如何在不同的方法java中打破标签

Question

这是我的源代码：

import java.io.BufferedReader;
import java.io.FileNotFoundException;
import java.io.FileReader;


public class PostfixConverter{



  static int top = 0;
  static String[] mainStack = new String[100];

  final static String asterisk = "*";
  final static String divisor = "/";
  final static String plus = "+";
  final static String minus = "-";
  final static String store = "ST TEMP";

  static int temp = 0;
  static int directionCounter = 0;
  static String[] directions = new String[100];
  static PostfixConverter s = new PostfixConverter();
  static String tempString = "TEMP";

  public static void main(String args[]) throws Exception {

      String string = null;

      String load = "LD ";
      String multiply = "ML ";
      String add = "AD ";
      String div = "DV ";
      String subtract = "SB ";

      String example = "AB+C-";

      try {
          // file reader code
          FileReader file = new     FileReader("ignore for now");
          BufferedReader reader = new BufferedReader(file);

          String line = "";

          while ((line = reader.readLine()) != null) {
              example = line;

              // for loop to print directions
              Characterloop:
              for (int i = 0; i < example.length(); i++) {

                  // get letter entered by user 1 by 1
                  char letter = example.charAt(i);

                  // convert char to string
                  String convertedChar = java.lang.String.valueOf(letter);

                  // finds operands in order or priority



                  // multiply character
                  if (convertedChar.equals(asterisk)) {

                      processOperand(PostfixConverter.multiply(string),  PostfixConverter.multiply(string), load,
                              multiply);
                  }

                  // division character
                  else if (convertedChar.equals(divisor)) {

                      processOperand(PostfixConverter.addition(string), PostfixConverter.addition(string), load, div);
                  }

                  // addition character
                  else if (convertedChar.equals(plus)) {

                      processOperand(PostfixConverter.addition(string), PostfixConverter.addition(string), load, add);
                  }

                  // subtraction character
                  else if (convertedChar.equals(minus)) {

                      processOperand(PostfixConverter.addition(string), PostfixConverter.addition(string), load,
                              subtract);

                  }
                  // letter character
                  else {
                      s.push(convertedChar);
                  }

              }

              // print out the instructions
              System.out.println("Assembly Directions are as follows: ");
              int printDirections = 0;
              for (int i = 0; i < directionCounter; i++) {
                  System.out.println(directions[printDirections]);
                  printDirections++;
              }
              printDirections = 0;
              directionCounter = 0;
              System.out.println("This is the end of the directions.");
              System.out.println("");
              directionCounter = 0;
              temp = 0;
              top = 0;

          }reader.close();

      } catch (FileNotFoundException exception) {
          System.out.println("The file was not found.");
      }
    }

    private static void processOperand(String postFileConverterOutput, String postFileConverterOutput2,
          String instruction1, String instruction2) {
      String outcome;
      String opReturn1 = postFileConverterOutput;
      String opReturn2 = postFileConverterOutput2;
      directions[directionCounter] = instruction1 + opReturn2;
      directionCounter++;
      directions[directionCounter] = instruction2 + opReturn1;
      directionCounter++;
      temp++;
      outcome = tempString + java.lang.String.valueOf(temp);
      directions[directionCounter] = store + java.lang.String.valueOf(temp);
      directionCounter++;
      s.push(outcome);
  }

  // multiply method
  public static String multiply(String a) {
     if(top == 0){
        System.out.println("Invalid Argument");
        System.out.println("Please resubmit a correct String");
        break Characterloop;
     }

      String multVariable = PostfixConverter.pop(mainStack[top]);
      top--;
      return multVariable;
  }

  // addition method
  public static String addition(String a) {
     if(top == 0){
        System.out.println("Invalid Argument");
        System.out.println("Please resubmit a correct String");
     }

      String addVariable = PostfixConverter.pop(mainStack[top]);
      top--;
      return addVariable;
  }

  // subtraction method
  public static String subtraction(String a) {
     if(top == 0){
        System.out.println("Invalid Argument");
        System.out.println("Please resubmit a correct String");
     }

      String subVariable = PostfixConverter.pop(mainStack[top]);
      top--;
      return subVariable;
  }

  // division method
  public static String division(String a) {
     if(top == 0){
        System.out.println("Invalid Argument");
        System.out.println("Please resubmit a correct String");
     }

      String divVariable = PostfixConverter.pop(mainStack[top]);
      top--;
      return divVariable;
  }

  public static boolean empty() {
     boolean check = false;

     if (top < 0){
          check = true;
     }
      else{
          check = false;
      }return check;
  }

  public static String pop(String j) {
     if (top < 0) {
         System.out.println("Stack is empty");
         System.exit(1);
     }
     return mainStack[top - 1];
 }

  public void push(String x) {
      if (top == 99) {
          System.out.println("Stack Overflow");
          System.exit(1);
      } else
          mainStack[top] = x;
         System.out.println("Top:" + top + "||" + " Array: " + mainStack[top]);
      top++;
   }// end push

   }

当给出无效字符串时，如何让此方法停止运行循环？ 我要问的项目在**问题**下方的星号中。 比如说，我得到了后缀字符串“AB+*AB+”和“AB+”。 显然这是无效的，因为不能有只有两个操作数的两个运算符。 如何跳过第一条语句并打印“无效参数”并继续下一个字符串（AB+）？ 我研究了 break 和 continue，但我不知道如何在另一种方法中实现这一点。 这是一个家庭作业，只是想指出正确的方向。

// multiply method
  public static String multiply(String a) {
     if(top == 0){
        System.out.println("Invalid Argument");
        System.out.println("Please resubmit a correct String");
        **break Characterloop;**
     }

Answer 1

像这样的事情可以使用异常和 try/catch 来完成。 然而，这样做应该谨慎使用。 抛出异常不应该用于向用户“返回”某些东西。 但是抛出异常有好处：

• 它们更容易重用
• 您可以将信息附加到异常
• 编译时检查（仅检查异常）。 对于标签，如果break语句按预期工作，则不会进行编译时检查来验证是否仅调用了该方法，如果堆栈中存在包含适当标签的方法调用
• 他们有更好的 IDE 支持（try / catch 块生成）
• 你可以捕获整组异常，如果它们有一个共同的超类

让我们简单化一下问题代码：

具有无效中断的代码

休息

void a() {
    label1: while (check()) {
        b();
    }
}

void b() {
    c();
}

void c() {
    if (check2()) {
        break label1;
    }
    doSomething();
}

继续

void d() {
    label2: while(check()) {
        e();
    }
    doSomething();
}

void e() {
    f();
}

void f() {
    if (check2()) {
        continue label2;
    }
    doSomething();
}

使用异常的解决方案

休息

class MyException1 extends Exception {
}    

void a() {
    try {
        while (check()) {
            b();
        }
    } catch (MyException1 ex) {
         // we're already outside the loop; nothing to be done here
    }
}

void b() throws MyException1 {
    c();
}

void c() throws MyException1 {
    if (check2()) {
        throw new MyException1();
    }
    doSomething();
}

继续

class MyException2 extends Exception {
} 

void d() {
    while(check()) {
        try {
            e();
            doSomething();
        } catch (MyException2 ex) {
            // nothing to do here; We're already at the end of the loop body
        }
    }
}

void e() throws MyException2 {
    f();
}

void f() throws MyException2 {
    if (check2()) {
        throw new MyException2();
    }
    doSomething();
}