The most straightforward way to write mixed-mode arithmetic is to follow Scott Meyers's Item 24 in Effective C++
template<class T>
class wrapper1
{
public:
wrapper1(T const& t): t_(t) {} // yes, no explicit here
friend wrapper1 operator+(wrapper1 const& lhs, wrapper1 const& rhs)
{
return wrapper1{ lhs.t_ + rhs.t_ };
}
std::ostream& print(std::ostream& os) const
{
return os << t_;
}
private:
T t_;
};
template<class T>
std::ostream& operator<<(std::ostream& os, wrapper1<T> const& rhs)
{
return rhs.print(os);
}
Note that you would still need to write operator+=
in order to provide a consistent interface ("do as the ints do"). If you also want to avoid that boilerplate, take a look at Boost.Operators
template<class T>
class wrapper2
:
boost::addable< wrapper2<T> >
{
public:
wrapper2(T const& t): t_(t) {}
// operator+ provided by boost::addable
wrapper2& operator+=(wrapper2 const& rhs)
{
t_ += rhs.t_;
return *this;
}
std::ostream& print(std::ostream& os) const
{
return os << t_;
}
private:
T t_;
};
template<class T>
std::ostream& operator<<(std::ostream& os, wrapper2<T> const& rhs)
{
return rhs.print(os);
}
In either case, you can then write
int main()
{
wrapper1<int> v{1};
wrapper1<int> w{2};
std::cout << (v + w) << "\n";
std::cout << (1 + w) << "\n";
std::cout << (v + 2) << "\n";
wrapper2<int> x{1};
wrapper2<int> y{2};
std::cout << (x + y) << "\n";
std::cout << (1 + y) << "\n";
std::cout << (x + 2) << "\n";
}
which will print 3 in all cases. Live example. The Boost approach is very general, e.g. you can derive from boost::arithmetic
to also provide operator*
from your definition of operator*=
.
NOTE: this code relies on implicit conversions of T
to wrapper<T>
. But to quote Scott Meyers:
Classes supporting implicit type conversions are generally a bad idea.
Of course, there are exceptions to this rule, and one of the most
common is when creating numerical types.