If all you need is a pointer to the
class a declaration is enough, for just about anything else you need the
definition too.

> b)
> class new_class
> {
> int a;
> public:
> new_class();
> };


This is the definition, it tells the compiler what the new_class is (how
much memory an instance needs, what members it has etc.). Notice that a
definition is also a declaration, you can declare a type as many times
as you want but you can only define it once.

> Because if i include (b) in my code as this one it works.
>
> using namespace std;
> #include <iostream>
>
> class new_class
> {
> int a;
> public:
> new_class();
>
> };
>
> int main(){
> new_class st1;
> return 0;
> }


You can also put the definition of the class in a header-file and incude
that in the .cpp-files that needs it.

> I assume class definition includes the function that you are defining
> something like
> class new_class
> {
> int a;
> public:
> new_class();
> };
>
> new_class::new_class
> {
> a = 10;
> }


No, that is the definition of the member function (the constructor in
this case), and it does not have to be included with the class
definition. Actually it is quite common to separate the class definition
from the definitions of its member-functions like this:

some_class.h:
-------------

class some_class
{
int a;
double b;
public:
some_class();
int getA();
double getB()
};

some_class.cpp:
---------------

#include "some_class.h"

some_class::some_class()
{
a = 1;
b = 2.5;
}

// Actually we should use an initialisation-list instead of setting the
// values of the class-members in the constructor, but that is for
// another day.

int some_class::getA()
{
return a;
}

double some_class::getB()
{
return b;
}

Any file that needs to use some_class can then include the some_class.h
file like this:

main.cpp:
---------

#include <iostream>

#include "some_class.h"

int main()
{
some_class c;
std::cout << c.getA() << "\n" << c.getB() << std::endl;
return 0;
}

--
Erik Wikström
 
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James Kanze
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      06-29-2008
On Jun 29, 1:10 pm, Peskov Dmitry <(E-Mail Removed)> wrote:
> On Jun 29, 3:04 pm, James Kanze <(E-Mail Removed)> wrote:


[...]
> Let me put my question differently ? which one below
>
> a)
> class new_class;


> or


> b)
> class new_class
> {
> int a;
> public:
> new_class();
> };


> is a class declaration ?


Both. The second is also a definition. (A definition is a
declaration, but the reverse is not necessarily true.)

> Because if i include (b) in my code as this one it works.


That's because the way you used the class required a definition.

> using namespace std;
> #include <iostream>


> class new_class
> {
> int a;
> public:
> new_class();
> };


> int main(){
> new_class st1;
> return 0;
> }


> I assume class definition includes the function that you are defining
> something like
> class new_class
> {
> int a;
> public:
> new_class();
> };


A class definition includes the contents (members) of the class.

> new_class::new_class
> {
> a = 10;
> }


> is a class definition ?


No, that's a syntax error.

A class definition is:
class <name> <baseclass-declarations>[opt] { ... } ;
A class declaration (which isn't also a definition) is just:
class <name> ;

--
James Kanze (GABI Software) email:(E-Mail Removed)
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
 
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James Kanze
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      06-29-2008
On Jun 29, 2:03 pm, Erik Wikström <(E-Mail Removed)> wrote:
> On 2008-06-29 13:10, Peskov Dmitry wrote:
> > Let me put my question differently ? which one below


> > a)
> > class new_class;


> This is a declaration, it tells the compiler that there exists
> a class called new_class, but nothing else. If all you need is
> a pointer to the class a declaration is enough, for just about
> anything else you need the definition too.


And references, of course.

Also, you can *declare* (but not define) objects of the type.
Thus, "extern new_class some_object ;" is legal with just a
class declaration. Or, within a class:

class Another
{
new_class illegal; // a non-static member is a definition
static new_class legal ;
// but a static member is just a
// declaration (which means that it
// must be defined elsewhere).
} ;

> > b)
> > class new_class
> > {
> > int a;
> > public:
> > new_class();
> > };


> This is the definition, it tells the compiler what the
> new_class is (how much memory an instance needs, what members
> it has etc.). Notice that a definition is also a declaration,
> you can declare a type as many times as you want but you can
> only define it once.


In general. Classes (and templates and inline functions) are an
exception. You can only define them once in each translation
unit, but you can define them in as many translation units as
you want, as long as the definitions are identical.

[...]
> > I assume class definition includes the function that you are
> > defining something like
> > class new_class
> > {
> > int a;
> > public:
> > new_class();
> > };


> > new_class::new_class
> > {
> > a = 10;
> > }


> No, that is the definition of the member function (the
> constructor in this case),


I don't think so. To be the definition of the constructor, he'd
need to have parentheses after the second new_class.

--
James Kanze (GABI Software) email:(E-Mail Removed)
Conseils en informatique orientée objet/
Beratung in objektorientierter Datenverarbeitung
9 place Sémard, 78210 St.-Cyr-l'École, France, +33 (0)1 30 23 00 34
 
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