c++ - 使用派生类型的C++ Mixin

Question

如何将typedef从类传递到其mixin？起初我以为可能是在命名冲突，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;
};

lang语语义问题:

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是一个不完整的类-编译器在class DerivedInt之外还没有看到任何其他类。这就是为什么DerivedInt::value_t无法识别的原因。
也许遵循以下思路:

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.
};