CODEWITHSUNDEEP
c++templatesspecialization
kiechant
kiechant

Reputation: 67

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

I'm using a variant to store a range of types for a syntax parser in C++. Each constituent of a syntax rule has a category (of type enum) and a value. The constituent stores a type of value according to the category. For the sake of example I've simplified the categories to 'String' => stores a string, and 'Number' => stores an int.

I would like to get the value of the constituent with the correct type according to its category enum. How can I do this?

I've written example code below, where I construct two constituents: strCon, storing a string, and intCon, storing an int, and attempt to get their values.

I want to assign the string in strCon into strVal, and the int from intCon into intVal.

#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++.
}

Upvotes: 4

Views: 344

Answers (3)

Igor Tandetnik
Igor Tandetnik

Reputation: 52471

I suspect something like this would work:

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

(might need to cast T to size_t). In other words, your getValue is a reinvention of std::get

Upvotes: 0

Aconcagua
Aconcagua

Reputation: 25526

You can go with one level of indirection by creating an intermediate traits 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;
}

You now can call the functions like this:

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

Upvotes: 3

Jarod42
Jarod42

Reputation: 217293

You need to add some traits to provide type from enum, for example reusing OfCategory:

template<Category T> struct OfCategory;

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

Then, without need of additional specialization:

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);
}

for call like: getValue(OfCategory<Category::String>{}, strCon).

or even:

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);
}

for call like getValue<Category::String>(strCon);

Upvotes: 1

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