输出不正确

Question

当我做了像d-c + a + b这样的不同组合时，它给了我一个像118.0这样的inccorect数字。谁能告诉我，在我的代码我的计算是错误的.. 谢谢

的ValVarPairs.txt包含这些numbers-> A = 100，B = 5，c = 10，d = 13 这是我编码。

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

让我们调试这一点，增加一些系统出局......

这是你会看到每一步 100.0 - 5.0 95.0 + 10.0 105.0 + 13.0 118.0

您值数组是{100,5,10,13}，而你的算子数组是{ - ，+，+}

你没有映射a = 100，b = 5，c = 10，d = 13，除非你然后映射那些使用基于非操作数输入键的映射来解析操作数的操作，它不会起作用。所以，如果我要使用字符的int值，我可以用这种方式翻译它。

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); 
    } 
} 

Answer 2

现在你会得到相同的输出，如果你的投入是-++

你永远不解析订单或A，B，c和d。你总是假定顺序a-> b-> c-> d。

所以d-C + A + B将是：A-B + C + d这与你提供的输出（100-5 + 10 + 13 = 118）

OP的CODE

一致

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

/OP的行为准则

在这个循环中，当k不是一个运营商，你应该读w ^这是它的信，并存储字母出现的顺序。或者建立一些其他类型的映射。在任何情况下，您都不能忽略非运算符字符，因为它们是输入的一部分。