向Java程序添加異常

Question

我無法添加String異常以防止輸入字符串而不是int時程序崩潰。 我確實環顧了一下，嘗試了try{}catch{}但是我的程序仍然會因字符串而崩潰。 我正在尋找修復getInt().

import java.util.*;
public class Binary{

  public static void main(String[]args){
    Scanner in  = new Scanner(System.in);
    String a = "";
    boolean cont  = true;
    while(cont){
      printb(convert(getInt(in, "Enter a number: ")));
      System.out.println("Do you want to continue (yes or no)?");
      a = in.next();
      if(a.equals("yes"))
        cont = true;
      else{
        System.out.println("You answerd no. Have a nice day.");
        cont = false;
      }
    }


  }

  public static int getInt( Scanner console, String prompt){
    System.out.print(prompt);
    while(!console.hasNext()){
        try{
           console.next();
           System.out.println("Not an integer, try again.");
           System.out.print(prompt);
        }
        catch(Input MismatchException exception){
           System.out.print("Not an integer, try again.");
           System.out.print(prompt);
        }

    }
    return console.nextInt();
  }

  public static int[] convert(int decimal){

    int decimalCopy = decimal;
    int len = 0;
    while(decimal != 0){
      decimal/=2;
      len++;
    }
    decimal = decimalCopy;

    int[] b = new int[len];
    int index = 0;
    while(decimal !=0){
      if(decimal%2 != 0)
        b[index] = 1;
      else{
        b[index] = 0;
      }
    decimal/=2;
    index++;
    }

    return b;

  }

  public static void printb(int[] b){
    for(int i = b.length-1; i>=0; i--){
      System.out.print(b[i]);
    }
    System.out.println();
  }

} 

Answer 1

import java.util.*;
public class Binary{

  public static void main(String[]args){
    try {
    Scanner in  = new Scanner(System.in);
    String a = "";
    boolean cont  = true;
    while(cont){
      printb(convert(getInt(in, "Enter a number: ")));
      System.out.println("Do you want to continue (yes or no)?");
      a = in.next();
      if(a.equals("yes"))
        cont = true;
      else{
        System.out.println("You answerd no. Have a nice day.");
        cont = false;
      } }
    } catch(Exception e) {
      System.out.println("Invalid input");
    }


  }

  public static int getInt( Scanner console, String prompt){
    System.out.print(prompt);
    while(!console.hasNext()){
        console.next();
        System.out.println("Not an integer, try again.");
        System.out.print(prompt);
    }
    return console.nextInt();
  }

  public static int[] convert(int decimal){

    int decimalCopy = decimal;
    int len = 0;
    while(decimal != 0){
      decimal/=2;
      len++;
    }
    decimal = decimalCopy;

    int[] b = new int[len];
    int index = 0;
    while(decimal !=0){
      if(decimal%2 != 0)
        b[index] = 1;
      else{
        b[index] = 0;
      }
    decimal/=2;
    index++;
    }

    return b;

  }

  public static void printb(int[] b){
    for(int i = b.length-1; i>=0; i--){
      System.out.print(b[i]);
    }
    System.out.println();
  }

} `enter code here`

Answer 2

try / catch / finally是處理此問題的方法，但是如果您對異常處理不熟悉，則很難弄清楚如何處理它們。 而且即使將它們放在正確的位置，也需要處理尚未正確“清理”的字符串輸入，可以這么說，因此，級聯到分配a位置並結束程序（因為所有是不是是否）。

關鍵是將可能引發異常的行放在try塊中，然后通過一些錯誤處理catch Exception ，然后繼續進行finally需要發生的事情。 （您可以在這里省略finally ，但是我想確保您理解它，因為它很重要。在try / catch之后finally出現，並且該代碼塊中的任何內容都會在兩種情況下執行（除非您過早退出程序）。）

應該這樣做：

while (cont) {

  // getInt() is the troublemaker, so we try it:
  // Notice I have changed 'number' to 'integer' here - this improves
  // UX by prompting the user for the data type the program expects:

  try {
    printb(convert(getInt(in, "Enter an integer: ")));
  }

  // we catch the Exception and name it. 'e' is a convention but you
  // could call it something else. Sometimes we will use it for info,
  // and in this case we don't really need it, but Java expects it
  // nonetheless.
  // We do our error handling here: (notice the call to in.next() -
  // this ensures that the string that was entered gets properly
  // handled and doesn't cascade down to the assignment of 'a') - if
  // this confuses you, try it without the in.next() and see what
  // happens!

  catch (Exception e) {
    System.out.println("\nPlease enter an integer.\n");
    in.next();
  }

  // Again, in this case, the finally isn't necessary, but it can be
  // very handy, so I'm using it for illustrative purposes:

  finally {
    System.out.println("Do you want to continue (yes or no)?");
    a = in.next();
    if (a.equals("yes")) {
      cont = true;
    } else {
      System.out.println("You answered no. Have a nice day.");
      cont = false;
    }
  }
}