C++ template and shadow parameter

Question

I have one simple question. What is the reason of compilation failure?

template <class T>
class test
{
    T varGoodForNothing;
public:
    test()
    { }
    test(test<T> & tt)
    {
        varGoodForNothing = tt.varGoodForNothing;
    }
    test<T> & operator=(const test<T> & tt)
    {
        if (this == &tt)
            return *this;
        test(tt);
        return *this;
    }
};

Compiler error is:

declaration of test tt shadows a parameter.

Answer 1

What does the standard say?

The standard says that a declaration of type Type (name) is the same has having used Type name, see the below standard quotation.

[dcl.meaning] / 6

In a declaration T D where D has the form

( D1 )

The type of the contained declarator-id is the same as that of the contained declarator-id in the declaration T D1.

Parentheses do not alter the type of the embedded declarator-id, but they can alter the binding of complex declarators.

^{With that said you are not calling the copy-constructor of test with an argument named tt, instead the compiler thinks that you are trying to declare a variable of type test with the name tt.}

---

How to get around the problem?

To circumvent the problem of T (D); being interpreted as T d; we will have to wrap T inside parentheses, such as in the below.

(test) (tt);

---

^{Note: Even if the code will compile after the proposed change it will not do what you want, nor think, it will.}

^{Instead of calling the copy-constructor of test for the given instance you will declare an anonymous instance of test initialized with the value of tt.}

^{Constructors can only be called from within other constructors (using a member initializer list).}