Re: template overloading of operator <<

From:
=?ISO-8859-1?Q?Daniel_Kr=FCgler?= <daniel.kruegler@googlemail.com>
Newsgroups:
comp.lang.c++.moderated
Date:
Tue, 7 Jul 2009 12:55:00 CST
Message-ID:
<c5df1bf9-0650-4872-9e45-6556f9d170ec@t21g2000yqi.googlegroups.com>
On 7 Jul., 16:02, Vicky <mail....@gmail.com> wrote:

#include<iostream>
using namespace std;

template<class T>
class A
{
        public:
                enum Status
                {
                        SUCCESS = 10000,
                        FAILURE = 10001,
                        UNKNOWN = 10002
                };
};

typedef A<int> IntClass;
typedef A<string> StringClass;

template<class T>
inline std::ostream & operator<<(std::ostream & s, typename
A<T>::Status const & state)


This operator is unusable without providing an explicit
template parameter, because typename A<T>::Status
is an non-deducible context. The reason is simple:
The compiler cannot extract T from the argument.

{
        switch (state)
        {
                case A<T>::SUCCESS: return s << "SUCCESS";
                case A<T>::FAILURE: return s << "FAILURE";
                case A<T>::UNKNOWN: return s << "UNKNOWN";
        }
}


You should definitively add a default case for the switch, because
a user can create an enumeration value (e.g. UNKNOWN + FAILURE)
that has a different value as any of the currently specified ones.

int main()
{
        IntClass::Status state = IntClass::SUCCESS;
        cout << "Status " << state << endl;}

============================================================================

Output of the above program:
Status 10000

While the expected output was:
Status SUCCESS

the overloaded operator << was not invoked.
Could anyone please suggest some solution for the above?


Two approaches come into my mind, both ensure that
the IO inserter (aka operator<<) will be a non-template
function:

1) Move the inserter as an inline friend into template A like this:

template<class T>
class A
{
public:
    enum Status
    {
        SUCCESS = 10000,
        FAILURE = 10001,
        UNKNOWN = 10002
    };

    friend std::ostream & operator<<(std::ostream & s,
        Status const & state)
    {
        switch (state)
        {
        case A<T>::SUCCESS: return s << "SUCCESS";
        case A<T>::FAILURE: return s << "FAILURE";
        case A<T>::UNKNOWN: return s << "UNKNOWN";
        }

    }
};

2) Move the enum into a non-template base class (or namespace)
and provide a free IO inserter like this:

struct B {
    enum Status
    {
        SUCCESS = 10000,
        FAILURE = 10001,
        UNKNOWN = 10002
    };
};

inline std::ostream & operator<<(std::ostream & s,
   B::Status const & state)
{
    switch (state)
    {
    case B::SUCCESS: return s << "SUCCESS";
    case B::FAILURE: return s << "FAILURE";
    case B::UNKNOWN: return s << "UNKNOWN";
    }
}

template<class T>
class A : public B
{
public:
};

HTH & Greetings from Bremen,

Daniel Kr?gler

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