如何将范围存储为类中的字段? [英] How to store a range as a field in a class?

问题描述

我想将范围存储为类中的字段，以便以后可以重用几次.但是，与局部变量不同，我不能简单地将其类型指定为auto.另一方面，库创建的范围类型非常复杂.我要花大量的时间才能手动找出正确的类型，而且如果我选择更改范围的获取方式，将来将很难维护.

I would like to store a range as a field in a class, so that I could reuse it several times later on. However, unlike local variables, I cannot simply specify it type as auto. On the other hand, the types of ranges that the library creates are very complex. It would take me disproportionately long time to figure out the correct type by hand + it would be unmaintainable in the future if I would choose to change how the range is obtained.

所以，我想，也许我可以使用decltype来帮助自己:

So, I thought, maybe I could use decltype to help myself:

class MyClass {
    public:
    using MyRange = decltype(std::declval<std::vector<int*>>() | ranges::v3::view::filter([=](int* elem) { return true; }));
    MyRange range;
}

(注意:我实际的std::declval实际上更复杂，但是我想使示例简短.)

(note: my actual std::declval is actually more complex, but I wanted to make the example brief.)

但是我得到一个错误: a lambda cannot appear in an unevaluated context

But I get an error: a lambda cannot appear in an unevaluated context

所以，我的问题是:

• 如何避免使用lambda并使我的decltype工作?
• 或者也许有一种更好/更干净的方法来获取范围的类型，以将其声明为类中的字段?

• How can I avoid using the lambda and get my decltype working?
• Or maybe there is a better/cleaner way to get a type of a range in order to declare it as a field in a class?

推荐答案

此处的语言很有帮助:如果decltype中允许使用lambda，则它对您无济于事，因为您将无法产生除默认初始化之外的类型为MyRange的值.由于最常见的lambda表达式具有唯一的类型，因此您无法生成正确类型的函数对象以传递给view::filter以使其返回可存储在MyRange中的内容.

The language is being helpful here: if the lambda were allowed in decltype, it wouldn't help you, because there would be no way for you to produce a value of type MyRange other than via default-initialization. Since very occurrence of a lambda expression has a unique type, you couldn't produce a function object of the correct type to pass to view::filter to get it to return something you could store in a MyRange.

解决方法是不要那样做"；例如，将函数对象存储在某处，并在decltype中引用它.在C ++ 17中，例如 :

The workaround is to "don't do that"; e.g., store your function object somewhere and refer to it in the decltype. In C++17, for example:

struct MyClass {
    static constexpr auto pred = [](int* elem) { return true; };
    using MyRange = decltype(std::declval<std::vector<int*>&>() | ranges::v3::view::filter(pred));
    MyRange range;
};

请注意使用std::vector&.如显式range-v3十进制类型评估为无效吗?

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