java中綴計算器

Question

我似乎無法得到嘗試塊內部的堆棧，我的意思是通過在try塊內使用堆棧來獲得一位數的計算器。 try塊的內部是給我的EmptyStackException

import java.util.EmptyStackException; 
import java.util.*; 

public class Infix 
{ 
    public static void main(String[] args){ 

     System.out.println(evaluateInfix("1*2*3")); 

    } 
public static double evaluateInfix(String infix) 
    { 
     try 
     { 
      Stack<Character> valueStack = new Stack<Character>(); 
      Stack<Character> operatorStack = new Stack<Character>(); 
      double result; 
      for(char ch: infix.toCharArray()){ 
       if(ch >= 0 && ch <= 9){ 
        valueStack.push(ch); 
       } else if(ch == '('){ 
        operatorStack.push(ch); 
       } else if(ch == '+' || ch == '-' || ch == '*' || ch == '/' || ch == '%' || ch == '^' || ch == '='){ 
        if(operatorStack.isEmpty()){ 
         operatorStack.push(ch); 
        } else if(precedence(ch) > precedence(operatorStack.peek())){ 
         operatorStack.push(ch); 
        } else { 
         while(!operatorStack.isEmpty() && precedence(ch) <= precedence(operatorStack.peek())){ 
          result = compute(valueOf(valueStack.pop()), valueOf(valueStack.pop()), operatorStack.pop()); 
          valueStack.push((char)result); 
         } 
         operatorStack.push(ch); 
        } 
       } else if(ch == ')'){ 
        while(operatorStack.peek() != '('){ 
         result = compute(valueOf(valueStack.pop()), valueOf(valueStack.pop()), operatorStack.pop()); 
         valueStack.push((char)result); 
        } 
        operatorStack.pop(); 
       } 
      } 

      while(!operatorStack.isEmpty()){ 
       result = compute(valueOf(valueStack.pop()), valueOf(valueStack.pop()), operatorStack.pop()); 
       valueStack.push((char)result); 
      } 
      result = valueStack.peek(); 

      System.out.println(result); 


      /* ********** 
      Task 3 
      complete this section to calculate the infix expression 
      */   



     } //end try 
     catch (EmptyStackException e) 
     { 
      /* ********** 
      Task 3 
      complete this to return Double.NaN 
      */    
     } 
     catch (ArithmeticException e) 
     { 
      /* ********** 
      Task 3 
      complete this to return Double.NEGATIVE_INFINITY 
      */    
     } 
     return 3.0; 

    } // end evaluateInfix 

    private static int precedence(char operator) 
    { 
     switch(operator){ 
      case '(': case ')': return 0; 
      case '+': case '-': return 1; 
      case '*': case '/': case '%': return 2; 
      case '^': return 3; 
      case '=': return 4; 
     } 
     return -1; 
    } 

    private static double valueOf(char variable) 
    { 
     switch (variable) 
     { 
      case '1': return 1.0; 
      case '2': return 2.0; 
      case '3': return 3.0; 
      case '4': return 4.0; 
      case '5': return 5.0; 
      case '6': return 6.0; 
      case '7': return 7.0; 
      case '8': return 8.0; 
      case '9': return 9.0; 
      case '0': return 0.0; 
     } // end switch 

     return 0; 
    } // end valueOf 

    private static Double compute(Double operandOne, Double operandTwo, char operator) 
    { 
     if(operator == '+'){ 
      return operandOne + operandTwo; 
     } else if(operator == '-'){ 
      return operandOne - operandTwo; 
     } else if(operator == '*'){ 
      return operandOne * operandTwo; 
     } else if(operator == '/'){ 
      return operandOne/operandTwo;   
     } else if(operator == '%'){ 
      return operandOne % operandTwo; 
     } else if(operator == '^'){ 
      return Math.pow(operandOne, operandTwo); 
     } else { 
      return null; 
     } 
    } // end compute 
} // end Infix 

Answer 1

你有一對冗餘的「}」和「{」，它把大量代碼到一個初始化塊 - 基本上是一箇中綴類的構造函數。

（根據您的評論） - 缺少方法聲明。

嘗試從改變前幾行：

public class Infix { 
    public static void main(String[] args){ 
    System.out.println(evaluateInfix("1*2*3")); 
    } 

    { 
    try { 
     ... 

到：

public class Infix { 
    public static void main(String[] args){ 
    System.out.println(evaluateInfix("1*2*3")); 
    } 

    public static double evaluateInfix(String infix) { 
    try { 
     ... 

Answer 2

不應該，如果含有'繞數第一？

if(ch >= '0' && ch <= '9'){