[英]May I use a constant number to choose a class at compile time, possibly using templates?
Let's say I have a constant value (possibly of some enum type). 假设我有一个常量值(可能是一些枚举类型)。 Let's say I have many classes A, B, D, etc.
假设我有很多A,B,D等课程。
Can I have something like this? 我能有这样的东西吗?
C<1> anInstanceOfA; //This will be of type A
C<2> anInstanceOfB; //This will be of type B
C<3> anInstanceOfD; //This will be of type D
So, is it possible to select a class based on a constant number at compile time? 那么,是否可以在编译时根据常数选择一个类?
The general problem is that I am trying to select a functor based on a table, in which the index is an enum. 一般的问题是我试图根据表选择一个仿函数,其中索引是一个枚举。 I would like to avoid polymorfism if possible.
如果可能的话,我想避免多态性。
Edit: For this project I cannot use C++11, thanks anyway to who answered in that context, very interesting to know anyway. 编辑:对于这个项目,我不能使用C ++ 11,无论如何,感谢谁在那个环境中回答,无论如何都知道非常有趣。
Edit 2: In general I can have more than 2 target classes, I have edited my question 编辑2:一般来说,我可以有两个以上的目标类,我编辑了我的问题
This isn't the only way to do this, but I hope acceptable for your purposes: 这不是唯一的方法,但我希望你的目的可以接受:
struct A { };
struct B { };
template <int N>
struct choices;
template <>
struct choices<1> { typedef A type; };
template <>
struct choices<2> { typedef B type; };
template <int N>
using C = typename choices<N>::type;
Update: To do the same without C++11 features, you should make C
a class with a typedef
member type equal to the corresponding type alias above: 更新:要在没有C ++ 11功能的情况下执行相同操作,您应该使
C
类成为类型与上面相应类型别名相同的typedef
成员:
template <int N>
struct C
{
typedef typename choices<N>::type type;
};
// ...
C<1>::type anInstanceOfA;
C<2>::type anInstanceOfB
Using the LSP and plain C++98: 使用LSP和普通C ++ 98:
template <int N> class C;
template <> class C<1> : public A {};
template <> class C<2> : public B {};
template <> class C<3> : public D {};
C<1> anInstanceOfA;
Since public inheritance in C++ satisfies the IS-A rule, anInstanceOfA
both IS-A C<1>
object and IS_AN A
object. 由于C ++中的公共继承满足IS-A规则,因此
anInstanceOfA
既是IS-A C<1>
对象又是IS_AN A
对象。
This is a fairly simple metafunction: 这是一个相当简单的元函数:
template <int N>
struct C {
typedef typename std::conditional<N == 1,A,B>::type type;
};
You would use this as C<1>::type foo;
你可以使用它作为
C<1>::type foo;
. 。
If your compiler supports C++11 template aliases, you can simplify to: 如果您的编译器支持C ++ 11模板别名,则可以简化为:
template <int N>
using C = typename std::conditional<N == 1,A,B>::type;
and have your preferred C<1> foo;
并拥有你喜欢的
C<1> foo;
syntax. 句法。
In pure C++03, implement std::conditional
as: 在纯C ++ 03中,将
std::conditional
实现为:
template <bool, typename A, typename>
struct conditional {
typedef A type;
};
template <typename A, typename B>
struct conditional<false, A, B> {
typedef B type;
};
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