Question

根据cppref，operator <<接受std::basic_ostream<wchar_t>的{{1}}重载。似乎转换操作只是将每个const char*扩展为char。也就是说，转换（插入）的宽字符数等于窄字符数。所以这是一个问题。窄字符串可能使用GB2312编码国际字符，例如中文字符。进一步假设wchar_t为sizeof(wchar_t)并使用UTF16编码。那么这种天真的字符转换方法应该如何运作？

Answer 1

我刚刚在Visual Studio 2015中检查过你是对的。 char只会在没有任何转化的情况下扩展为wchar_t秒。在我看来，你必须自己将窄字符串转换为宽字符串。有几种方法可以做到，有些方法已经提出过了。

这里我建议您可以使用纯C ++工具来完成它，假设您的C ++编译器和标准库足够完整（Visual Studio，或Linux上的GCC（仅在那里））：< / p>

void clear_mbstate (std::mbstate_t & mbs);

void
towstring_internal (std::wstring & outstr, const char * src, std::size_t size,
    std::locale const & loc)
{
    if (size == 0)
    {
        outstr.clear ();
        return;
    }

    typedef std::codecvt<wchar_t, char, std::mbstate_t> CodeCvt;
    const CodeCvt & cdcvt = std::use_facet<CodeCvt>(loc);
    std::mbstate_t state;
    clear_mbstate (state);

    char const * from_first = src;
    std::size_t const from_size = size;
    char const * const from_last = from_first + from_size;
    char const * from_next = from_first;

    std::vector<wchar_t> dest (from_size);

    wchar_t * to_first = &dest.front ();
    std::size_t to_size = dest.size ();
    wchar_t * to_last = to_first + to_size;
    wchar_t * to_next = to_first;

    CodeCvt::result result;
    std::size_t converted = 0;
    while (true)
    {
        result = cdcvt.in (
            state, from_first, from_last,
            from_next, to_first, to_last,
            to_next);
        // XXX: Even if only half of the input has been converted the
        // in() method returns CodeCvt::ok. I think it should return
        // CodeCvt::partial.
        if ((result == CodeCvt::partial || result == CodeCvt::ok)
            && from_next != from_last)
        {
            to_size = dest.size () * 2;
            dest.resize (to_size);
            converted = to_next - to_first;
            to_first = &dest.front ();
            to_last = to_first + to_size;
            to_next = to_first + converted;
            continue;
        }
        else if (result == CodeCvt::ok && from_next == from_last)
            break;
        else if (result == CodeCvt::error
            && to_next != to_last && from_next != from_last)
        {
            clear_mbstate (state);
            ++from_next;
            from_first = from_next;
            *to_next = L'?';
            ++to_next;
            to_first = to_next;
        }
        else
            break;
    }
    converted = to_next - &dest[0];

    outstr.assign (dest.begin (), dest.begin () + converted);
}

void
clear_mbstate (std::mbstate_t & mbs)
{
    // Initialize/clear mbstate_t type.
    // XXX: This is just a hack that works. The shape of mbstate_t varies
    // from single unsigned to char[128]. Without some sort of initialization
    // the codecvt::in/out methods randomly fail because the initial state is
    // random/invalid.
    std::memset (&mbs, 0, sizeof (std::mbstate_t));
}

此功能是log4cplus库的一部分，它可以工作。它使用codecvt构面进行转换。您必须正确设置locale。

Visual Studio可能会出现问题，为您提供适当的GB2312语言环境设置。您可能必须使用_setmbcp()才能使其正常工作。有关详细信息，请参阅“double byte character sequence conversion issue in Visual Studio 2015”。

将窄字符串插入std :: basic_ostream <wchar_t>

1 个答案: