使用派生类型的 C++ Mixin

Question

如何将 typedef 从 class 传递到它的 mixin？起初我以为它可能是命名冲突，但是在 mixin 中重命名 value_t 也无济于事。

template <typename Derived>
class Mixin
{
public:
    using value_t = typename Derived::value_t;
    
    Derived * self()
    {
        return static_cast<Derived *>(this);
    }
    
    value_t x() const
    {
        return self()->x;
    }
};

class DerivedInt : public Mixin<DerivedInt>
{
public:
    using value_t = int;
    value_t x = 0;
};

class DerivedDouble : public Mixin<DerivedDouble>
{
public:
    using value_t = double;
    value_t x = 0.0;
};

clang 语义问题：

file.h:14:39: error: no type named 'value_t' in 'DerivedInt'
file.h:27:27: note: in instantiation of template class 'Mixin<DerivedInt>' requested here

file.h:14:39: error: no type named 'value_t' in 'DerivedDouble'
file.h:34:30: note: in instantiation of template class 'Mixin<DerivedDouble>' requested here

Answer 1

在实例化 Mixin<DerivedInt> 时，DerivedInt 是一个不完整的 class - 编译器还没有看到 class DerivedInt 之外的任何内容。这就是 DerivedInt::value_t 无法识别的原因。

也许是这样的：

template <typename Derived, typename ValueType>
class Mixin
{
public:
    using value_t = ValueType;
};

class DerivedInt : public Mixin<DerivedInt, int> {
  // doesn't need its own `value_t` typedef.
};

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