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Template feature

 
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Felipe Magno de Almeida
Guest
PostPosted: Wed Sep 15, 2004 11:57 pm    Post subject: Template feature Reply with quote


is there somehow to make a variable name be a template parameter?
something like this:
template <T>
class Test
{
int T;
};

and then:
int main(void)
{
Test<abc> MyClass;

MyClass.abc = 0;
return 0;
}
Unleast it doesnt compile in my VC++
the message it gives is that abc in the "Test<abc> MyClass" line isnt
defined.
There is other way to make this?
Thanks in advance.

--
Felipe Magno de Almeida
Ciencia da Computacao - Unicamp
[email]felipe.almeida (AT) ic (DOT) unicamp.br[/email] - UIN: 2113442
Cause Rock and Roll can never die.
"if you want to learn something really well, teach it to a computer."
What is Communism?
Answer: "Communism is the doctrine of the conditions of the liberation
of the proletariat." (by Karl Marx)

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Alberto Barbati
Guest
PostPosted: Thu Sep 16, 2004 11:05 am    Post subject: Re: Template feature Reply with quote


Felipe Magno de Almeida wrote:
Quote:
is there somehow to make a variable name be a template parameter?

That's not possible. Template parameters can be:

1) types
2) non-types such as integer, enumeration, pointers or references
3) templates

"names" cannot be used as template parameters. If you have this kind of
doubts I suggest reading a good book about templates, for example "C++
Templates: The complete guide" by Vandevoorde and Josuttis.

Quote:
There is other way to make this?

The only way is by using macro preprocessing, but I wouldn't suggest
that if you care about code readability.

Alberto

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Jonathan Turkanis
Guest
PostPosted: Thu Sep 16, 2004 11:18 am    Post subject: Re: Template feature Reply with quote



"Felipe Magno de Almeida" <felipe.almeida (AT) ic (DOT) unicamp.br> wrote:

Quote:
is there somehow to make a variable name be a template parameter?
something like this:
template <T
class Test
{
int T;
};

and then:
int main(void)
{
Test
MyClass.abc = 0;
return 0;
}

The short answer is no. Below, for fun, I'll suggest three ways to approximate
the above syntax. However, I'm sure they won't satisfy you, and I don't
recommend you actually use them.

The real clue that you are doing something wrong is the template declaration:

template<T>
class Test;

A template parameter must be delared as a type, non-type or template. E.g.

template<typename T>
class Test;

template<int T>
class Test;

template< template
class Test;

You've done none of these. You seem to want to be able to do this:

template<identifier T>
class Test
{
int T;
};

But the language doesn't allow this. Macros are the principle language
constructs which can take identifiers as arguments (See method #3, below).

So, let me ask, what problem are you trying to solve?

Now, for fun.

------------------------------

Method #1. Use enumerators for variable names, and operator[] instead of
operator. ("operator dot"):

enum varname {
cat, dog, ant, rat, horse
};

template<varname Name>
class MyClass {
public:
int& operator[] (varname name)
{
assert(Name == name);
return n;
}
const int& operator[] (varname name) const
{
assert(Name == name);
return n;
}
private:
int n;
};

int main()
{
MyClass<cat> c1;
c1[cat] = 0;

// print value of cat variable:
std::cout << c1[cat] << "n";

// Attempt to access non-existing variable:
c1[rat] = 0; // Assertion failure.

MyClass c2[horse] = 6;

etc.
}

------------------------------

Method #2. Use c-style strings with external linkage for variable names:

template<const char* Name>
class MyClass2 {
public:
int& operator[] (const char* name)
{
assert(strcmp(Name, name) == 0);
return n;
}
const int& operator[] (const char* name) const
{
assert(strcmp(Name, name) == 0);
return n;
}
private:
int n;
};

extern const char cat[] = "cat";
extern const char dog[] = "dog";
extern const char ant[] = "ant";
extern const char rat[] = "rat";
extern const char horse[] = "horse";

int main()
{
MyClass<cat> c1;
c1[cat] = 0;
// or: c1["cat"] = 0;

// print value of cat variable:
std::cout << c1["cat"] << "n";

// Attempt to access non-existing variable:
c1["sloth"] = 0; // Assertion failure.

MyClass c2["horse"] = 6;

etc.
}

------------------------------

Method #3. Macros

#define DEFINE_MYCLASS(varname)
struct MyClass_##varname { int varname; };
#define MYCLASS(varname) MyClass_##varname

DEFINE_MYCLASS(cat)
DEFINE_MYCLASS(dog)

int main()
{
MYCLASS(cat) c1;
c1.cat = 0;
std::cout << c1.cat << "n"; // prints '0'
//c1.rat = 0; // Error!

MYCLASS(dog) c2;
c2.dog = 6;

etc.
}

------------------------

Jonathan




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jakacki
Guest
PostPosted: Thu Sep 16, 2004 11:21 am    Post subject: Re: Template feature Reply with quote
Quote:
is there somehow to make a variable name be a template
parameter? something like this:
template <T
class Test
{
int T;
};

and then:
int main(void)
{
Test
MyClass.abc = 0;
return 0;
}

No, but you can create "data" class:

template <class Tag>
struct Data { int d; };

and "variable names" classes:

struct abc;
struct foo;

then build your class from "data" classes:

class Test : public Data<abc>, public Data<foo>
{
};

and "write/read" your "variables" like this:

Test t;
(Data<abc>&)(t).d = 1;
cout << (Data
Syntax can be made nicer, see Andrei Alexandrescu's "Modern C++ Design"
for more details.

BR
Grzegorz



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Michiel Salters
Guest
PostPosted: Fri Sep 17, 2004 2:00 am    Post subject: Re: Template feature Reply with quote
Felipe Magno de Almeida <felipe.almeida (AT) ic (DOT) unicamp.br> wrote

Quote:
is there somehow to make a variable name be a template parameter?
something like this:
template <T
class Test
{
int T;
};

and then:
int main(void)
{
Test
MyClass.abc = 0;
return 0;
}

Yes, with some problems (on older compilers)

template< template < typename > B >
class Test : public B < int >
{
}

template< typename T >
struct Member_abc {
T abc;
}

Test<Member_abc> MyObject;
MyObject.abc = 0;

Regards,
Michiel Salters

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Felipe Magno de Almeida
Guest
PostPosted: Fri Sep 17, 2004 2:18 am    Post subject: Re: Template feature Reply with quote
Quote:
Syntax can be made nicer, see Andrei Alexandrescu's "Modern C++ Design"
for more details.

That's the one I'm reading.

Quote:

BR
Grzegorz


--
Felipe Magno de Almeida
Ciencia da Computacao - Unicamp
[email]felipe.almeida (AT) ic (DOT) unicamp.br[/email] - UIN: 2113442
Cause Rock and Roll can never die.
"if you want to learn something really well, teach it to a computer."
What is Communism?
Answer: "Communism is the doctrine of the conditions of the liberation
of the proletariat." (by Karl Marx)

[ See http://www.gotw.ca/resources/clcm.htm for info about ]
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Felipe Magno de Almeida
Guest
PostPosted: Fri Sep 17, 2004 2:21 am    Post subject: Re: Template feature Reply with quote
Alberto Barbati wrote:

Quote:
Felipe Magno de Almeida wrote:
is there somehow to make a variable name be a template parameter?

1) types
2) non-types such as integer, enumeration, pointers or references
enumaration isnt considered a type? or you mean an instance of an

enumaration;
Quote:
Alberto

--
Felipe Magno de Almeida
Ciencia da Computacao - Unicamp
[email]felipe.almeida (AT) ic (DOT) unicamp.br[/email] - UIN: 2113442
Cause Rock and Roll can never die.
"if you want to learn something really well, teach it to a computer."
What is Communism?
Answer: "Communism is the doctrine of the conditions of the liberation
of the proletariat." (by Karl Marx)

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Francis Glassborow
Guest
PostPosted: Fri Sep 17, 2004 2:24 am    Post subject: Re: Template feature Reply with quote
In article <Chc2d.5365$2T6.169569 (AT) twister1 (DOT) libero.it>, Alberto Barbati
<AlbertoBarbati (AT) libero (DOT) it> writes
Quote:
Felipe Magno de Almeida wrote:
is there somehow to make a variable name be a template parameter?

That's not possible. Template parameters can be:

1) types
OK
2) non-types such as integer, enumeration, pointers or references
But a reference is a type and I think you meant values of integer type,

enumeration type and pointer type for the rest.


--
Francis Glassborow ACCU
Author of 'You Can Do It!' see http://www.spellen.org/youcandoit
For project ideas and contributions: http://www.spellen.org/youcandoit/projects


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Terje Slettebų
Guest
PostPosted: Fri Sep 17, 2004 3:02 am    Post subject: Re: Template feature Reply with quote
"Felipe Magno de Almeida" <felipe.almeida (AT) ic (DOT) unicamp.br> wrote

Quote:
is there somehow to make a variable name be a template parameter?
something like this:
template <T
class Test
{
int T;
};

and then:
int main(void)
{
Test
MyClass.abc = 0;
return 0;
}

In addition to the suggestions given by the other posters, you might try
tuples in combination with enum, not at least if you want to access more
than one element:

#include <iostream>
#include <string>
#include <boost/tuple/tuple.hpp>
#include <boost/tuple/tuple_io.hpp>

int main()
{
enum { abc, def, ghi };

boost::tuple<int,double,std::string> value;

value.get<abc>()=123;
value.get<def>()=1.23;
value.get<ghi>()="Test";

std::cout << value << "n";
}

This prints: "(123 1.23 Test)"

Regards,

Terje



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Maxim Yegorushkin
Guest
PostPosted: Fri Sep 17, 2004 10:58 am    Post subject: Re: Template feature Reply with quote
Felipe Magno de Almeida <felipe.almeida (AT) ic (DOT) unicamp.br> wrote:

Quote:
is there somehow to make a variable name be a template parameter?
something like this:
template <T
class Test
{
int T;
};

and then:
int main(void)
{
Test
MyClass.abc = 0;
return 0;
}
Unleast it doesnt compile in my VC++
the message it gives is that abc in the "Test<abc> MyClass" line isnt
defined.
There is other way to make this?

It could be helpful if you elaborated what you need this for.

Meanwhile, I've sketched for fun some machinery to allow for the following
syntax:

struct a {};
struct b {};
struct c {};

struct with_named_members
: members_as_hierarchy<mpl::vector<
named_member , named_member<b, int>
, named_member<c, double>
Quote:
::type
{

};

int main(int ac, char** av)
{
with_named_members t;
std::cout
<< typeid(t.member << typeid(t.member << typeid(t.member ;

t.member t.member<b>() = 0xdead;
t.member<c>() = 22/7.0;
}

If it suffices, here is the machinery:

#include "boost/mpl/identity.hpp"
#include "boost/mpl/fold.hpp"
#include "boost/mpl/find_if.hpp"
#include "boost/mpl/same_as.hpp"
#include "boost/mpl/vector.hpp"
#include "boost/mpl/deref.hpp"

namespace mpl = boost::mpl;

template<class Name, class T>
struct named_member
{
typedef Name name;
typedef T type;
};

namespace aux
{

class named_member_base
{
private:
struct no;

protected:
~named_member_base() {}

public:
void member(mpl::identity<no>);
};

struct make_hierarchy_unit
{
template<class T, class Member> struct apply
{
struct type : T
{
using T::member;
typename Member::type& member(mpl::identity<typename
Member::name>) { return member_; }
typename Member::type member_;
};
};
};

template<class Name>
struct same_name
{
template<class T> struct apply : mpl::same_as<Name>::template
apply<typename T::name>
{};
};

} // namespace aux

template<class MemberSeq>
struct members_as_hierarchy
{
class type : mpl::fold<MemberSeq, aux::named_member_base,
aux::make_hierarchy_unit>::type
{
private:
typedef typename mpl::fold<MemberSeq, aux::named_member_base,
aux::make_hierarchy_unit>::type base;
template<class Name> struct type_by_name
{
typedef typename mpl::deref<typename
mpl::find_if::type>::type t;
typedef typename t::type type;
};

public:
template<class Name>
typename type_by_name<Name>::type& member()
{
return this->base::member(mpl::identity<Name>());
}
};
};

--
Maxim Yegorushkin

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Alberto Barbati
Guest
PostPosted: Fri Sep 17, 2004 5:33 pm    Post subject: Re: Template feature Reply with quote
Francis Glassborow wrote:
Quote:
In article <Chc2d.5365$2T6.169569 (AT) twister1 (DOT) libero.it>, Alberto Barbati
[email]AlbertoBarbati (AT) libero (DOT) it[/email]> writes

2) non-types such as integer, enumeration, pointers or references

But a reference is a type and I think you meant values of integer type,
enumeration type and pointer type for the rest.


Sure. I thought it should have been clear from the context, but clearly
it wasn't. Thanks for pointing that out.

Alberto

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