针对类的enable_if和is_arithmetic模板专业化

Question

I am trying to implement a Expression class with 2 specialisation for arithmetic types. This is the default class:

template<typename Left, typename Op, typename Right, typename std::enable_if<!std::is_arithmetic<Left>::value, Left>::type* = nullptr>
class Expression { /* ... */ }

And those are the two specialisations:

template<typename Left, typename Op, typename Right, typename std::enable_if<std::is_arithmetic<Left>::value, Left>::type* = nullptr>
class Expression { /* ... */ };

template<typename Left, typename Op, typename Right, typename std::enable_if<std::is_arithmetic<Right>::value, Right>::type* = nullptr>
class Expression { /* ... */ };

If I now compile my code I get this error:

Error C3855 'Expression': template parameter '__formal' is incompatible with the declaration Vector

How can I solve my problem with templates and specialisation or dummy types as I used them.

Answer 1

You have multiple primary class templates and these can't be replaced. You need to have one primary template followed by multiple specializations. A simple approach is to do it differently:

template<typename Left, typename Op, typename Right,
         int = std::is_arithmetic_v<Left> + 2 * std::is_arithmetic_v<Right>>
class Expression;

template <typename Left, typename Op, typename Right>
class Expression<Left, Op, Right, 0> {
    // base case
};
template <typename Left, typename Op, typename Right>
class Expression<Left, Op, Right, 1> {
    // only the left operand is arithmetic
};
template <typename Left, typename Op, typename Right>
class Expression<Left, Op, Right, 2> {
    // only the right operand is arithmetic
};
template <typename Left, typename Op, typename Right>
class Expression<Left, Op, Right, 3> {
    // both operands are arithmetic
};

If you have multiple cases which can be handled together you can make these primary template and only specialize the remaining special cases.