Question

我正在阅读 Advanced C++ Metaprogramming 这本书，我在第2.2.3节遇到了一段代码。（第49页）没有编译（我当前的编译器是Visual Studio 2013）。分配时会出现错误，但书中说明了作业＆＃34; ok：它最终调用basic_inner :: operator = ＆＃34;。我错过了什么吗？

template <typename X, typename T>
struct basic_inner
{
    template <typename T2>
    basic_inner& operator=(const basic_inner<X, T2>&)
    {
        // do something...
        return *this;
    }
};

template <typename T>
struct outer
{
    template <typename X>
    struct inner : public basic_inner<X, T>
    {
        inner& operator=(const basic_inner<X, T> &that)
        {
            static_cast<basic_inner<X, T>&>(*this) = that;
            return *this;
        }
    };
};

template <>
struct outer<int>
{
    template <typename X>
    struct inner : public basic_inner<X, int>
    {
        inner& operator=(const basic_inner<X, int> &that)
        {
            static_cast<basic_inner<X, int>&>(*this) = that;
            return *this;
        }
    };
};

客户端代码是：

outer<double>::inner<void> x1;
outer<int>::inner<void> x2;

x1 = x2;    // <-- error: no operator found which takes right-hand operand of type 'outer<int>::inner<void>'(or there is no acceptable conversion)

Answer 1

我猜测由于名称隐藏，basic_inner类的赋值运算符在派生类inner中不可见。您可以将using basic_inner<X, T>::operator=;添加到outer课程中并获得此类内容。

template <typename T>
struct outer {
  template <typename X>
  struct inner : public basic_inner<X, T> {
    using basic_inner<X, T>::operator=;
    inner& operator=(const basic_inner<X, T>& that) {
      static_cast<basic_inner<X, T>&>(*this) = that;
      return *this;
    }
  };
};

template <>
struct outer<int> {
  template <typename X>
  struct inner : public basic_inner<X, int> {
    using basic_inner<X, int>::operator=;
    inner& operator=(const basic_inner<X, int>& that) {
      static_cast<basic_inner<X, int>&>(*this) = that;
      return *this;
    }
  };
};

高级C ++元编程中的代码无法编译

1 个答案: