Herb Sutter's Guru of the Week #4, "Class Mechanics", teaches that the "a op b" form of an overloaded operator should be implemented in terms of the "a op= b" form (see point #4 in the solutions).
As an example, he shows how do to this for the + operator:
T& T::operator+=( const T& other ) {
//...
return *this;
}
T operator+( T a, const T& b ) {
a += b;
return a;
}
He points out that first parameter in operator+ is intentionally being passed by value, so that it can be moved if the caller passes a temporary.
Notice that this requires that the operator+ be a non-member function.
My question is, how can I apply this technique to an overloaded operator in a CRTP base class?
So say this is my CRTP base class with its operator+=:
template <typename Derived>
struct Base
{
//...
Derived operator+=(const Derived& other)
{
//...
return static_cast<Derived&>(*this);
}
};
I can see how to implement operator+ in terms of operator+= as a member function if I dispense with the "pass the first argument by value" optimization:
template <typename Derived>
struct Base
{
//...
Derived operator+(const Derived& other) const
{
Derived result(static_cast<const Derived&>(*this);
result += other;
return result;
}
};
but is there a way to do this while using that optimization (and therefore making the operator+ a nonmember)?
The normal way to implement Herb's advice is as follows:
struct A {
A& operator+=(cosnt A& rhs)
{
...
return *this;
}
friend A operator+(A lhs, cosnt A& rhs)
{
return lhs += rhs;
}
};
Extending this to CRTP:
template <typename Derived>
struct Base
{
Derived& operator+=(const Derived& other)
{
//....
return *self();
}
friend Derived operator+(Derived left, const Derived& other)
{
return left += other;
}
private:
Derived* self() {return static_cast<Derived*>(this);}
};
If you try to avoid the use of friend here, you realize it's almost this:
template<class T>
T operator+(T left, const T& right)
{return left += right;}
But is only valid for things derived from Base<T>, which is tricky and ugly to do.
template<class T, class valid=typename std::enable_if<std::is_base_of<Base<T>,T>::value,T>::type>
T operator+(T left, const T& right)
{return left+=right;}
Additionally, if it's a friend internal to the class, then it's not technically in the global namespace. So if someone writes an invalid a+b where neither is a Base, then your overload won't contribute to the 1000 line error message. The free type-trait version does.
T operator+(T&&, T) //left side can't bind to lvalues, unnecessary copy of right hand side ALWAYS
T operator+(T&&, T&&) //neither left nor right can bind to lvalues
T operator+(T&&, const T&) //left side can't bind to lvalues
T operator+(const T&, T) //unnecessary copies of left sometimes and right ALWAYS
T operator+(const T&, T&&) //unnecessary copy of left sometimes and right cant bind to rvalues
T operator+(const T&, const T&) //unnecessary copy of left sometimes
T operator+(T, T) //unnecessary copy of right hand side ALWAYS
T operator+(T, T&&) //right side cant bind to lvalues
T operator+(T, const T&) //good
//when implemented as a member, it acts as if the lhs is of type `T`.
If moves are much faster than copies, and you're dealing with a commutative operator, you may be justified in overloading these four. However, it only applies to commutative operators (where A?B==B?A, so + and *, but not -, /, or %). For non-commutative operators, there's no reason to not use the single overload above.
T operator+(T&& lhs , const T& rhs) {return lhs+=rhs;}
T operator+(T&& lhs , T&& rhs) {return lhs+=rhs;} //no purpose except resolving ambiguity
T operator+(const T& lhs , const T& rhs) {return T(lhs)+=rhs;} //no purpose except resolving ambiguity
T operator+(const T& lhs, T&& rhs) {return rhs+=lhs;} //THIS ONE GIVES THE PERFORMANCE BOOST