解析 – 在scala解析器组合中回溯？

copy in to out .

这应该很容易用回溯解析：

stmt: (to out(copy in))

还是我错过了什么？

分析器：

type ExprP = Parser[Expr]
type ValueP = Parser[ValExpr]
type CallP = Parser[Call]
type ArgsP = Parser[Seq[Expr]]

val ident     = "[a-zA-Z+-*/%><\=]+".r
val sqstart   = "["                          .r
val sqend     = "]"                          .r
val del       = ","                            .r
val end       = "."                          .r

def stmt: ExprP      = expr <~ end
def expr: ExprP      = ucall | call | value
def value: ValueP    = ident ^^ {str => IdentExpr(str)}
def call: CallP      = (args ~ ident ~ expr) ^^ {case args ~ method ~ upon => Call(args,method,upon)}
def ucall: CallP     = (ident ~ expr) ^^ {case method ~ upon => Call(Seq(),upon)}
def args: ArgsP      = advargs | smplargs
def smplargs: ArgsP  = expr ^^ {e => Seq(e)}
def advargs: ArgsP   = (sqstart ~> repsep(expr,del) <~ sqend) ^^ {seq => seq}