Scala中的语义规则

问题描述

在这里，我根据涉及Scala中处理环境的两个规则编写了代码。 一切都可以在代码中完美地完成，但是我对我为解释幕后发生的事情而编写的定义不太自信。我不确定是否有人可以查看我的规则是否正确？我对翻译语义规则没有信心。对于第二条规则，我了解到无效变量将导致评估错误。我不清楚应该在哪个规则1中定义。

规则1：

规则2：

sealed trait Environment 
sealed trait Value

case object EmptyEnv extends Environment
case class Extend(x: String,v: Value,sigma: Environment) extends Environment
case class ExtendRec(f: String,x: String,e: Expr,sigma: Environment ) extends Environment
case class ExtendMutualRec2(f1: String,x1: String,e1: Expr,f2: String,x2: String,e2: Expr,sigma: Environment) extends Environment 

/* -- We need to redefine values to accomodate the new representation of environments --*/
case class NumValue(d: Double) extends Value
case class BoolValue(b: Boolean) extends Value
case class Closure(x: String,pi: Environment) extends Value
case object ErrorValue extends Value


/*2. Operators on values */

def valueToNumber(v: Value): Double = v match {
    case NumValue(d) => d
    case _ => throw new IllegalArgumentException(s"Error: Asking me to convert Value: $v to a number")
}

def valueToBoolean(v: Value): Boolean = v match {
    case BoolValue(b) => b
    case _ => throw new IllegalArgumentException(s"Error: Asking me to convert Value: $v to a boolean")
}

def valueToClosure(v: Value): Closure = v match {
    case Closure(x,e,pi) => Closure(x,pi)
    case _ =>  throw new IllegalArgumentException(s"Error: Asking me to convert Value: $v to a closure")
}


/*-- Operations on environments --*/

def lookupEnv(sigma: Environment,x: String): Value = sigma match {
    case EmptyEnv => throw new IllegalArgumentException(s"Error could not find string $x in environment")
    case Extend(y,v,_) if y == x => v
    case Extend(_,_,pi) => lookupEnv(pi,x)
    case ExtendRec(f,y,pi) => if (x == f) 
                                          Closure(y,sigma)
                                   else
                                          lookupEnv(pi,x)
    case ExtendMutualRec2(f1,x1,e1,f2,x2,e2,pi ) => 
    {
        if (x == f1)
            Closure(x1,sigma)
        else if (x == f2)
            Closure(x2,sigma)
        else 
            lookupEnv(pi,x)
    }
}

case class Seq(e1: Expr,e2: Expr) extends Expr
....
....
case Seq(e1,e2) => {
            val (v1,store1) = evalExpr(e1,env,store)
            val (v2,store2) = evalExpr(e2,store1)
            (v2,store2)
            
        }

解决方法

关于规则1，看来您应该再向评估人员添加一个案例

def evalExpr(e: Expr,env: Environment,store: ???): (Value,???) = e match {
  //...
  case Seq(e1,e2) =>
    val (v1,store1) = evalExpr(e1,env,store)
    val (v2,store2) = evalExpr(e2,store1) // what if v1 = error?
    (v2,store2)

  case LetRec2(f1,x1,e1,f2,x2,e2,env) =>
    evalExpr(e,ExtendMutualRec2(f1,env),store /*???*/)
}