Question

Here is a repro case:

#include <iostream>

template< class MessageType >
class Augmented
{
public:
    Augmented( const MessageType& message ) 
        : m_message( message )
    {}

    const MessageType* operator->() const { return &m_message; }

private:
    const MessageType& m_message;
};

template< class MessageType >
Augmented<MessageType> augmented( MessageType&& message )
{
    return Augmented<MessageType>( std::forward<MessageType>(message) );
}

class Test
{
public:
    void print() const {  std::cout << "Hello World" << std::endl; }
};


int main()
{
    Test test;

    auto augmented_test = augmented( test );
    augmented_test->print();

    return 0;
}

I’m using VS2011 (update 1).
When I try to use code that use the -> operator, I get this error:

error C2528: ‘->’ : pointer to reference is illegal

I understand the error, but what I don’t find is how to avoid it in this specific case.
I just need a pointer to the object inferred by the member reference. I tried several different syntaxes which all resulted in the same error.

Any idea how to write this operator correctly?

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Note to keep the focus on the question: I’m voluntarly using a reference to avoid a copy that should never occur in the very specific and isolated context this helper code is used; the question isn’t about the design of the class.

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LAST EDIT: replaced the question code with a full repro case. DeadMG is spot on so I accept his answer. To be more precise: the helper function (augmented()) forward MessageType as Test& instead of Test, which is wrong. There are several ways to fix this, the simplest being to make the helper function not forwarding the type but just take a const MesssageType&.

Answer 1

This code is quite legal for any particular value type. However, I suspect that you have instantiated it with a reference. This would lead to const MessageType* expanding to const (T&)*, which is not legal. You need to check that the template parameter is not a reference.