我有一个模板类，它由另一个具有深层层次结构的类参数化。我想传递给另一个基类参数化的函数基模板类。这是示例：
// Base template class test
#include <stdio.h>
// Base classes
template<class T>
class Method {
public:
Method<T> (T * t) {
this->t = t;
}
virtual int solve() = 0;
protected:
T * t;
};
class Abstract {
public:
virtual int get() = 0;
virtual void set(int a) = 0;
};
// Concrete classes, there might be a few of them
template<class T>
class MethodImpl : public Method<T> {
public:
MethodImpl<T> (T * t) : Method<T>(t) {}
int solve() {
return this->t->get() + 1;
}
};
class Concrete : public Abstract {
public:
int get() {
return this->a;
}
void set(int a) {
this->a = a;
}
protected:
int a;
};
// Uses methods of Base classes only
class User {
public:
int do_stuff(Abstract & a, Method<Abstract> & ma) {
a.set(2);
return ma.solve();
}
};
// Example usage
int main () {
Concrete * c = new Concrete();
MethodImpl<Concrete> * mc = new MethodImpl<Concrete>(c);
User * u = new User();
int result = u->do_stuff(*c, *mc);
printf("%i", result);
return 0;
}
我在编译期间收到此错误：
btc.cpp: In function 'int main()':
btc.cpp:62: error: no matching function for call to 'User::do_stuff(Concrete&, MethodImpl<Concrete>&)'
btc.cpp:50: note: candidates are: int User::do_stuff(Abstract&, Method<Abstract>&)
但是，如果我使用相同的逻辑创建局部变量，它就可以正常工作：
int main () {
Abstract * a = new Concrete();
Method<Abstract> * ma = new MethodImpl<Abstract>(a);
a->set(2);
int result = ma->solve();
printf("%i", result);
return 0;
}