递归表达式求值

Question

我写这段代码是为了递归地计算一个表达式(我还没有完成，只是在括号和括号上工作)，我刚刚完成了乘法/除法/加法/减法的递归。我得到一个StringOutOfBoundsException为String sub1 = s.substring(0,i);，有什么想法吗?我放了一些print语句来检查I的值，它从来没有作为一个字符串索引不可能的值出现，那么我的问题是什么呢?

    public float evaluate() { 
    String s = expr; 
    float answer = 0;
    //one single variable or just a number
    if(s.contains("+") == false && s.contains("-") == false && s.contains("*") == false && s.contains("/") == false && s.contains("[") == false &&s.contains("]") == false && s.contains("(") == false && s.contains(")") == false){ 
        if(scalars.size() == 0){
            answer = Float.parseFloat(s);
            return answer;
        }
        answer = this.scalars.get(0).value;
        System.out.println("one var/number loop");
        return answer;
    }
    //no parentheses/brackets
    if(s.contains("(") == false && s.contains(")") == false && s.contains("[") == false && s.contains("]") == false && (s.contains("+") == true || s.contains("-") == true || s.contains("*") == true || s.contains("/") == true)){
        answer = evalNoPB(s);
        System.out.println("no parens loop");
        return answer;
    }
    //make compiler happy 
    System.out.println("no loop");
    return 0;
    }
    private float evalNoPB(String s){
        float tempAns = 0;
    if(s.contains("(") == false && s.contains(")") == false && s.contains("[") == false && s.contains("]") == false){
        int i;
        for(i=s.length()-1; i>=0; i--){
            if(s.charAt(i) == '+' || s.charAt(i) == '-'){
                System.out.println(i);
                break; // keep value of i for substrings 
            }
        } if (i<0) { // for loop went through and did not find + or -
            for(i=s.length()-1; i>=0; i--){
                if(s.charAt(i) == '*' || s.charAt(i) == '/'){
                    System.out.println(i);
                    break; // keep value of i for substrings
        }
    }
    }
    String sub1 = s.substring(0,i);
    String sub2 = s.substring(i+1, s.length());
    if(s.charAt(i) == '+'){
        tempAns = evalNoPB(sub1) + evalNoPB(sub2);
    } else if(s.charAt(i) == '-'){
        tempAns = evalNoPB(sub1) - evalNoPB(sub2);
    }else if(s.charAt(i) == '*'){
        tempAns = evalNoPB(sub1) * evalNoPB(sub2);
    }else if (s.charAt(i) == '/'){
        float divisorCheck = evalNoPB(sub2);
        if(divisorCheck!= 0){
        tempAns = evalNoPB(sub1) / evalNoPB(sub2);
        }else { // cannot divide by 0 
            throw new IllegalArgumentException("cannot divide by 0");
        }
}
 }
    return tempAns;
}

测试

Enter the expression, or hit return to quit => 3*3
1 (this is the i value)
Exception in thread "main" java.lang.StringIndexOutOfBoundsException: String       
index out of range: -1
 at java.lang.String.substring(Unknown Source)
 at apps.Expression.evalNoPB(Expression.java:306)
 at apps.Expression.evalNoPB(Expression.java:314)
 at apps.Expression.evaluate(Expression.java:280)
 at apps.Evaluator.main(Evaluator.java:36)

Answer 1

for(i=s.length()-1; i>=0; i--){
   if(s.charAt(i) == '*' || s.charAt(i) == '/'){
        System.out.println(i);
        break; // keep value of i for substrings
   }
}

在此代码段中，变量i最终变为-1并继续执行。即使实际的设计应该改变，对于这一步，只需在for语句之后添加另一个if语句就可以解决这个问题。

if (i < 0) { // for loop went through and did not find + or -
  for (i = s.length() - 1; i >= 0; i--) {
    if (s.charAt(i) == '*' || s.charAt(i) == '/') {
      System.out.println(i);
      break; // keep value of i for substrings
    }
  }
}
if(i < 0)
  return tempAns;
String sub1 = s.substring(0, i);

但只有这种情况下，你会有其他的问题。