RecursionError：在python中使用百灵鸟时超出了最大递归深度

Question

我已经写在cs143规定的无咖啡因的语法课程。这是我的代码。import sysfrom lark import Lark, Transformer, v_argsdecaf_grammar = r"""    start : PROGRAM    PROGRAM : DECL+    DECL : VARIABLEDECL | FUNCTIONDECL | CLASSDECL | INTERFACEDECL    VARIABLEDECL : VARIABLE ";"    VARIABLE : TYPE  "ident"    TYPE : "int" | "double" | "bool" | "string" | "ident" | TYPE "[]"    FUNCTIONDECL : ( TYPE "ident"  "(" FORMALS ")" STMTBLOCK ) | ( "void" "ident" "(" FORMALS ")"  STMTBLOCK )    FORMALS : VARIABLE ("," VARIABLE)*    CLASSDECL : "class" "ident" ["extends" "ident"] ["implements" "ident" ("," "ident")*] "{" FIELD* "}"    FIELD : VARIABLEDECL | FUNCTIONDECL    INTERFACEDECL : "interface" "ident" "{" PROTOTYPE* "}"    PROTOTYPE : (TYPE "ident" "(" FORMALS ")" ";") | ("void" "ident" "(" FORMALS ")" ";")    STMTBLOCK : "{" VARIABLEDECL* STMT* "}"    STMT : ( EXPR? ";") | IFSTMT | WHILESTMT | FORSTMT | BREAKSTMT | RETURNSTMT | RETURNSTMT | PRINTSTMT | STMTBLOCK    IFSTMT : "if" "(" EXPR ")" STMT ["else" STMT]    WHILESTMT : "while" "(" EXPR ")" STMT    FORSTMT : "for" "(" EXPR? ";" EXPR ";" EXPR? ")" STMT    RETURNSTMT :  "return" EXPR? ";"    BREAKSTMT : "break" ";"    PRINTSTMT : "print" "(" EXPR ("," EXPR)* ")" ";"    EXPR : (LVALUE "=" EXPR) | CONSTANT | LVALUE | "this" | CALL | "(" EXPR ")" | (EXPR "+" EXPR) | (EXPR "-" EXPR) | (EXPR "*" EXPR) | (EXPR "/" EXPR) | (EXPR "%" EXPR) | ("-" EXPR) | (EXPR "<" EXPR) | (EXPR "<=" EXPR) | (EXPR ">" EXPR) | (EXPR ">=" EXPR) | (EXPR "==" EXPR) | (EXPR "!=" EXPR) | (EXPR "&&" EXPR) | (EXPR "||" EXPR) | ("!" EXPR) | ("ReadInteger" "(" ")") | ("ReadLine" "(" ")") | ("new" "ident") | ("NewArray" "(" EXPR "," TYPE ")")    LVALUE : "ident" | (EXPR "." "ident") | (EXPR "[" EXPR "]")    CALL : ("ident" "(" ACTUALS ")") | (EXPR "." "ident" "(" ACTUALS ")")    ACTUALS : EXPR ("," EXPR)* | ""    CONSTANT : "intConstant" | "doubleConstant" | "boolConstant" | "stringConstant" | "null" 它抛出RecursionError：超出最大递归深度。我是第一次使用百灵

Answer 1

你遇到的问题是你感觉不到云雀的规则和终端之间的区别。终端（它们只能以大写字母命名）应该匹配字符串，而不是语法结构。

您必须支持的主要终端属性是，与规则不同，它们不是“递归的”。由于lark努力构建您的语法并进入无限递归和堆栈溢出。