下面的代码创建一个记忆所有操作数和操作符的计算器。“Evaluate”函数接收一个Ops数组并返回一个包含结果和剩余操作的元组。但我无法理解交换机出现了什么问题。return语句如何在switch中工作?
如果情况是opStack有4个x,5,+,6最后一个元素6在开关中被删除并检查。由于它是一个操作数,它将返回6(操作数),剩下的元素4,x,5,+。
我的理解是否正确? “6”和“4,x,5,+”去哪里以及整个表达如何评估?
private enum Op {
case Operand (Double)
case UnaryOperation(String , Double -> Double)
case BinaryOperation(String , (Double,Double) -> Double)
}
private var opStack = [Op]()
private var knownOps = [String : Op]()
init() {
knownOps["×"] = Op.BinaryOperation("×" , *)
knownOps["÷"] = Op.BinaryOperation("÷"){ $1/$0 }
knownOps["+"] = Op.BinaryOperation("+" , +)
knownOps["−"] = Op.BinaryOperation("−"){ $1 - $0 }
knownOps["√"] = Op.UnaryOperation("√" ,sqrt)
}
private func evaluate(ops : [Op]) -> (result : Double? , remainingOps: [Op]){
if !ops.isEmpty {
var remainingOps = ops
let op = remainingOps.removeLast()
switch op{
case .Operand(let operand):
return (operand , remainingOps)
case .UnaryOperation(_ , let operation):
let operandEvaluation = evaluate(remainingOps)
if let operand = operandEvaluation.result{
return (operation(operand), operandEvaluation.remainingOps) }
}
case .BinaryOperation(_, let operation):
let op1Evaluation = evaluate(remainingOps)
if let operand1 = op1Evaluation.result {
let op2Evaluation = evaluate(op1Evaluation.remainingOps)
if let operand2 = op2Evaluation.result {
return (operation(operand1, operand2), op2Evaluation.remainingOps)
}
}
}
}
return(nil , ops)
}
函数是递归的;请注意'UnaryOperation'和'BinaryOperation'事件调用'evaluate'。你可以在函数中设置一个断点和一步,看看它是如何工作的,但是当你说操作数堆栈里有'4,x,5,+,6'时,我怀疑你是不正确的。它看起来会有两个操作“4x5”和“+6” – Paulw11