返回类型std :: optional< std :: variant< ...> [英] Return type std::optional<std::variant<...>>

问题描述

我遇到一种情况，即函数必须返回从表中获取的值.该表中的一个单元格(假设该表可以正常工作...)可能包含一个值，也可能没有.此值也可以是以下几种类型之一:int, double, string, date(但没有其他类型).

I have a situation where a function must return a value taken from a table. A cell in this table (let's assume the table just works...) may contain a value, or it might not. This value can also be one of several types: int, double, string, date (but no other type).

这样的函数会返回什么?返回std::optional<std::variant<std::string, int, double, std::chrono::time_point>>是个好主意吗?

What would such a function return? Is it a good idea to return std::optional<std::variant<std::string, int, double, std::chrono::time_point>>?

optional和variant会很好用吗?

推荐答案

我认为这是std::monostate的有用用法.具体来说，是variant<std::monostate, int, double, std::string, std::chrono::time_point>. monostate对于variant可能不包含值的情况很有用.

I would consider this to be a useful use of std::monostate. Specifically, variant<std::monostate, int, double, std::string, std::chrono::time_point>. monostate is useful for cases where a variant may not contain a value.

关于使用实际类型而不是optional<variant>的好处是，访问可以在其上正常进行.您可以编写一个可以使用monostate参数的函子，从而甚至可以将visit用于空"变体.

The nice thing about using an actual type rather than optional<variant> is that visitation works normally on it. You can write a functor that can take a monostate parameter, thus allowing you to use visit for even "empty" variants.

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