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C++ 中的链式选项

[英]Chain Optionals in C++

How to avoid nested if statements with chained optionals in C++?如何避免 C++ 中带有链式选项的嵌套 if 语句?

For example, if type A contains an std::optional<B> b and type B an std::optional<C> c , I would like to be able to write something like:例如,如果类型 A 包含一个std::optional<B> b并且类型 B 一个std::optional<C> c ,我希望能够编写如下内容:

const auto v = if_exists(if_exists(a->b)->c);

And v would get the value from c or an empty optional if either b or c are empty optionals.如果 b 或 c 是空的可选项,则 v 将从 c 或空的可选项中获取值。

I think this would be nicer that nested ifs like this:我认为像这样嵌套 if 会更好:

if (a->b) {
 const auto b = *(a->b);
 if (b->c) {
  const auto c = *(b->c);
 }
}

The following question seems to go in this direction but I am not sure how to adapt it to my use-case: Haskell style "Maybe" type & *chaining* in C++11以下问题似乎朝着这个方向发展,但我不确定如何使其适应我的用例: Haskell style "Maybe" type & *chaining* in C++11

You can do something like this (pseudocode-ish; link to buildable code is provided below):你可以做这样的事情(pseudocode-ish;下面提供了可构建代码的链接):

// wrap std::optional for chaining
template <class T> class Maybe {
  std::optional<T> t;

  // ... constructors etc

  // Maybe chaining 
  // If A has a member named m of type M, 
  // then Maybe<A>.fetch(&A::m) returns a Maybe<M>

  template <class M>
  Maybe<M> fetch(M T::*mem_ptr) {
     return (bool(t)) ? Maybe<M>((*t).*mem_ptr) : Maybe<M>() ;
  }

  // Maybe chaining special case 
  // If A has a member named m, which is itself a Maybe<M>,
  // then return it without wrapping it in an additional Maybe

  template <class M>
  Maybe<M> fetch(Maybe<M> T::*mem_ptr) {
     return (bool(t)) ? ((*t).*mem_ptr) : Maybe<M>() ;
  }

};

Now if you have this:现在,如果你有这个:

 struct C { int d ; }
 struct B { C c; }
 struct A { B b; }
 A a;
 Maybe<A> ma;

and you can do this你可以这样做

 int d = a.b.c.d;

you cannot do the same with ma , but you can use the next best thing, namely:你不能对ma做同样的事情,但你可以使用下一个最好的东西,即:

 Maybe<int> md = ma.fetch(&A::b).fetch(&B::c).fetch(&C::d);

And you can still use this if you Maybe -ify any or all struct members above:如果你Maybe -ify 上面的任何或所有struct成员,你仍然可以使用它:

 struct C { Maybe<int> d ; }
 struct B { Maybe<C> c; }
 struct A { Maybe<B> b; }

Live example (not production quality but it builds) .现场示例(不是生产质量,但可以构建)

You might use你可能会使用

template <typename T, typename F>
auto convert_optional(const std::optional<T>& o, F&& f)
-> std::optional<std::decay_t<decltype(std::invoke(std::forward<F>(f), *o))>>
{
    if (o)
        return std::invoke(std::forward<F>(f), *o);
    else
        return std::nullopt;
}

template <typename T, typename F>
auto convert_optional(std::optional<T>& o, F&& f)
-> std::optional<std::decay_t<decltype(std::invoke(std::forward<F>(f), *o))>>
{
    if (o)
        return std::invoke(std::forward<F>(f), *o);
    else
        return std::nullopt;
}

template <typename T, typename F>
auto convert_optional(std::optional<T>&& o, F&& f)
-> std::optional<std::decay_t<decltype(std::invoke(std::forward<F>(f), *std::move(o)))>>
{
    if (o)
        return std::invoke(std::forward<F>(f), *std::move(o));
    else
        return std::nullopt;
}

or或者

template <typename> struct is_optional : std::false_type {};
template <typename T> struct is_optional<std::optional<T>> : std::true_type {};

template <typename O, typename F>
auto convert_optional(O&& o, F&& f)
-> std::enable_if_t<
    is_optional<std::decay_t<O>>::value,
    std::optional<std::decay_t<decltype(std::invoke(std::forward<F>(f),
                                                    *std::forward<O>(o)))>>>
{
    if (o)
        return std::invoke(std::forward<F>(f), *o);
    else
        return std::nullopt;
}

and your example becomes:你的例子变成:

auto c = convert_optional(convert_optional(a, &A::b).value_or(std::nullopt),
                          &B::c).value_or(std::nullopt);

convert_optional(a, &A::b) will return std::optional<std::optional<B>> convert_optional(a, &A::b)将返回std::optional<std::optional<B>>

You might even simplify by additional function:您甚至可以通过附加功能进行简化:

template <typename O, typename F>
auto convert_optional_fact(O&& o, F&& f)
-> decltype(convert_optional(std::forward<O>(o),
                             std::forward<F>(f)).value_or(std::nullopt))
{
    return convert_optional(std::forward<O>(o),
                            std::forward<F>(f)).value_or(std::nullopt);
}

and then接着

auto c = convert_optional_fact(convert_optional_fact(a, &A::b), &B::c);

Demo演示

This can be achieved with a simple macro.这可以通过一个简单的宏来实现。

#define CHAIN(OPTIONAL, MEMBER)                                                 \
  ([](auto &&opt) {                                                             \
    return opt ? std::optional{opt->MEMBER} : std::nullopt;                     \
  }(OPTIONAL))

const auto v = CHAIN(CHAIN(a, b), c);

Note that there is a proposal to allow this optional chaining in C++:请注意,有一个提议允许在 C++ 中使用此可选链接:

https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2019/p0798r3.html https://www.open-std.org/jtc1/sc22/wg21/docs/papers/2019/p0798r3.html

I like this part:我喜欢这部分:

This is common in other programming languages.这在其他编程语言中很常见。 Here is a list of programming languages which have a optional-like type with a monadic interface or some similar syntactic sugar:下面是一个编程语言列表,它们具有类似于可选类型的单子接口或一些类似的语法糖:

  • Java: Optional Java:可选
  • Swift: Optional斯威夫特:可选
  • Haskell: Maybe哈斯克尔:也许
  • Rust: Option锈:选项
  • OCaml: option OCaml:选项
  • Scala: Option斯卡拉:选项
  • Agda: Maybe阿格达:也许
  • Idris: Maybe伊德里斯:也许
  • Kotlin: T?科特林:T?
  • StandardML: option StandardML:选项
  • C#: Nullable C#:可以为空

Here is a list of programming languages which have a optional-like type without a monadic interface or syntactic sugar:以下是一个没有单子接口或语法糖的类可选类型的编程语言列表:

  • C++ C++

  • I couldn't find any others我找不到其他人

The proposal would allow for code like that:该提案将允许这样的代码:

std::optional<image> get_cute_cat (const image& img) {
    return crop_to_cat(img)
        .and_then(add_bow_tie)
        .and_then(make_eyes_sparkle)
        .transform(make_smaller)
        .transform(add_rainbow);
}

Using the code given in referenced question, You can do something like this使用参考问题中给出的代码,您可以执行以下操作

maybe_do(a->b, [](B &b){ 
    return maybe_do(b.c, [](C &c){ 
        //do what You want to do with the C optional
    })
});

C++23 introduces and_then or or_else to address this inconvenience. C++23 引入了and_thenor_else来解决这个不便。

Here is some paper with proposal.这是一些带有提案的论文

Before we can use C++23 you can try write some template which could resolve this.在我们可以使用 C++23 之前,您可以尝试编写一些可以解决此问题的模板。

My attempt:我的尝试:

namespace detail {
template <auto Field, class T>
struct field_from_opt;

template<typename T, typename FieldType, FieldType T::*ptr>
struct field_from_opt<ptr, T>
{
    static auto get(const std::optional<T>& x) -> std::optional<FieldType>
    {
        if (x) return (*x).*ptr;
        return {};
    }
};
}

template<auto Field, typename T>
auto if_exists(const std::optional<T>& x)
{
    return detail::field_from_opt<Field, T>::get(x);
}

https://godbolt.org/z/dscjYqrx1 https://godbolt.org/z/dscjYqrx1

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