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
});
这个步骤会遍历一系列的tokens,根据编码点的内容,将其与相关枚举值进行对比并赋予对应值。如果未找到,则将其赋值为"Identifier"。但是,此处存在编译错误。
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
这是完整的错误信息,没有警告,也没有其他的错误。怎么办?我该如何修复这个错误?
编辑:我进行了一些重构,并且在一个更简单的lambda中遇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();
}
};
我已清理并重建了该项目。
我修复了这个问题。
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,真的。现在它已经编译并且运行得非常好了。
Wide::Lexer::TokenType::Identifier
是什么意思? - Nicol Bolasthis
的引用,所以我认为它并不重要。 - Puppy