Re: virtual function template workaround

Greg Herlihy <>
Mon, 2 Apr 2007 11:53:39 CST
On 4/2/07 6:07 AM, in article, "John Moeller"
<> wrote:

        // Note that including the type in the read_ function name
        // is done primarily for the purposes of exposition.
        // It would probably be considered better form to overload
        // a single name (say, "read_value") instead.

I think that you must actually name them differently, because if you
want to do this:

No, overloading the various "read_type()" methods with a single
"read_value()" method (as shown below) works just as well.

    class data_access
        virtual void read_value(const string& query, int& outInt ) = 0;
        virtual void read_value(const string& query, string& outString)= 0;
        virtual void read_value(const string& query, timeval& outDate) = 0;

Specializations establish the actual pairings:

     // for an int value_type call data_access::read_int()

     template <>
     const data_traits< int >::MF
     data_traits< int >::read_method = &data_access::read_int;

     template <>
     const data_traits< string >::MF
     data_traits< sint >::read_method = &data_access::read_string;

You couldn't do it if they were each named read_value. You can't get
a pointer-to-member to an overloaded member function, can you?

Yes, you can - in fact in this case, just replace each of the various
&data_access::read_int, &data_access::read_string and the other
individualized member pointers with a common "&data_access::read_value"
member pointer expression:

    template <>
    const data_traits< int >::MF
    data_traits< int >::read_method = &data_access::read_value;

    template <>
    const data_traits< string >::MF
    data_traits< string >::read_method = &data_access::read_value;

Since the definition of the data_traits "MF" typedef varies according to the
template's parameterized type, the compiler is able to pick the correct
read_value overload on the right because its type must match the MF typedef
type that appears on the left-hand side of the expression.

Note that even in contexts in which no member pointer typedef has been
defined, a program may always use an explicit cast to disambiguate a member
pointer to an overloaded method.


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