Question

当我做d-c + a + b这样的不同组合时，它会给我一个像118.0这样的内含号。有人能告诉我在我的代码中我的计算是错误的.. 谢谢

ValVarPairs.txt包含这些数字 - ＆gt;一个= 100，B = 5，C = 10，d = 13 这就是我编码的内容。

package com.ecsgrid;

import java.io.*;

public class testC {

public static void main(String[] args) {
  int i = 0,j = 0;
  double result, values[] = new double[4];
  char k, operators[] = new char[3];
  for (i = 0; i <= 2; i++) 
    operators[i] = '+';      // default is to add the values

  File myfile;
  StreamTokenizer tok;
  BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
  String InputText;

  i = 0;
  try {
    myfile = new File("C:\\VarValPairs.txt");
    tok = new StreamTokenizer(new FileReader(myfile));  
    tok.eolIsSignificant(false);

    while ((tok.nextToken() != StreamTokenizer.TT_EOF) && (i <= 3)){
      if ((tok.ttype == StreamTokenizer.TT_NUMBER))
        values[i++] = tok.nval;
      }
  }
  catch(FileNotFoundException e) { System.err.println(e);  return; }
  catch(IOException f) { System.out.println(f); return; }

  System.out.println("Enter letters and operators:");

  try {
    InputText = in.readLine(); 
  }  
  catch(IOException f) { System.out.println(f); return; }

  for (i = 0; i < InputText.length(); i++)
  {
     k = InputText.charAt(i);
     if ((k == '+') || (k == '-'))
     {
        if (j <= 2) operators[j++] = k;   
     }
  } 

  result = values[0];
  for (i = 0; i <= 2; i++){
   if (operators[i] == '+')
     result = result + values[i+1];  
   else
     result = result - values[i+1];
  }
  System.out.println(result);  
 }
}

Answer 1

如果输入为-++

，您现在可以获得相同的输出

您永远不会解析订单或a，b，c和d。您总是假设订单a-> b>＆gt; c-＆gt; d。

所以d-c + a + b将是：a-b + c + d，它与你提供的输出一致（100-5 + 10 + 13 = 118）

OP的代码

  for (i = 0; i < InputText.length(); i++)
  {
     k = InputText.charAt(i);
     if ((k == '+') || (k == '-'))
     {
        if (j <= 2) operators[j++] = k;   
     }
  }

/ OP'代码

在这个循环中，当k不是操作符时，你应该读取它是哪个字母，并存储字母出现的顺序。或者构建一些其他类型的映射。在任何情况下，您都不能忽略非操作符，因为它们是输入的一部分。

Answer 2

让我们稍微调试一下，添加一些系统输出......

这是您将在每个步骤中看到的内容 100.0 - 5.0 95.0 + 10.0 105.0 + 13.0 118.0

您的值数组是{100,5,10,13}，您的运算符数组是{ - ，+，+}

你没有映射a = 100，b = 5，c = 10，d = 13，除非你映射那些然后使用基于非操作数输入键的映射解析操作数，它不会起作用。< / p>

所以，如果我要使用字符的int值，我就可以这样翻译。

import java.io.*;

public class TestC {

    public static void main(String[] args) {
        int i = 0, j = 0;
        double result, values[] = new double[4];
        char k, operatorsAndOperands[] = new char[3];
        for (i = 0; i <= 2; i++)
            operatorsAndOperands[i] = '+'; // default is to add the values

        File myfile;
        StreamTokenizer tok;
        BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
        String InputText;

        i = 0;
        try {
            myfile = new File("C:\\VarValPairs.txt");
            tok = new StreamTokenizer(new FileReader(myfile));
            tok.eolIsSignificant(false);

            while ((tok.nextToken() != StreamTokenizer.TT_EOF) && (i <= 3)) {
                if ((tok.ttype == StreamTokenizer.TT_NUMBER))
                    values[i++] = tok.nval;
            }
            for (int l = 0; l < values.length; l++) {
                System.out.println(values[l]);
            }
        } catch (FileNotFoundException e) {
            System.err.println(e);
            return;
        } catch (IOException f) {
            System.out.println(f);
            return;
        }

        System.out.println("Enter letters and operators:");

        try {
            InputText = in.readLine().toUpperCase();
        } catch (IOException f) {
            System.out.println(f);
            return;
        }

        if(InputText.length() > 0){
            operatorsAndOperands = new char[InputText.length()];
        } else {
            System.out.println("No Operations specified");
            return;
        }
        for (i = 0; i < InputText.length(); i++) {
            k = InputText.charAt(i);
            operatorsAndOperands[j++] = k;
        }

        result = 0; 
        for (i = 0; i < operatorsAndOperands.length; i++) {
            System.out.println(operatorsAndOperands[i] + " " + (int)operatorsAndOperands[i]);
            if(i+1<operatorsAndOperands.length)
                System.out.println(operatorsAndOperands[i+1]);
            switch(operatorsAndOperands[i]){
            case '+':
                if(operatorsAndOperands[i+1] != '+' && operatorsAndOperands[i+1] != '-'){
                    result+=values[(int)operatorsAndOperands[i+1] - (int)'A'];
                    i++;
                }
                break;
            case '-':
                if(operatorsAndOperands[i+1] != '+' && operatorsAndOperands[i+1] != '-'){
                    result-=values[(int)operatorsAndOperands[i+1] - (int)'A'];
                    i++;
                }
                break;
            default:
                result = values[(int)operatorsAndOperands[i] - (int)'A'];
                break;
            };
            System.out.println(result);
        }
        System.out.println(result);
    }
}

输出不正确

2 个答案: