Question

我正在尝试进行编码/解码程序，并且在这里遇到了各种异常！

由多个/单个扫描仪引起的弹出问题：

InputMismatchException | NumberFormatException（ATTEMPT 2）
NoSuchElementException（ATTEMPT 3）

在进行遍历之前，我想指出这不是重复的，我已经在此类StackOverFlow上查找了多个问题，但没有一个对我有太大帮助。我看过的类似问题：link1 link2

请注意，期望的最终结果与第一次尝试的结果类似，但是在某种程度上可以更好地清除异常处理程序并关闭扫描程序。

第一次尝试

现在，该程序为我提供了理想的结果，但是使用两个扫描仪，其中一个（输入法scanner）从未关闭，这是一个糟糕的编程：

public static void main(String[] args) {
Scanner sc=new Scanner (System.in);
int choice = 0;
do {
    System.out.println("This program to encode or decode a byte array " +
            "\n (o_O) Choices are: " +
            "\n 1: Press 1 to enter the encode mode" +
            "\n 2: Press 2 to enter the decode mode" +
            "\n 3: Press 3 to Exit!");
    try {
        //it has to be parseInt because if you used sc.nextInt() the program will go nuts even with try catch.
        choice=Integer.parseInt(sc.next());
        //choice=sc.nextInt();
        /*Question: why when i use this with the existing try catch i the program work for ever but when i use Integer.parseInt(sc.nextLine())
         * the program would normally ask for another value?
         */
    } catch (InputMismatchException | NumberFormatException e) {
        System.out.println("invalid type or format!");
    } catch (NoSuchElementException e) {
        System.out.println("no such");
        //break; if i uncomment this the programm will work For Ever
    }
    switch(choice){

    case 1 :
        System.out.println("entering the encode mode!");
        countAndEncode( input() );
        break;
    case 2 :
        countAndDecode( input() );
        break;
    case 3 :
        System.out.println("exiting...");
        break;
    default :
        System.out.println("please enter a valid option and valid format!");
    }

} while (choice!=3);
sc.close();
 }

 public static byte [] input() {
//arrayList because we dont know the size of the array its like StringBuilder
//ArrayList<Byte> inArray = new ArrayList<Byte>(); 
//according to StackOverflow using ArrayList to store bytes is inefficient
Scanner inScanner=new Scanner (System.in);

ByteArrayOutputStream inArray= new ByteArrayOutputStream();

System.out.println("enter a sequence of ints please! ");
System.out.println("non-int will terminate the input!");

while (inScanner.hasNext()) {
    byte i;
    try {
        i = inScanner.nextByte();
        inArray.write(i);
    } catch (InputMismatchException e) {
        System.out.println("input terminated!");
        break;
    }
}
//System.out.println(Arrays.toString(inArray.toByteArray()));
//inScanner.close();
return inArray.toByteArray();
 }

第一次尝试的输出：

This is a program to encode or decode bytes based on RLE ALgorithm
(o_O) Choices are: 
 1: Press 1 to enter the encode mode
 2: Press 2 to enter the decode mode
 3: Press 3 to Exit!
 1
 entering the encode mode!
 enter a sequence of bytes please! 
 non-int will terminate the input!
 1
 1
 3
 e
 input terminated!
 [1, 1, 3]
 the encoded list is [-1, 1, 2, 3]
 This is a program to encode or decode bytes based on RLE ALgorithm
 (o_O) Choices are: 
 1: Press 1 to enter the encode mode
 2: Press 2 to enter the decode mode
 3: Press 3 to Exit!
 At it goes forever without errors.

第二次尝试

所以在你们中的一个家伙建议看这个问题link之后，我做了什么：

现在我没有关闭输入扫描仪，而是给输入法一个扫描仪作为参数：

public static void main(String[] args) {
    Scanner sc=new Scanner (System.in);
    int choice = 0;
    do {
        System.out.println("This is a program to encode or decode bytes based on RLE ALgorithm" +
                "\n (o_O) Choices are: " +
                "\n 1: Press 1 to enter the encode mode" +
                "\n 2: Press 2 to enter the decode mode" +
                "\n 3: Press 3 to Exit!");
        try {
            //it has to be parseInt because if you used sc.nextInt() the program will go nuts even with try catch.
            choice=Integer.parseInt(sc.next());
            //choice=sc.nextInt();
            /*Question: why when i use this with the existing try catch i the program work for ever but when i use Integer.parseInt(sc.nextLine())
             * the program would normally ask for another value?
             */
        } catch (InputMismatchException | NumberFormatException e) {
            System.out.println("invalid type or format!");
        } catch (NoSuchElementException e) {
            System.out.println("no such");//TODO SOLVE IT PLEASE ITS DRIVING ME CRAZYYYYYYYYYYY!!!!!!!
            break;
        }
        switch(choice){

        case 1 :
            System.out.println("entering the encode mode!");
            countAndEncode( input(sc) );
            break;
        case 2 :
            //countAndDecode( input(sc) );
            break;
        case 3 :
            System.out.println("exiting...");
            break;
        default :
            System.out.println("please enter a valid option and valid format!");
        }

    } while (choice!=3);
    sc.close();
}
/**
 * with this method user will be able to give the desired sequence of bytes. 
 * @return a byte array to be encoded.
 */
public static byte [] input(Scanner inScanner) {
    //arrayList because we dont know the size of the array its like StringBuilder
    //ArrayList<Byte> inArray = new ArrayList<Byte>(); 
    //according to StackOverflow using ArrayList to store bytes is inefficient
    //Scanner   inScanner=new Scanner (System.in);

    ByteArrayOutputStream inArray= new ByteArrayOutputStream();

    System.out.println("enter a sequence of bytes please! ");
    System.out.println("non-int will terminate the input!");

    while (inScanner.hasNext()) {//TODO THIS MIGHT BE THE REASON FOR THE above "SUCH"
        byte i;
        try {
            i = inScanner.nextByte();   
            inArray.write(i);   
        } catch (InputMismatchException e) {
            System.out.println("input terminated!");
            break;
        }
    }
    System.out.println(Arrays.toString(inArray.toByteArray()));
    //inScanner.close();  dont close it because it cant be re-opened
    return inArray.toByteArray();
}

这样做根本无法给我想要的结果：

选择一个编码并接收编码后的字节后，我将永远陷入编码模式，并且InputMismatchException | NumberFormatException子句将被激活，所以我没有机会选择新的输入！

这是一个基于RLE算法对字节进行编码或解码的程序（o_O）选择是： 1：按1进入编码模式 2：按2进入解码模式 3：按3退出！ 1个进入编码模式！请输入字节序列！非整数将终止输入！ 1个 Ë 输入已终止！ 1 编码列表为1 这是一个基于RLE算法对字节进行编码或解码的程序（o_O）选择是： 1：按1进入编码模式 2：按2进入解码模式 3：按3退出！类型或格式无效！进入编码模式！请输入字节序列！非整数将终止输入！

注意：
1。在main中注释sc.close()导致与上述完全相同的错误。
2。将扫描仪移至main上方并将其称为全局静态变量，这确实是上面失败的结果所致。

第三次尝试

现在我把所有关闭的扫描仪都关闭了，这激活了NoSuchElementException主菜单中的内容。

public static void main(String[] args) {
    Scanner sc=new Scanner (System.in);
    int choice = 0;
    do {
        System.out.println("This is a program to encode or decode bytes based on RLE ALgorithm" +
                "\n (o_O) Choices are: " +
                "\n 1: Press 1 to enter the encode mode" +
                "\n 2: Press 2 to enter the decode mode" +
                "\n 3: Press 3 to Exit!");
        try {
            //it has to be parseInt because if you used sc.nextInt() the program will go nuts even with try catch.
            choice=Integer.parseInt(sc.next());
            //choice=sc.nextInt();
            /*Question: why when i use this with the existing try catch i the program work for ever but when i use Integer.parseInt(sc.nextLine())
             * the program would normally ask for another value?
             */
        } catch (InputMismatchException | NumberFormatException e) {
            System.out.println("invalid type or format!");
        } catch (NoSuchElementException e) {
            System.out.println("no such");//TODO SOLVE IT PLEASE ITS DRIVING ME CRAZYYYYYYYYYYY!!!!!!!
            break;
        }
        switch(choice){

        case 1 :
            System.out.println("entering the encode mode!");
            countAndEncode( input() );
            break;
        case 2 :
            //countAndDecode( input() );
            break;
        case 3 :
            System.out.println("exiting...");
            break;
        default :
            System.out.println("please enter a valid option and valid format!");
        }

    } while (choice!=3);
    sc.close();
}
/**
 * with this method user will be able to give the desired sequence of bytes. 
 * @return a byte array to be encoded.
 * @throws IOException 
 */
public static byte [] input() {
    //arrayList because we dont know the size of the array its like StringBuilder
    //ArrayList<Byte> inArray = new ArrayList<Byte>(); 
    //according to StackOverflow using ArrayList to store bytes is inefficient
    Scanner inScanner=new Scanner (System.in);

    ByteArrayOutputStream inArray= new ByteArrayOutputStream();

    System.out.println("enter a sequence of bytes please! ");
    System.out.println("non-int will terminate the input!");

    while (inScanner.hasNext()) {//TODO THIS MIGHT BE THE REASON FOR THE above "SUCH"
        byte i;
        try {
            i = inScanner.nextByte();   
            inArray.write(i);   
        } catch (InputMismatchException e) {
            System.out.println("input terminated!");
            break;
        }
    }
    System.out.println(Arrays.toString(inArray.toByteArray()));
    inScanner.close(); 
    return inArray.toByteArray();
}

在这种尝试中，我至少可能知道是什么原因导致NoSuchElementException跳起来，并且我认为是因为关闭一台扫描仪将关闭整个代码的输入流。（如果我错了，请纠正我！）

第三次尝试的输出是：

This is a program to encode or decode bytes based on RLE ALgorithm
(o_O) Choices are: 
 1: Press 1 to enter the encode mode
 2: Press 2 to enter the decode mode
 3: Press 3 to Exit!
 1
 entering the encode mode!
 enter a sequence of bytes please! 
 non-int will terminate the input!
-1
-1
 e
 input terminated!
 [-1, -1]
 the encoded list is [-1, -1, -1, -1]
 This is a program to encode or decode bytes based on RLE ALgorithm
 (o_O) Choices are: 
 1: Press 1 to enter the encode mode
 2: Press 2 to enter the decode mode
 3: Press 3 to Exit!
no such

@Villat的解决方案

首先，非常感谢您帮助和投入时间和精力。现在，我对这些行有一个小问题：

 if(sc.hasNextInt()) choice=sc.nextInt();
            else {
                sc.next();
                continue;
            }
            error = false;

所以让我看看我是否正确，这些行起了预防作用，如果我错了，请纠正我，以防止异常弹出。

因此，编写try-catch块下面的沟渠是不够的，因为NoSuchElementException没有机会出现，InputMismatchException被else块对待并阻止了：

             while (error){
             if(sc.hasNextInt()) choice=sc.nextInt();
             else {
                 sc.next();
                 continue;
             }
             error = false;
             }

仅出于培训目的，如果我想通过try-catch来处理此错误，如果我这样写，您会认为它是干净的并且不受异常影响：（放弃NumberFormatException）

-如此证明您的答案的Handle variant就是这样吗？

                while (error){
                try {
                    choice=sc.nextInt();
                    error = false;                
                } catch (InputMismatchException /*| NumberFormatException*/ e) {
                    error = false;
                    //System.out.println("invalid type or format!");    
                    sc.next();
                    continue;
                }
            }

Answer 1

我对您的代码进行了一些更改（并删除了注释以使其更具可读性）。基本上，我现在只使用一个Scanner，在出现sc.nextInt()之前，我不会继续使用这些选项。

public static void main(String[] args){
    Scanner sc=new Scanner (System.in);
    int choice = 0;
    do {
        System.out.println("This is a program to encode or decode bytes based on RLE ALgorithm" +
                "\n (o_O) Choices are: " +
                "\n 1: Press 1 to enter the encode mode" +
                "\n 2: Press 2 to enter the decode mode" +
                "\n 3: Press 3 to Exit!");
        boolean error = true;
        while (error){
            try {
                if(sc.hasNextInt()) choice=sc.nextInt();
                else {
                    sc.next();
                    continue;
                }
                error = false;
            } catch (InputMismatchException | NumberFormatException e) {
                System.out.println("invalid type or format!");
            } catch (NoSuchElementException e) {
                System.out.println("no such");
            }
        }
        switch(choice){

            case 1 :
                System.out.println("entering the encode mode!");
                System.out.println(input(sc));
                break;
            case 2 :
                //countAndDecode(input(sc));
                break;
            case 3 :
                System.out.println("exiting...");
                break;
            default :
                System.out.println("please enter a valid option and valid format!");
        }

    } while (choice!=3);
    sc.close();
}

输入法：

public static byte [] input(Scanner sc) {
    ByteArrayOutputStream inArray= new ByteArrayOutputStream();

    System.out.println("enter a sequence of bytes please! ");
    System.out.println("non-int will terminate the input!");

    while (sc.hasNext()) {
        byte i;
        try {
            i = sc.nextByte();
            inArray.write(i);
        } catch (InputMismatchException e) {
            System.out.println("input terminated!");
            break;
        }
    }
    System.out.println(Arrays.toString(inArray.toByteArray()));
    return inArray.toByteArray();
}

处理输入扫描程序引起的异常

1 个答案: