help partial sepecialization

[wakun]

Hi there,
I am learning template programming. When testing the partial
specialization, I have some probelms

Here is a full templated class
template <typename T, int n>
class CT
{
public:
T data[n][100][100];

CT() {...}

const T& get(int k, int i, int j)
{
return data[k][i][j];
}
};

I have a partial specialization verion as follow

template <typename T>
class CT<T, 1>
{
public:
T data[1][100][100];

const T& get(int i, int j)
{
return data[0][i][j];
}
};

In main, I have

int main(void)
{
CT<double, 2> ct1;
CT<double, 1> ct2;

cout << ct1.get(1, 5, 8) << endl; // OK! return what I want
cout << ct2.get(5, 8) << endl; // also OK
cout << ct2.get(0, 5, 8) << endl; // ERROR !!!
}

I wonder how this error come ! Both ct1 and ct2 come from a same class
except the different template parameter, why compiler complian no
get(k, i, j) is defined in partial specialization class? If I want a
partial specialization class with all members available, should I
rewrite all the code ?

Thanks in advance

[puzzlecracker]

wa***@wakun.com wrote:

Hi there,
I am learning template programming. When testing the partial
specialization, I have some probelms

Here is a full templated class
template <typename T, int n>
class CT
{
public:
T data[n][100][100];

CT() {...}

const T& get(int k, int i, int j)
{
return data[k][i][j];
}
};

I have a partial specialization verion as follow

template <typename T>
class CT<T, 1>
{
public:
T data[1][100][100];

const T& get(int i, int j)
{
return data[0][i][j];
}
};

In main, I have

int main(void)
{
CT<double, 2> ct1;
CT<double, 1> ct2;

cout << ct1.get(1, 5, 8) << endl; // OK! return what I want
cout << ct2.get(5, 8) << endl; // also OK
cout << ct2.get(0, 5, 8) << endl; // ERROR !!!
}

I wonder how this error come ! Both ct1 and ct2 come from a same class
except the different template parameter, why compiler complian no
get(k, i, j) is defined in partial specialization class? If I want a
partial specialization class with all members available, should I
rewrite all the code ?

Thanks in advance
They are different classes; in fact very differentcout << ct2.get(0, 5, 8) << endl; // ERROR !!!

you specialized class doesn't have an overload of this method that
takes 3 params.
look at faq for more clarification

[Jonathan Mcdougall]

wa***@wakun.com wrote:

Hi there,
I am learning template programming. When testing the partial
specialization, I have some probelms
If I want a
partial specialization class with all members available, should I
rewrite all the code ?

Yes.
Jonathan

[wakun]

Jonathan Mcdougall wrote:

wa***@wakun.com wrote:

Hi there,
I am learning template programming. When testing the partial
specialization, I have some probelms
If I want a
partial specialization class with all members available, should I
rewrite all the code ?

Yes.

OK. This time I consider inheritance

template <typename T, int n>
class Base
{
public:
T data[n][100][100];

Base() {...}

void show(void)
{
cout << "Hi!" << endl;
}

const T& get(int k, int i, int j)
{
return data[k][i][j];
}

const T& get2(int k, int i, int j)
{
return data[k][i][j];
}

};

template <typename T, int n>
class CT : public Base<T, n>
{
public:
CT() :Base<T, n> {...}
};

template <typename T>
class CT<T, 1> : public Base<T, 1>
{
public:

CT<T,1>() :Base<T, 1> {...}

const T& get2(int i, int j)
{
return data[0][i][j];
}

};

// main
int main(void)
{
CT<double, 2> ct1;
CT<double, 1> ct2;

cout << ct1.get(1, 2, 3) << endl; // OK, of course
cout << ct1.show() << endl; // OK, of course

cout << ct2.show() << endl; // No problem, show is defined
in Base
cout << ct2.get(0, 1, 2) << endl; // OK. get is kept in both
CT<double, n> and CT<double, 1>
cout << ct2.get2(3, 2) << endl; // OK. overload function
ctou << ct2.get2(0, 1, 2) << endl; // ERROR!!! No such function!?
return 0;
}

Why compiler cannot find get2?

[Greg]

wa***@wakun.com wrote:

Hi there,
I am learning template programming. When testing the partial
specialization, I have some probelms

I wonder how this error come ! Both ct1 and ct2 come from a same class
except the different template parameter, why compiler complian no
get(k, i, j) is defined in partial specialization class? If I want a
partial specialization class with all members available, should I
rewrite all the code ?

While you certainly have the partially-specialized class template be
more-or-less a duplicate of the general class template, doing so
creates a maintenance hassle, since the two class templates need to be
maintained in parallel.

A simpler approach is to declare the get method that accepts three
parameters in the general template, but provide an implementation only
in a full specialization:

#include <iostream>

using std::cout;
using std::endl;

template <class T, int N>
class CT
{
public:
T data[N][100][100];

CT() {}

const T& get(int k, int i, int j)
{
return data[k][i][j];
}

const T& get(int i, int j); // not implemented here
};

template <>
const double& CT<double, 1>::get(int i, int j)
{
return data[0][i][j];
}

int main()
{
CT<double, 2> ct1;
CT<double, 1> ct2;

cout << ct1.get(1, 5, 8) << endl; // OK! return what I want
cout << ct2.get(5, 8) << endl; // also OK
cout << ct2.get(0, 5, 8) << endl; // OK
}

Greg