我正在为一种小型语言编写编译器,而我的 Parser 类目前负责构建 AST 以供以后使用。但是,递归表达式不能正常工作,因为每个 AST 节点中包含子节点的向量不能正常工作。目前我的 AST 的头文件是这样的:
class AST
{
public:
enum ASTtype {nil, fdecl, pdecl, vdecl, rd, wr, set, rdLV, setLV, exprLV, add, sub, mul, fcall,
divide, mod, lt, gt, lte, gte, eq, ne, aAnd, aOr, aNot, aNeg, nConst, t, f, vs, dl, loop,
cond, ss};
enum scalarType {tNA, tINVALID, tINT, tLONG, tBOOL};
AST ();
AST (AST const&);
AST (ASTtype);
AST (ASTtype, std::string);
void addChild(AST);
ASTtype getNodeType();
std::string text;
ASTtype nodeType;
int size;
scalarType evalType;
std::vector<AST> children;
};
这是导致问题的表达式解析代码:
void Parser::e(AST& parent)
{
AST expr;
AST::ASTtype check = AST::nil;
bool binOp = false;
switch (lookahead.type)
{
case Lexer::AND : check = AST::aAnd ; binOp = true; break;
case Lexer::OR : check = AST::aOr ; binOp = true; break;
case Lexer::NOT : check = AST::aNot ; break;
case Lexer::NEG : check = AST::aNeg ; break;
case Lexer::PLUS : check = AST::add ; binOp = true; break;
case Lexer::MINUS : check = AST::sub ; binOp = true; break;
case Lexer::SPLAT : check = AST::mul ; binOp = true; break;
case Lexer::FSLASH : check = AST::divide; binOp = true; break;
case Lexer::MOD : check = AST::mod ; binOp = true; break;
case Lexer::EQ : check = AST::eq ; binOp = true; break;
case Lexer::LT : check = AST::lt ; binOp = true; break;
case Lexer::GT : check = AST::gt ; binOp = true; break;
case Lexer::GTE : check = AST::gte ; binOp = true; break;
case Lexer::LTE : check = AST::lte ; binOp = true; break;
case Lexer::NE : check = AST::ne ; binOp = true; break;
}
if (check != AST::nil && binOp)
{
match(lookahead.type);
expr = AST(check);
e(expr);
e(expr);
} else if (check != AST::nil && !binOp) {
match(lookahead.type);
expr = AST(check);
} else if (lookahead.type == Lexer::IDENT) {
if (symbols.resolve(lookahead.text).sym_type == symbol::FUNC)
{
expr = AST(AST::fcall, lookahead.text);
match(Lexer::IDENT);
while (lookahead.type != Lexer::BANG)
{
e(expr);
}
match(Lexer::BANG);
} else {
expr = AST(AST::exprLV);
lv(expr);
}
} else if (lookahead.type == Lexer::T) {
match(Lexer::T); //true
expr = AST(AST::t);
} else if (lookahead.type == Lexer::F) {
match(Lexer::F); //false
expr = AST(AST::f);
} else {
expr = AST(AST::nConst, lookahead.text);
match(Lexer::NUM);
}
parent.children.push_back(expr);
}
一个不起作用的示例表达式是 + 1 + 2 + 3 4。它应该解析成这样的 AST:+ [1, + [2, + [3, 4]],但我得到的是:+ [ 1, + []]
关于我做错了什么有什么建议吗?