使用派生类型的 C++ Mixin答案

【问题标题】：C++ Mixin using derived types使用派生类型的 C++ Mixin
【发布时间】：2020-10-25 16:54:24
【问题描述】：

如何将 typedef 从一个类传递给它的 mixin？起初我以为可能是命名冲突，但在 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

【问题讨论】：

标签： c++ mixins crtp

【解决方案1】：

在Mixin<DerivedInt> 被实例化时，DerivedInt 是一个不完整的类——编译器在class DerivedInt 之外没有看到任何它。这就是无法识别DerivedInt::value_t 的原因。

大概是这样的：

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.
};

【讨论】：