Question

I have a method foo in class C which either calls foo_1 or foo_2.
This method foo() has to be defined in C because foo() is pure virtual in BaseClass and I actually
have to make objects of type C. Code below:

template <class T>
class C:public BaseClass{

  void foo() {
    if (something()) foo_1;
    else foo_2;

  }
  void foo_1() {
  ....
  }

  void foo_2() {
    ....
     T t;
     t.bar(); // requires class T to provide a method bar() 
     ....
   }
};

Now for most types T foo_1 will suffice but for some types foo_2 will be called
(depending on something()). However the compiler insists on instantiating both foo_1
and foo_2 because either may be called.

This places a burden on T that it has to provide
a bar method.

How do I tell the compiler the following:

• if T does not have bar(), still allow it as an instantiating type?

Answer 1

you could use boost.enable_if. something like this:

#include <boost/utility/enable_if.hpp>
#include <iostream>

struct T1 {
    static const bool has_bar = true;
    void bar() { std::cout << "bar" << std::endl; }
};

struct T2 {
    static const bool has_bar = false;
};

struct BaseClass {};

template <class T>
class C: public BaseClass {
public:
    void foo() {
        do_foo<T>();
    }

    void foo_1() {
        // ....
    }

    template <class U>
    void foo_2(typename boost::enable_if_c<U::has_bar>::type* = 0) {
        // ....
        T t;
        t.bar(); // requires class T to provide a method bar() 
        // ....
    }

private:

    bool something() const { return false; }


    template <class U>
    void do_foo(typename boost::enable_if_c<U::has_bar>::type* = 0) {
        if (something()) foo_1();
        else foo_2<U>();
    }

    template <class U>
    void do_foo(typename boost::disable_if_c<U::has_bar>::type* = 0) {
        if (something()) foo_1();
        // I dunno what you want to happen if there is no T::bar()
    }
};

int main() {
    C<T1> c;
    c.foo();
}