Can I use template typedef without clarification

Question

I have BadArray class in Array class for exeptions.

template <size_t n, typename T>
class Array<n, T>::BadArray {
private:
    string _reason; 
    size_t _index; 
public: 
    BadArray(const string& reason = "", const size_t index = 0); 
    ~BadArray();  
    const string reason() const; 
    const size_t index() const;
};

And this operator<<

template <size_t n, typename T>
ostream& operator<<(ostream& os, const typename Array<n, T>::BadArray& seq) {
    os << seq.reason() << " " << seq.index();
    return os;
}

So does exist some way to use BadArray in try catch without clarification of size_t and typename ?

int main()
{
    try {
        Array<3, double> array2{};
        array2[3] = 4;
    }
    catch (Array<size_t n, typename T>::BadArray ba) {
        cout << ba;
    }
}

Answer 1

A simple solution is to define the class outside of the template:

class BadArray {
    // ...
};

// ...
catch (BadArray& ba)

P.S. As shown in the example, I recommend catching a reference to avoid unnecessary copying.

P.P.S I recommend following the common convention of inheriting the std::exception hierarchy. With the inheritance and the virtual functions that come with it, it will also become important to catch reference so that you don't accidentally slice the base off a derived exception.