指定容器类型的迭代器类型的部分专业化 [英] partial specialization for iterator type of a specified container type

问题描述

我有一个模板结构，该模板结构接受Iterator类型的模板参数. 现在，我需要将该类专门用于不同容器的迭代器. 我已经尝试过std :: vector

I have a template struct, which accepts a Iterator type for the template argument. now I need to specialize that class for iterators of different containers. I have tried with std::vector

template<typename Iterator>
struct AC {

};

template<typename T, typename Alloc>
struct AC<typename std::vector<T, Alloc>::iterator> { //this doesn't work

};

但是我得到了这个编译器错误(VS11): 'T':模板参数未在部分专业化中使用或推导

but I got this compiler error(VS11): 'T' : template parameter not used or deducible in partial specialization

有人可以告诉我为什么这不起作用吗?以及如何使其工作?

Can someone please tell me why this doesn't work? And how to make it work?

推荐答案

您无法推断出嵌套::剩余的类型.确实，您的问题毫无道理.考虑以下更简单的反例:

You can't deduce types left of a nesting ::. Indeed, your question makes no sense. Consider this simpler counter-example:

template <typename> struct Foo;
template <> struct Foo<bool> { typedef float type; };
template <> struct Foo<char> { typedef float type; };

template <typename> struct DoesntWork;

template <typename T> struct DoesntWork<typename Foo<T>::type> { };

现在，如果我说DoesntWork<float>，T应该是什么?

Now if I say DoesntWork<float>, what should T be?

重点是，没有理由任何 T应该存在，而Foo<T>::type是您要匹配的东西，即使有一个，也没有理由会是唯一的.

The point is that there is no reason that any T should exist for which Foo<T>::type is a thing you want to match, and even if there were one, there's no reason why it would be unique.

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