How to use the template member function of a template in the memberfunction of another template class?

[Peng Yu]

Hi,

I have the following program. The error line is commented. I want to
call _a's member function 'doit' with the template argument of
function<T>. Would you please what is the correct way to do it?

Thanks,
Peng

#include <iostream>

template <typename T>
struct function {
T operator()() const {
return T();
}
};

template <typename T>
struct A {
template <typename F>
T doit() {
return F()();
}
};

template <typename T>
struct B {
public:
T doit() {
return _a.doit<function<T();//error
}
private:
A<T_a;
};
int main() {
A<doublea;
std::cout << a.doit<function<double() << std::endl;
}

[Kai-Uwe Bux]

Peng Yu wrote:

I have the following program. The error line is commented. I want to
call _a's member function 'doit' with the template argument of
function<T>.

[snip]

>
#include <iostream>

template <typename T>
struct function {
T operator()() const {
return T();
}
};

template <typename T>
struct A {
template <typename F>
T doit() {
return F()();
}
};

template <typename T>
struct B {
public:
T doit() {
return _a.doit<function<T();//error

Try:

return _a.template doit< function<T();

}
private:
A<T_a;
};
int main() {
A<doublea;
std::cout << a.doit<function<double() << std::endl;
}

Best

Kai-Uwe Bux

[annamalai]

On Oct 24, 9:54 pm, Peng Yu <[e-mail]:

template <typename T>
struct B {
public:
T doit() {
return _a.doit<function<T();//error
}
private:
A<T_a;

};

The only way I was able to make it work was to make the B::doit() as a
template member function. I don't know enough to explain why the
compiler is not able to instantiate function<Tas used above. Maybe
someone with more knowledge will be able to explain it. Here is the
complete program.

#include <iostream>

template <typename T>
struct function {
T operator()() const {
std::cout << "T function::operator() called" << std::endl;
return T();
}

};

template <typename T>
struct A {
template <typename F>
T doit() {
std::cout << "T A::doit() called" << std::endl;
return F()();
}

};

template <typename T>
struct B {
public:
template <typename P>
T doit() {
std::cout << "T B::doit() called" << std::endl;
return _a.doit<P>();
}
private:
A<T_a;

};

int main() {
A<doublea;
double d = a.doit< function<double();
B<doubleb;
double e = b.doit< function<double();
}

--
Copyright of Wedding Album Must Belong to Family (The Producers)
http://tinyurl.com/d0253082-copyright

[Thomas J. Gritzan]

On Oct 24, 9:54 pm, Peng Yu <[e-mail]:

>template <typename T>
struct B {
public:
T doit() {
return _a.doit<function<T();//error

doit is a dependent name, so you have to tell the compiler, when it is a
typename or template. Otherwise, the compiler assumes a non-typename and
non-template member. Insert the template keyword before the function name:

return _a.template doit<function<T();

The FAQ explains this for the typename keyword, but it also applies to
templates:
http://www.parashift.com/c++-faq-lit...html#faq-35.18

> }
private:
A<T_a;

};

--
Thomas