如何在 C++ 中使用枚举专门化函数的 return 类型？

Question

我正在使用一个变体来为 C++ 语法分析器存储一系列类型。语法规则的每个组成部分都有一个类别（枚举类型）和一个值。成分根据类别存储一种类型的值。为了举例，我将类别简化为 'String' => 存储一个字符串，'Number' => 存储一个整数。

我想根据其类别枚举获取具有正确类型的成分的值。我该怎么做？

我在下面编写了示例代码，其中我构造了两个组成部分：strCon，存储一个字符串，以及 intCon，存储一个 int，并尝试获取它们的值。

我想将strCon中的字符串赋值给strVal, 以及从 intCon 到 intVal 的 int。

#include <variant>

struct Constituent
{
    enum class Category {String, Number};
    using Value = std::variant<std::string, int>;

    Category cat;
    Value val;

    // Using a struct ideally to allow partial specialisation of the template,
    // so I can pass the enum without the return type.
    template<Category T>
    struct OfCategory {};

    template<Category T, typename U>
    friend U const& getValue(OfCategory<T>, Constituent const&);
}

using Category = Constituent::Category;

// Template to return the value as the correct type
// for the constituent's category.
template<Category T, typename U>
U const& getValue(OfCategory<T> type, Constituent const& constituent)
{
    // Uses the variant's get function.
    return std::get<U>(constituent.val);
}

// Specialisation to return string from Category::String.
template<>
string const& getValue(OfCategory<Category::String> type,
    Constituent const& constituent)
{
    return getValue<Category::String, string>(constituent);
}

// Specialisation to return int from Category::Number.
template<>
int const& getValue(OfCategory<Category::Number> type,
    Constituent const& constituent)
{
    return getValue<Category::Number, int>(constituent);
}

int main()
{
    Constituent strCon = {Category::String, "This is a string!"};
    Constituent intCon = {Category::Number, 20};

    // In my current implementation, I want this to work with
    // the type wrapper as an overload for the function.
    string strVal = getValue(OfCategory<Category::String>{}, strCon);
    int intVal = getValue(OfCategory<Category::Number>{}, intCon);

    // But it would be better to directly use the template.
    strVal = getValue<Category::String>(strCon);
    intVal = getValue<Category::Number>(intCon);

    // The only way I can get it to work, is to explicitly provide
    // the return type, which defeats the point.
    strVal = getValue<Category::String, string>(
        OfCategory<Category::String>{}, strCon);
    intVal = getValue<Category::Number, int>(
        OfCategory<Category::Number>{}, intCon);

    // Ideally, I could use the cat parameter in Constituent to dynamically
    // infer the return type, but I don't believe something like this is
    // possible in C++.
}

Answer 1

您需要添加一些特征以提供枚举类型，例如重用 OfCategory:

template<Category T> struct OfCategory;

template<> struct OfCategory<Category::String> { using type = std::string; };
template<> struct OfCategory<Category::Number> { using type = int; };

然后，不需要额外的专业化：

template <Category T>
const typename OfCategory<T>::type&
getValue(OfCategory<T> type, Constituent const& constituent)
{
    // Uses the variant's get function.
    return std::get<typename OfCategory<T>::type>(constituent.val);
}

如调用：getValue(OfCategory<Category::String>{}, strCon).

甚至：

template <Category T>
const typename OfCategory<T>::type&
getValue(Constituent const& constituent)
{
    // Uses the variant's get function.
    return std::get<typename OfCategory<T>::type>(constituent.val);
}

像getValue<Category::String>(strCon);

这样的电话

Answer 2

您可以通过创建一个中间特征来进行一级间接 class:

enum E
{
    X,
    Y
};

template <E e>
struct Traits;

template <>
struct Traits<X>
{
    using type = std::string;
};

template <>
struct Traits<Y>
{
    using type = int;
};

template <E e>
typename Traits<e>::type get();

template <>
typename Traits<X>::type get<X>()
{
    return "";
}

template <>
// actually, using the appropriate type directly works as well...
int get<Y>()
{
    return 7;
}

您现在可以像这样调用函数：

std::string s = get<X>();
int n = get<Y>();

Answer 3

我怀疑这样的事情会奏效：

template<Category T>
auto getValue(OfCategory<T> type, Constituent const& constituent)
    -> decltype(std::get<T>(constituent.val))
{
    return std::get<T>(constituent.val);
}

（可能需要将 T 转换为 size_t）。换句话说，您的 getValue 是对 std::get

的改造

如何在 C++ 中使用枚举专门化函数的 return 类型？

How to specialise the return type of a function with an enum in C++?

c++

templates

specialization