Passing template class name without type arguments

Hello, I have a problem. I would like to use standart C++ <functional> function wrappers, but have a problem. At least in MS VC 7.0 a binary_function derivate (like plus) is defined: template<class Type> struct plus : public binary_function <Type, Type, Type> { Type operator()( const Type& _Left, const Type& _Right ) const; }; So if I am goint to add int and double value I will get int back. How can find a work around? First I was thinking about smth like: template<class _Arg1, class _Arg2> struct _ResultType { //some general implementation like typedef typename _Arg1 result_type; }; //specializations like: template<> struct _ResultType<long, double> { typedef double result_type; }; template<class _BinaryFct, class _ResultType> struct Expression { typedef typename _ResultType::result_type result_t; typedef typename _BinaryFct<result_t> binary_fct; //other definitions etc... }; somewhere in the code: #include <functional> Expression<std::plus, _ResultType<long, double> > myExpr; A C++ compiler awaits a real type which causes an error. How else can I change the type of plus<long> to plus<double>. Probably I could write smth like this: typedef std::plus<_ResultType<long, double> > myBinaryFunction; Expression<myBinaryFunction> myExpr; But somehow I don't like it, since the _ResultType class will contain more information as simple result_type. And I will need to pass it twice like: typedef std::plus<_ResultType<long, double> > myBinaryFunction; Expression<myBinaryFunction, _ResultType<long, double> > myExpr; This make the code not very readable... Is there any way using boost to overcome this? With Kind Regards, Ovanes Markarian