C++ 使用派生類型的 Mixin

Question

我如何將 class 中的 typedef 傳遞給它的 mixin？ 起初我以為可能是命名沖突，但是在 mixin 中重命名value_t也無濟於事。

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 語義問題：

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Mixin<DerivedInt>被實例化時， DerivedInt是一個不完整的 class - 編譯器沒有看到任何超出class DerivedInt的。 這就是無法識別DerivedInt::value_t的原因。

也許沿着這些路線：

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.
};