[英]partially specialized template as a trait member type
I want to choose a template from a trait class, like in the following: 我想从特征类中选择一个模板,如下所示:
template<typename T>
class JobTypeA { };
template<typename T>
class JobTypeB { };
template<typename T>
class JobTraits
{
/* nothing */
};
class A { };
template<>
class JobTraits<A>
{
typedef JobTypeA Type;
};
class B {};
template<>
class JobTraits<B>
{
typedef JobTypeB Type;
};
class JobTarget1 { };
class JobTarget2 { };
template<typename T, typename U>
class JobUser
{
public:
typedef typename JobTraits<T>::Type<U> JobType;
void doSomething (void)
{
JobType j;
/*... */
}
};
int
main (void)
{
JobUser<B, JobTarget1> j;
}
The above wont compile because of the "typedef of an incomplete type" in the specialized traits classes. 由于特殊特征类中的“不完整类型的typedef”,因此不会编译以上内容。 I got this working using 'alias templates' in std=c++11 with g++ 4.7.2.
我在带有g ++ 4.7.2的std = c ++ 11中使用“别名模板”来工作。 However VS2010 does not support it yet.
但是VS2010还不支持它。 Are there any workarounds to achieve the same without 'alias templates'.
没有“别名模板”,是否有任何变通办法可以实现相同的目的。
How about 怎么样
template<typename U>
struct Type
{
typedef JobTypeA<U> type;
};
and 和
template<typename U>
struct Type
{
typedef JobTypeB<U> type;
};
Usage: 用法:
typedef typename JobTraits<T>::template Type<U>::type JobType;
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