我的编解码器(也称为构面)从未被调用. [英] My codecvt (a.k.a. facet) never gets called.

问题描述

问题是无法调用utf16_codecvt 方法，因此结果是错误的.我已经在网上搜索了，但是我所能找到的只是应该起作用的示例.不幸的是，他们都没有成功.我在网上也看到过其他海报，也有同样的问题，但是没有人给他们回答.

我已经测试过，以确保它具有刻面(utf16_codecvt)及其它.做.因此，我认为没有理由永远不会调用其虚拟方法.相反，它会继续调用codecvt<wchar_t,char, mbstate>方法.

有什么想法吗?

The problem is that utf16_codecvt methods never get called and, therefore, the result is wrong. I have search the net, but all I can find is examples of what is supposed to work. Unfortunately none of them has worked. I have also seen other posters, on the net, with the same problem, but no one gave them and answer to it.

I have tested to make sure that it has the facet (utf16_codecvt) and it does. So I see no reason why its virtual methods are never called. Instead it keeps calling the codecvt<wchar_t,char, mbstate> methods.

Any ideas?

class utf16_codecvt : public std::codecvt<char16_t, char16_t, std::mbstate_t>
{
    ...//
};

void MyTestFunc()
{
    ... //
    std::wifstream myFile;
    std::locale myLoc = std::locale(myFile.getloc(), new utf16_codecvt);
    myFile.imbue(myLoc);
    myFile.open(pFileName, std::ios::in | std::ios::binary);
    ... //
    myFile.read(bom_buffer, 1);
    ... //
}

以下链接提供了我要尝试执行的操作类型的示例:
1999年4月1日-Unicode文件-PJ Plauger 推荐答案

据我所知，C ++流系统假定该文件是字节序列，而不是字符序列-即使在使用宽流时-宽流(AFAICT)的宽"部分表示流对象如何与C ++交互，而不是与基础文件或其他对象交互.因此，您的编解码器方面必须包含字符.

通过将编解码器方面的声明更改为以下所示，我可以在设置的替换方面中获得断点.

From what I can tell, the C++ stream system presumes that files are sequences of bytes, not characters - even when you use wide streams - the ''wide'' part of wide stream (AFAICT) indicates how the stream object interacts with C++, not the underlying file or whatever. Thus, your codecvt facet has to take in characters.

By changing the declaration of your codecvt facet to that shown below, I was able to get breakpoints in the replacement facet being set.

class utf16_codecvt : public std::codecvt<char16_t, char, std::mbstate_t>
{
   typedef std::codecvt<char16_t, char, std::mbstate_t> Base;
   typedef char16_t ElemT;
   typedef char ByteT;
   virtual result __CLR_OR_THIS_CALL do_in(std::mbstate_t& s,
      const ByteT *_First1, const ByteT *_Last1, const ByteT *& _Mid1,
      ElemT*_First2, ElemT* _Last2, ElemT *& _Mid2) const
   {	// convert bytes [_First1, _Last1) to [_First2, _Last)
      return Base::do_in(s, _First1, _Last1, _Mid1, _First2, _Last2, _Mid2);
   }

   virtual result __CLR_OR_THIS_CALL do_out(std::mbstate_t& s,
      const ElemT*_First1, const ElemT*_Last1, const ElemT*& _Mid1,
      ByteT*_First2, ByteT*_Last2, ByteT*& _Mid2) const
   {	// convert [_First1, _Last1) to bytes [_First2, _Last)
      return Base::do_out(s, _First1, _Last1, _Mid1, _First2, _Last2, _Mid2);
   }

   virtual result __CLR_OR_THIS_CALL do_unshift(std::mbstate_t& s,
      ByteT*_First2, ByteT*_Last2, ByteT*&_Mid2) const
   {	// generate bytes to return to default shift state
      return Base::do_unshift(s, _First2, _Last2, _Mid2);
   }

   virtual int __CLR_OR_THIS_CALL do_length(const std::mbstate_t& s, const ByteT*_First1,
      const ByteT*_Last1, size_t _Count) const
   {	// return min(_Count, converted length of bytes [_First1, _Last1))
      return Base::do_length(s, _First1, _Last1, _Count);
   }
};

因此，替换方面必须知道每个字符需要读取两个字节(显然，反之亦然).此类信息的最佳参考可能是 Standard C ++ IOStreams和语言环境Angelika Langer和Klaus Kreft [ ^ ]-但是即使如此，在C ++中，语言环境和构面也很繁琐:(

So, your replacement facet will have to know it needs two bytes read for every character (and vice versa, obviously). The best reference for that sort of information is probably Standard C++ IOStreams and Locales by Angelika Langer and Klaus Kreft[^] - but even then, locales and facets are heavy going in C++ :(

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