Java：掃描儀輸出不佳

Question

我最近發佈了一個關於掃描儀沒有提供預期結果的問題，並瞭解到我的問題是我沒有用.nextLine()來刷掃描儀。我對我正在處理的一個程序感到困惑，因爲我正在沖洗掃描儀，但是當我測試我的程序時，如果 - 當提示輸入數字時 - 我輸入一個字符串，我會得到錯誤的輸出。它重複兩次相同的循環。

循環的頂部有對nextLine（）的調用，而處理無效輸入的else塊（例如我輸入的字符串）也有對nextLine（）的調用。但不知何故還是我越來越壞輸出

所以要具體，這裏與用戶輸入的壞輸出的樣品中大膽和斜體問題的輸出

進入左邊的值：

輸入操作者：-

輸入右手值：噸

無效的輸入

輸入操作（+ - * /或：

無效操作

輸入操作（+ - * /或：

的上面四行會自動吐出到控制檯。

這是代碼片斷，其中有一個很大的評論，其中的代碼是錯誤的。我只是發佈了問題所在的while塊，但由於while塊是程序的大部分，整個程序只比這個段大一些，所以我認爲最好是發佈它。

import java.util.Scanner; 
import javax.swing.JOptionPane; 

public class Calculator{ 

public static void main(String[] args){ 

    double leftHandVal = 0.0; 

    //Output Title & Instructions 
    System.out.print("!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n"); 
    System.out.print("!\t\t\t\t!\n"); 
    System.out.print("!\t INSTRUCTIONS\t\t!\n"); 
    System.out.print("!\t\t\t\t!\n"); 
    System.out.print("! INPUT\t\tOUTPUT\t\t!\n"); 
    System.out.print("! *******\t   *********\t\t!\n"); 
    System.out.print("! c or C\t\tClear\t\t!\n"); 
    System.out.print("! q or Q\t\tQuit\t\t!\n"); 
    System.out.print("!  +\t\tAddition\t\t!\n"); 
    System.out.print("!  -\t\tSubtraction\t!\n"); 
    System.out.print("!  *\t\tMultiplication\t!\n"); 
    System.out.print("!  /\t\tDivision\t\t!\n"); 
    System.out.print("!\t\t\t\t!\n"); 
    System.out.print("!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!\n\n"); 

    while(true){ 

     Scanner input = new Scanner(System.in); 
     char op = '\n';//(+, -, *, or /) will use in switch statement for their ascii decimal values 

     System.out.print("Enter the left-hand value: "); 

     //these blocks allow the code at the very bottom to not erroneously ask the user for extra input with hasNext() calls 
     if(input.hasNext("c") || input.hasNext("C")){//even though its unlikely for a user to clear so early...just in case 
      leftHandVal = 0.0; 
     } 
     else if(input.hasNext("q") || input.hasNext("Q")){//even though its unlikely for a user to quit so early...just in case 
      op = 113;//assign q for quit code 
     } 
     else if(input.hasNextDouble()){ 
      leftHandVal = input.nextDouble(); 


      /* 
      * 
      *BAD CODE INSIDE WHILE BELOW 
      *BAD CODE INSIDE WHILE BELOW 
      *BAD CODE INSIDE WHILE BELOW 
      *BAD CODE INSIDE WHILE BELOW 
      * 
      */ 

      while(true){ 

       input.nextLine(); 
       double rightHandVal = 0.0; 

       System.out.print("\nEnter operator (+ - * or/: "); 

       if(input.hasNext()){ 
        op = input.next().charAt(0); 
       } 

       //if user wishes to cancel or quit on operator prompt, break out of inner while to access the clear and quit code 
       if(op == 99 || op == 67){ 
        op = 99; 
        break; 
       } 
       else if(op == 113 || op == 81){ 
        op = 113; 
        break; 
       } 
       else if((op != 43) && (op != 45) && (op != 42) && (op != 47)){//if invalid operator, restart inner while 
        System.out.print("Invalid Operator"); 

        continue; 
       } 

       System.out.print("Enter the right-hand value: "); 

       if(input.hasNextDouble()){ 
        rightHandVal = input.nextDouble(); 

        switch(op){ 
         case 43: 
          System.out.printf("%.3f + %.3f = %.3f", leftHandVal, rightHandVal, (leftHandVal + rightHandVal)); 
          leftHandVal += rightHandVal; 
          break; 
         case 45: 
          System.out.printf("%.3f - %.3f = %.3f", leftHandVal, rightHandVal, (leftHandVal - rightHandVal)); 
          leftHandVal -= rightHandVal; 
          break; 
         case 42: 
          System.out.printf("%.3f * %.3f = %.3f", leftHandVal, rightHandVal, (leftHandVal * rightHandVal)); 
          leftHandVal *= rightHandVal; 
          break; 
         case 47: 
          System.out.printf("%.3f/%.3f = %.3f", leftHandVal, rightHandVal, (leftHandVal/rightHandVal)); 
          leftHandVal /= rightHandVal; 
          break; 

        } 
       } 

       //if clear or quit requested from prompt for right-hand value, break to reach the clear and quit code 
       else if(input.hasNext("c") || input.hasNext("C")){ 
        op = 99; 
        break; 
       } 
       else if(input.hasNext("q") || input.hasNext("Q")){ 
        op = 113; 
        break; 
       } 
       else{ 
        System.out.print("Invalid Input"); 

       } 

      } 
     } 

     //if c || C reset op to null and restart outer while 
     if(op == 99 || op == 67){ 
      op = '\n'; 
      leftHandVal = 0.0; 
      continue; 
     } 
     //else if q || Q, prompt user with a popup to confirm. 
     if(op == 113 || op == 81){ 
      int response = JOptionPane.showConfirmDialog(null, "QUIT CALCULATOR?", null, JOptionPane.YES_NO_OPTION); 
      if(response == 0){ 
       System.exit(0); 
      } 
      continue; 
     } 



    } 
} 
} 

Answer 1

else if(input.hasNext("c") || input.hasNext("C")) 
{ 
    op = 99; 
    break; 
} 
else if(input.hasNext("q") || input.hasNext("Q")){ 
    op = 113; 
    break; 
} 
else{ 
    System.out.print("Invalid Input"); 

} 

在這段代碼

，當你輸入「T」爲右手值，你需要檢查hasNext();最後談到其他並打印輸入無效。

但是輸入仍然具有值 「t」 的，所以它進入第二而循環再次開始和

   System.out.print("\nEnter operator (+ - * or/: "); 

       if(input.hasNext()){ 
        op = input.next().charAt(0); 
       } 

檢查input.hasNext（）已經具有 「T」，所以需要「t」並繼續。

解決方案是在進入while循環之前刷新「t」。