c++ - 错误 : 'int_type' does not name a type - How to inherit typedefs and usings

Question

我正在编写从 std::basic_streambuf 派生的层 streambuf:

/// Basic Socket Buffer
template<
        typename CharT_,
        typename Traits_ = std::char_traits< CharT_ > >
class BasicSocketBuffer //BasicSocketStreamBuffer
        : public std::basic_streambuf< CharT_, Traits_ >
{
...
    virtual int_type
    overflow(int_type __c = traits_type::eof());
...

错误:

error: 'int_type' does not name a type
    virtual int_type
             ^~~~~~~~
note: (perhaps 'typename std::basic_streambuf<_CharT, _Traits>::int_type' was intended)

哈哈。编译器知道我想要什么，但不会做。为什么？

这里是父标准类的第一行:

template<typename _CharT, typename _Traits>
class basic_streambuf
{
public:
    //@{
    /**
     *  These are standard types.  They permit a standardized way of
     *  referring to names of (or names dependent on) the template
     *  parameters, which are specific to the implementation.
    */
    typedef _CharT char_type;
    typedef _Traits traits_type;
    typedef typename traits_type::int_type int_type;
    typedef typename traits_type::pos_type pos_type;
    typedef typename traits_type::off_type off_type;
...

解决方法

几乎在派生类中再次重新声明所有使用的类型；

class Derived : class Parent {
  using Parent::int_type;  // for non-templated
  using typename Parent::char_type;
...

你知道更好的方法吗？欢迎使用 C++11,14。

Answer 1

Ha-ha. Compiler knows what I want, but won't do. Why?

C++ 标准规定您必须在那里使用 typename 才能使代码有效，因为类模板中的名称查找不会查找依赖基。参见 https://gcc.gnu.org/wiki/VerboseDiagnostics#dependent_base及其链接的解释。

Do you know better way?

这才是正确的做法。