我有以下代码:
std::for_each(tokens.begin(), tokens.end(), [&](Token& t) {
static const std::unordered_map<std::wstring, Wide::Lexer::TokenType> mapping([]() -> std::unordered_map<std::wstring, Wide::Lexer::TokenType>
{
// Maps strings to TokenType enumerated values
std::unordered_map<std::wstring, Wide::Lexer::TokenType> result;
// RESERVED WORD
result[L"namespace"] = Wide::Lexer::TokenType::Namespace;
result[L"for"] = Wide::Lexer::TokenType::For;
result[L"while"] = Wide::Lexer::TokenType::While;
result[L"do"] = Wide::Lexer::TokenType::Do;
result[L"type"] = Wide::Lexer::TokenType::Type;
// PUNCTUATION
result[L"{"] = Wide::Lexer::TokenType::OpenCurlyBracket;
result[L"}"] = Wide::Lexer::TokenType::CloseCurlyBacket;
return result;
}());
if (mapping.find(t.Codepoints) != mapping.end()) {
t.type = mapping.find(t.Codepoints)->second;
return;
}
t.type = Wide::Lexer::TokenType::Identifier; // line 121
});
这将遍历一个令牌列表,并根据代码点的内容进行判断,从相关的枚举中为它们分配一个值.如果找不到,则给它一个“标识符”值.但这无法编译.
1>Lexer.cpp(121): error C2065: '__this' : undeclared identifier
1>Lexer.cpp(121): error C2227: left of '->Identifier' must point to class/struct/union/generic type
这是完整错误,没有警告,没有其他错误.什么?我该如何解决这个错误?
编辑:我做了一些重要的重构,我在一个更简单的lambda中得到了完全相同的问题.
auto end_current_token = [&] {
if (current != Wide::Lexer::Token()) {
current.type = Wide::Lexer::TokenType::Identifier; // error line
if (reserved_words.find(current.Codepoints) != reserved_words.end())
current.type = reserved_words.find(current.Codepoints)->second;
if (punctuation.find(current.Codepoints[0]) != punctuation.end())
current.type = punctuation.find(current.Codepoints[0])->second;
tokens.push_back(current);
current = Wide::Lexer::Token();
}
};
我已经清理并重建了这个项目.
我解决了这个问题.
auto end_current_token = [&] {
if (current != Wide::Lexer::Token()) {
// WORKAROUND compiler bug- dead code
struct bug_workaround_type {
int Identifier;
};
bug_workaround_type bug;
bug_workaround_type* __this = &bug;
current.type = Wide::Lexer::TokenType::Identifier;
if (reserved_words.find(current.Codepoints) != reserved_words.end())
current.type = reserved_words.find(current.Codepoints)->second;
if (punctuation.find(current.Codepoints[0]) != punctuation.end())
current.type = punctuation.find(current.Codepoints[0])->second;
tokens.push_back(current);
current = Wide::Lexer::Token();
}
};
不完全是.现在它编译并运行得很好.
最佳答案 FWIW我试图编写一个最小的工作样本,以便在VS2010上编译并编译以下内容而不会出错.
#include <string>
#include <vector>
#include <algorithm>
#include <unordered_map>
namespace Wide { namespace Lexer {
enum TokenType
{
OpenCurlyBracket,
CloseCurlyBacket,
Namespace,
For,
While,
Do,
Type,
Identifier,
};
} }
struct Token
{
std::wstring Codepoints;
Wide::Lexer::TokenType type;
};
int main()
{
std::vector<Token> tokens;
std::for_each(tokens.begin(), tokens.end(), [&](Token& t) {
static const std::unordered_map<std::wstring, Wide::Lexer::TokenType> mapping([]() -> std::unordered_map<std::wstring, Wide::Lexer::TokenType>
{
// Maps strings to TokenType enumerated values
std::unordered_map<std::wstring, Wide::Lexer::TokenType> result;
// RESERVED WORD
result[L"namespace"] = Wide::Lexer::TokenType::Namespace;
result[L"for"] = Wide::Lexer::TokenType::For;
result[L"while"] = Wide::Lexer::TokenType::While;
result[L"do"] = Wide::Lexer::TokenType::Do;
result[L"type"] = Wide::Lexer::TokenType::Type;
// PUNCTUATION
result[L"{"] = Wide::Lexer::TokenType::OpenCurlyBracket;
result[L"}"] = Wide::Lexer::TokenType::CloseCurlyBacket;
return result;
}());
if (mapping.find(t.Codepoints) != mapping.end()) {
t.type = mapping.find(t.Codepoints)->second;
return;
}
t.type = Wide::Lexer::TokenType::Identifier; // line 121
});
}
你可以从这段代码开始,将显示问题的最小编辑一分为二吗?