Question

我正在为解析器编写一个“字符串拆分器”，我希望能够同时使用char和wchar_t。我有以下方法：

static void tokenize(const basic_string<T> &, vector<CToken> &);

我想给它一个宽字符串来解析：

vector<CToken> tTokens;
CTokenizer<wstring>::tokenize(L"if(i == 0) { print i + 2; } else { return; }", tTokens);

我也尝试过：

vector<CToken> tTokens;
const wstring sCode = L"if(i == 0) { print i + 2; } else { return; }";
CTokenizer<wstring>::tokenize(sCode, tTokens);

所以我假设有一些我对模板不了解的东西。我该怎么办？

感谢您的帮助！

编辑：这是我的构建日志：

1>------ Build started: Project: c_parser, Configuration: Debug Win32 ------
1>  Parsercpp.cpp
1>c:\users\virus\documents\visual studio 2012\projects\c_parser\c_parser\parsercpp.cpp(44): error C2958: the left parenthesis '(' found at 'c:\users\virus\documents\visual studio 2012\projects\c_parser\c_parser\parsercpp.cpp(38)' was not matched correctly
1>c:\program files (x86)\microsoft visual studio 11.0\vc\include\xstring(529): error C2621: member 'std::_String_val<_Val_types>::_Bxty::_Buf' of union 'std::_String_val<_Val_types>::_Bxty' has copy constructor
1>          with
1>          [
1>              _Val_types=std::_Simple_types<std::basic_string<wchar_t,std::char_traits<wchar_t>,std::allocator<wchar_t>>>
1>          ]
1>          c:\program files (x86)\microsoft visual studio 11.0\vc\include\xstring(532) : see reference to class template instantiation 'std::_String_val<_Val_types>::_Bxty' being compiled
1>          with
1>          [
1>              _Val_types=std::_Simple_types<std::basic_string<wchar_t,std::char_traits<wchar_t>,std::allocator<wchar_t>>>
1>          ]
1>          c:\program files (x86)\microsoft visual studio 11.0\vc\include\xstring(627) : see reference to class template instantiation 'std::_String_val<_Val_types>' being compiled
1>          with
1>          [
1>              _Val_types=std::_Simple_types<std::basic_string<wchar_t,std::char_traits<wchar_t>,std::allocator<wchar_t>>>
1>          ]
1>          c:\program files (x86)\microsoft visual studio 11.0\vc\include\xstring(700) : see reference to class template instantiation 'std::_String_alloc<_Al_has_storage,_Alloc_types>' being compiled
1>          with
1>          [
1>              _Al_has_storage=false,
1>              _Alloc_types=std::_String_base_types<std::wstring,std::allocator<std::wstring>>
1>          ]
1>          c:\users\virus\documents\visual studio 2012\projects\c_parser\c_parser\parsercpp.cpp(104) : see reference to class template instantiation 'std::basic_string<_Elem>' being compiled
1>          with
1>          [
1>              _Elem=std::wstring
1>          ]
1>c:\users\virus\documents\visual studio 2012\projects\c_parser\c_parser\parsercpp.cpp(104): error C2664: 'nsParser::CTokenizer<T>::tokenize' : cannot convert parameter 1 from 'const std::wstring' to 'const std::basic_string<_Elem> &'
1>          with
1>          [
1>              T=std::wstring
1>          ]
1>          and
1>          [
1>              _Elem=std::wstring
1>          ]
1>          Reason: cannot convert from 'const std::wstring' to 'const std::basic_string<_Elem>'
1>          with
1>          [
1>              _Elem=std::wstring
1>          ]
1>          No user-defined-conversion operator available that can perform this conversion, or the operator cannot be called
========== Build: 0 succeeded, 1 failed, 0 up-to-date, 0 skipped ==========

编辑：

template<class T>
class CTokenizer
{
public:
    static vector<T> s_tKeywords;
public:
    static void tokenize(const basic_string<T> &, vector<CToken> &);
};

parser.cpp中的代码是上面的代码：

vector<CToken> tTokens;
const wstring sCode = L"if(i == 0) { print i + 2; } else { return; }";
CTokenizer<wstring>::tokenize(sCode, tTokens);

Answer 1

您应该实例化CTokenizer<wchar_t>，而不是CTokenizer<std::wstring>。你这样做的方式，tokenize()的签名变为

static void tokenize(const basic_string<wstring> &, vector<CToken> &);

Answer 2

将tokenize的声明更改为

static void tokenize(const T &, vector<CToken> &);

或致电

CTokenizer<wchar_t>::tokenize(L"if(i == 0) { print i + 2; } else { return; }", tTokens);

评论后

编辑：然后使用“或”部分（而不是“或”）......

由于T是“char类型”（不是字符串类型），我建议你将它重命名为有意义的东西。

获取您的第二个编辑的代码，在这些更改后，该类看起来像：

template<class CharT>
class CTokenizer
{
public:
    static vector< basic_string<CharT> > s_tKeywords;
public:
    static void tokenize(const basic_string<CharT> &, vector<CToken> &);
};

然后你按照我上面的说法（tokenize而非wchar_t作为模板类型参数调用wstring。（wstring只是basic_string<wchar_t>的类型别名。）

C ++ L“不管” - ＆gt; wstring - ＆gt; basic_string的<T> </T>

2 个答案: