专门化模板别名的最佳方法(或解决方法) [英] Best way (Or workaround) to specialize a template alias
问题描述
我目前正在实现一个很小的基于元编程的编译时计算库.
Im currently implementing a tiny metaprogramming-based compile-time computations library.
如果已经为运算符定义了一个基类，那么它的结果将是typedef(我决定使用像std::integral_constant这样的整数包装器作为值而不是原始整数值，以沿库提供统一的接口)，以及一个n -ary运算符基类，用于检查运算符是否具有至少一个操作数:
If have defined a base class for operators, wich has a result typedef (I have decided to use integral wrappers like std::integral_constant as values instead of raw integral values, to provide an uniform interface along the library), and a n-ary operator base class, that checks if the operators has at least one operand:
template<typename RESULT>
struct operator
{
using result = RESULT;
};
template<typename RESULT , typename... OPERANDS>
struct nary_operator : public operator<RESULT>
{
static_assert( sizeof... OPERANDS > 0 , "An operator must take at least one operand" );
};
因此，我为一元和二进制运算符定义了别名:
So I defined alias for unary and binary operators:
template<typename OP , typename RESULT>
using unary_operator = nary_operator<RESULT , OP>;
template<typename LHS , typename RHS , typename RESULT>
using binary_operator = nary_operator<RESULT , LHS , RHS>;
该运算符接口用于将自定义运算符定义为别名，例如下面的比较运算符:
That operator interfaces are used to define custom operators as alias, like comparison operators bellow:
template<typename LHS , typename RHS>
using equal = binary_operator<LHS,RHS,bool_wrapper<LHS::value == RHS::value>>;
template<typename LHS , typename RHS>
using not_equal = logical_not<equal<LHS,RHS>>;
template<typename LHS , typename RHS>
using less_than = binary_operator<LHS,RHS,bool_wrapper<LHS::value < RHS::value>>;
template<typename LHS , typename RHS>
using bigger_than = less_than<RHS,LHS>;
template<typename LHS , typename RHS>
using less_or_equal = logical_not<bigger_than<LHS,RHS>>;
template<typename LHS , typename RHS>
using bigger_or_equal = logical_not<less_than<LHS,RHS>>;
现在假设我们要为自己的类实现自定义相等运算符.例如:
Now suppose we want to implement our custom equality operator for our own class. For example:
template<typename X , typename Y , typename Z>
struct vec3
{
using x = X;
using y = Y;
using z = Z;
};
如果将相等运算符设为uppon继承，而不是使用别名，则可以通过模板专用化轻松完成此操作:
If the equality operator was made uppon inheritance, instead of aliasing, this could be easily done through template specialitation:
//Equaity comparator implemented through inheritance:
template<typename LHS , typename RHS>
struct equal : public binary_operator<LHS,RHS,bool_wrapper<LHS::value == RHS::value>> {};
//Specialitation of the operator for vec3:
template<typename X1 , typename Y1 , typename Z1 , typename X2 , typename Y2 , typename Z2>
struct equal<vec3<X1,Y1,Z1>,vec3<X2,Y2,Z2>> : public binary_operator<vec3<X1,Y1,Z1>,vec3<X2,Y2,Z2> , bool_wrapper<X1 == X2 && Y1 == Y2 && Z1 == Z2>> {};
我知道模板别名不能专门化.
我的问题是:有没有一种方法，不是使用继承设计代替模板别名来专门化这种模板别名?
I know that template alias cannot be specialized.
My quesion is: Is there a way, that is not to use inheritance dessign instead of template aliases, to specialize this kind of template aliases?
推荐答案
我用来专门化模板别名(或提供递归别名)的模式是具有一个对应的_impl结构.例如:
The pattern I use to specialize template aliases (or provide recursive aliases) is to have a corresponding _impl struct. For example:
template <typename T>
struct my_alias_impl { /* def'n */ };
template <>
struct my_alias_impl<int> { /* alternate def'n */ };
template <typename T>
using my_alias = my_alias_impl<T>;
用户必须专门研究my_alias_impl，但其余公共接口仍然保持清洁.
Users would have to specialize on my_alias_impl instead, but the rest of the public interface remains clean.
这篇关于专门化模板别名的最佳方法(或解决方法)的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
