如何计算运算符和操作数的字符串混合

Question

当遇到操作符时发生错误。我知道运算符不能转换为
转换后的int或其他格式。我使用运算符通过读取字节代码来计算和传递给枚举defined.But作为我的字符串有运营商，所以我有prob在处理这些。请帮助我这一点。 ---- My Inputs is 1 + 2 ---- Expected Output-- 1 + 2 = 3 ---

错误行---- b = Integer.parseInt（st.nextToken（） ）;

--error在exceution期间----- 输入系列 - 1 + 2

no of tokens:3 
yo 
1 
go 
1 
available 

byte info:10 
....... 
Exception in thread "main" java.lang.NumberFormatException: For input string: "+" 
    at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) 
    at java.lang.Integer.parseInt(Integer.java:484) 
    at java.lang.Integer.parseInt(Integer.java:527) 
    at Abc.main(Abc.java:42) 



I am not able to rectify it. Below is my code 

import java.io.*; 
import java.util.StringTokenizer; 

public class Abc{ 
public static void main(String[] args) throws Exception 
{ 
System.out.println("Hello World"); 
System.out.println("Enter the series"); 
BufferedReader br=new BufferedReader(new InputStreamReader(System.in)); 
String s=br.readLine(); 
int a=0; 
int b=0; 
System.out.println(s); 
while ((br.readLine()) != null) 
{ 
StringTokenizer st=new StringTokenizer(s); 

while (st.hasMoreTokens()) 
{ 
int i=0; 
i=st.countTokens(); 
System.out.println("no of tokens:"+i); 
String token = st.nextToken(); 
System.out.println("yo"); 
System.out.println(token); 
System.out.println("go"); 


a=Integer.parseInt(token); 
System.out.println(a); 

if (st.hasMoreTokens()) // before consuming another token, make sure 
     { 
     System.out.println("available"); 
     byte b1=(byte)br.read(); 
     System.out.println("byte info:"+b1); 
         // there's one available 
         if (st.hasMoreTokens()){ 
       System.out.println("......."); 
     b = Integer.parseInt(st.nextToken()); 
     System.out.println("///////"); 

System.out.println(a); 
System.out.println("reached"); 
System.out.println(b); 
} 
if (b1==43) 
{ 
System.out.println("go"); 
int foo = Integer.parseInt(calculate(operator.ADDITION, a, b)); 
} 
else if (b1==45) 
{ 
int foo = Integer.parseInt(calculate(operator.SUBTRACTION, a, b)); 
} 
else if (b1==42) 
{ 
int foo = Integer.parseInt(calculate(operator.MULTIPLY, a, b)); 
} 
else if (b1==47) 
{ 
int foo = Integer.parseInt(calculate(operator.DIVIDE, a, b)); 
} 

} 
} 
} 
} 

public enum operator 
{ 
    ADDITION("+") { 
     public int apply(int x1, int x2) { 
      return x1 + x2; 
     } 
    }, 
    SUBTRACTION("-") { 
     public int apply(int x1, int x2) { 
      return x1 - x2; 
     } 
    }, 
MULTIPLY("*") { 
     public int apply(int x1, int x2) { 
      return x1 * x2; 
     } 
    }, 
    DIVIDE("/") { 
     public int apply(int x1, int x2) { 
      return x1/x2; 
     } 
    }; 

// You'd include other operators too... 
private final String text; 

    private operator(String text) { 
     this.text = text; 
    } 

    // Yes, enums *can* have abstract methods. This code compiles... 
    public abstract int apply(int x1, int x2); 

    public String toString() { 
     return text; 
    } 

} 

public static String calculate(operator op, int x1, int x2) 
{ 
    return String.valueOf(op.apply(x1, x2)); 
} 
} 

Answer 1

夫妇的问题：

• 你只是在字符串要求输入但不处理它，因此会导致变量和引用的地方。

• 定义一个变量String line;修改及更新您的while循环：

  ，而（（行= br.readLine（））！= NULL）

• 你不需要这行byte b1=(byte)br.read();，因为它只会有换行即输入您输入线时按键

• while循环应该是：

  declare operand1, operand2, count as int 
  declare operator as char 
  while tokenizer has more tokens 
  do 
      optional validate String with token count as 3 with middle token as operator. 
      read token 
      if count == 0 then operand1 = int(token) 
      else if count == 1 then operator = char(token) 
      else operand2 = int(token) 
  done 
  

Answer 2

如果你能做到这一点，那将是更容易使用Java的ScriptEngine类来评估用户给定的字符串，像这样：

ScriptEngineManager engine = new ScriptEngineManager(); 
ScriptEngine javaScript = engine.getEngineByName("JavaScript"); 

System.out.print("Please enter a mathematical operation: "); 
String op = new Scanner(System.in).nextLine(); 

try { 
    Object a = javaScript.eval(op); 
    System.out.println("Result: " + a.toString()); 
} catch (ScriptException ex) { 
    System.out.println("Error in the input!"); 
} 

我测试过它的正常工作。