Boost signals - type passed as parameter

Question

I found this C++ code that uses boost signals and I'm trying to understand it.

// A boost::signal wrapper structure 
template <typename Signature>
struct SignalBase : public boost::noncopyable
{
    typedef boost::function_traits< Signature >        SignatureTraits;

    typedef boost::signal<Signature>                   SignalType;
    typedef typename SignalType::slot_function_type    SlotFunctionType;
    typedef typename SignalType::result_type           ResultType;
    typedef boost::signals::connection                 ConnectionType;

    SignalBase() : m_signal() {};
    virtual ~SignalBase() {};

protected:
    SignalType m_signal;
};


// I use a specialization of this template for an arity of 1.
// The template generates the correct function call operator for the arity of the signal.
template<int Arity, typename Signature>
struct SelArity : public SignalBase<Signature> {};

// Specialization
template<typename Signature>
struct SelArity< 1, Signature> : public SignalBase<Signature>
{
    typedef SignalBase<Signature> BaseType;
    inline typename BaseType::ResultType operator()(typename BaseType::SignatureTraits::arg1_type arg )
    {
        return BaseType::m_signal( arg );
    }
};

I can't figure out what SelArity functors will return. As far as I understand m_signal is a type which can declare signals that will be able to connect to functions with Signature signature. How is it possible to have types as parameters?( see return BaseType::m_signal( arg ); ) What is the type represented by ResultType ? And how will I be able to use the object returned by SelArity functor?

Answer 1

No, m_signal is not a type, it's an instance of class SignalType, which is boost::signal<Signature>.

SelArity functor actually invokes m_signal with 1 argument and returns its return value.

(What all these wrappers are needed for I don't know.)