继承std::tuple的自定义类型为何触发std::tuple_size不完整类型错误？

尝试获取继承自std::tuple的自定义类型type_descriptor<foo>的元组大小时，编译器报错称std::tuple_size<type_descriptor<foo>>是不完整类型，尽管struct foo和type_descriptor<foo>均已完整定义。

代码示例 #

#include <utility>
#include <tuple>
#include <cstdio>

struct foo
{
    int a_;
    int b_;
};

template <typename T>
struct type_descriptor
{};

template<>
struct type_descriptor<foo>
    : std::tuple<int, bool>
{
};

int main(){
    printf("Size of tuple = %zu\n", std::tuple_size_v<type_descriptor<foo>>);
}

报错信息 #

<source>:89:27:   required from here
/opt/compiler-explorer/gcc-12.2.0/include/c++/12.2.0/bits/utility.h:75:61: error: incomplete type 'std::tuple_size<type_descriptor<foo>>' used in nested name specifier
   75 |     inline constexpr size_t tuple_size_v = tuple_size<_Tp>::value;
      |   

原因分析 #

std::tuple_size的标准特化仅针对std::tuple<Ts...>本身，不会自动推导或继承到派生类。你的type_descriptor<foo>是std::tuple<int, bool>的派生类，但并非std::tuple类型本身，因此标准库中没有为type_descriptor<foo>提供对应的tuple_size特化，编译器会将其视为不完整类型。

解决方法 #

手动为type_descriptor<foo>特化std::tuple_size，复用基类的tuple_size定义即可：

#include <utility>
#include <tuple>
#include <cstdio>

struct foo
{
    int a_;
    int b_;
};

template <typename T>
struct type_descriptor
{};

template<>
struct type_descriptor<foo>
    : std::tuple<int, bool>
{
};

// 手动特化std::tuple_size
namespace std {
    template<>
    struct tuple_size<type_descriptor<foo>> : tuple_size<std::tuple<int, bool>> {};
}

int main(){
    printf("Size of tuple = %zu\n", std::tuple_size_v<type_descriptor<foo>>);
}

如果需要适配多个继承std::tuple的type_descriptor特化，可编写通用模板特化：

namespace std {
    template<typename T>
    requires std::is_base_of_v<std::tuple<>, type_descriptor<T>>
    struct tuple_size<type_descriptor<T>> : tuple_size<std::remove_cvref_t<decltype(std::declval<type_descriptor<T>>())>> {};
}

内容的提问来源于stack exchange，提问作者glades