Template class inside class template in c++

Question

noob here still experimenting with templates. Trying to write a message processing class template

template <typename T> class MessageProcessor {

  //constructor, destructor defined
  //Code using t_ and other functions
foo( void ) {

//More code in a perfectly fine method
}
  private:  T *t_

};

All defined in a header file. I've built and tested my class and all is well. Now, I'm trying to do this:

template <typename T> class MessageProcesor {

  //Same stuff as before

foo(void) {
//Same code as before in foo, but one new line:
  t_->getMessageSender<MessageType>();

}

private: T *t_;
};

However, this line gives me an error of bad expression-type before '>' token.

I've added the necessary header files to define what a MessageType is. I've used this function many time before, just not in this context.

I suspect that the compiler doesn't like the fact that the template function is fully defined (specialized?) within an undefined class template (unspecialized?). I'm not fully grokking what makes a template 'specialized'. Most explanations center on the concepts of 'full' or 'partial', but not what makes it specialized in the first place.

Apologies if you'd like to see more code. I have no internet access at work and that's where I'm doing this, so I have to put everything into my mental 'scratchpad' and bring it home.

Solution 1

Your member function 'foo' needs a return type and you need to use the keyword 'template' when you use member templates in dependent expressions (expressions whose meanings rely directly or indirectly on a generic template parameter)

t_->template getMessageSender<MessageType>();  // ok
t_->getMessageSender<MessageType>(); // not ok

Perhaps this example will help you appreciate when a member template needs to be prefixed by the 'template' keyword [Note: in the interest of symmetry you may always use the 'template' prefix on member templates, but it is optional when used on a non-dependent expression.

struct MyType
{  
  template<class T> void foo() { }
};

template<class U>
struct S
{
  template<class T>
  void bar()
  {
    MyType mt;  // non-dependent on any template parameter
    mt.template foo<int>(); // ok
    mt.foo<int>();  // also ok

    // 't' is dependent on template parameter T
    T t;
    t.template foo<int>();    // ok
    t.foo<int>(); // not ok

    S<T> st; // 'st' is dependent on template parameter T
    st.template foo<int>();    // ok
    st.foo<int>(); // not ok


    S<MyType> s;  // non-dependent on any template parameter
    s.bar<int>(); // ok
    s.template bar<int>(); // also ok

  }

};

Hope that helps.

Solution 2

Add the keyword template between -> and the name of the template method:

t_->template getMessageSender<MessageType>();

Comments

[Edouard A.]

On which compiler do you need to do that?

[Faisal Vali]

@Edouard - The standard requires the 'template' prefix only when accessing member templates from an identifier that depends on a template parameter - most compilers should get this right since this has been part of the standard since '98 - in '03 they modified the rules for the sake of simplicity so that you could use it on all member template accesses - and asides from the EDG compilers (that do mostly get this right), i haven't checked it on any of the other compilers - please let me know if you have.