使用字符串选择模板类的专用化
select specialization of template class using a string
我使用策略创建了几种类型,即
template <typename PolicyA, typename PolicyB>
class BaseType : PolicyA, PolicyB
{};
struct MyPolicyA {};
struct MyPolicyB {};
struct OtherPolicyB {};
using SpecializedTypeX = BaseType<MyPolicyA, MyPolicyB>;
using SpecializedTypeY = BaseType<MyPolicyA, OtherPolicyB>;
现在我想介绍一些机制,它允许我根据命令行等输入优雅地选择应该使用哪个 SpecializedType。理想情况下,它将是一个工厂方法,创建正确类型的对象,例如:
auto CreateSelectedSpecializedType(const std::string &key);
// selected has type SpecializedTypeX
auto selected = CreateSelectedSpecializedType("SpecializedTypeX");
我将不胜感激任何建议。谢谢!
C++类型不可能依赖于运行时数据,因为类型在编译时是静态固定的。因此,不可能使函数的返回类型依赖于输入参数的值。因此,您可以做的最好的事情可能是为所有策略创建一个公共基类,例如:
struct CommonBase {};
template <typename PolicyA, typename PolicyB>
class BaseType : CommonBase, PolicyA, PolicyB {};
struct MyPolicyA {};
struct MyPolicyB {};
struct OtherPolicyB {};
using SpecializedTypeX = BaseType<MyPolicyA, MyPolicyB>;
using SpecializedTypeY = BaseType<MyPolicyA, OtherPolicyB>;
CommonBase * createObjectOfType(std::string const & type) {
if (type == "SpecializedTypeX")
return new SpecializedTypeX();
if (type == "SpecializedTypeY")
return new SpecializedTypeY();
// etc...
return nullptr;
}