模板化递归数据类型 [英] templated recursive data types

问题描述

我有一个像这样的递归数据类型:

I have a recursive data type like this:

template<typename T>
struct SomeType {
    std::map<T, SomeType<T>> mapping;
};

SomeType<int> foo;

这可以正常工作，但是由于类型不完整，将 std :: map 替换为 std :: unordered_map 会导致编译错误.我(或gcc)在某处出错吗?还是这只是标准的一部分?

This works fine, but replacing std::map with std::unordered_map results in a compile error due to an incomplete type. Am I (or gcc) making an error somewhere? or is this just part of the standard?

我还希望通过模板参数(例如 std :: stack 和 std :: queue )确定内部容器，但我无法确定一种可行的方法，因为这将需要已定义SomeType.

I would also like to have the internal container determined by a template parameter (like std::stack and std::queue), but I can't figure out a way to do it since that would require SomeType to already be defined.

不完整的示例:

template<typename T, typename C = std::map<T, SomeType<[???]>>>
struct SomeType {
    C mapping;
};

SomeType<int, [???]> foo;

我知道这可以通过运行时间接操作来完成，但这不是我想要的.

I know this can be done with runtime indirection, but that's not what I'm looking for.

推荐答案

您的课程在其定义的最后} 之前的任何地方都不完整.因此， mapping 成员在其类型的模板参数中使用了不完整的类型 SomeType .

Your class is incomplete anywhere before the final } of its definition. So the mapping member is using incomplete type SomeType in its type's template arguments.

该标准不允许这样做，而且很幸运，它可以与某些STL容器一起使用.

您的第二个问题属于相同的答案-首先这样做是违法的.

Your second questions falls under the same answer -- it is illegal to do that in the first place.

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