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处理由输入扫描仪引起的异常

javajava.util.scannernumberformatexceptionnosuchelementexceptioninputmismatchexception
3

我正在尝试编写编码/解码程序，但是我遇到了各种异常！

出现问题的原因是由多个/单个扫描器引起的：

  • InputMismatchException | NumberFormatException（尝试2）

  • NoSuchElementException（尝试3）

在继续之前，我想说明这不是重复的问题，我已经在StackOverFlow上查看了多个类似的问题，但没有真正帮助我解决。

我查看的类似问题：链接1 链接2

请注意，期望的最终结果类似于第一次尝试的结果，但具有更好的清晰的异常处理和关闭的扫描器。

第一次尝试

  • Now this program gives me the desired results but its a bad programming to have two scanners and one of them ( input method scanner ) is never closed: 

    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();
 }

第一次尝试的输出结果：

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

在有位同仁建议我查看这个问题的链接后，我做了以下操作：

现在我没有关闭输入扫描器，并给了输入方法一个扫描器作为参数：

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();
}

这样做根本没有给我想要的结果：
- 在选择一个字节进行编码并接收编码后的字节后，我将永远陷入编码模式中，而且InputMismatchException | NumberFormatException子句将被激活，所以我无法有机会选择新输入！ - 这是一个基于RLE算法对字节进行编码或解码的程序(o_O) 选项如下： 1. 按1进入编码模式 2. 按2进入解码模式 3. 按3退出！ 1 进入编码模式！ 请输入一系列字节！ 非int类型将终止输入！ 1 e 输入终止！ 1 编码后的列表为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) 选项如下： 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！ 请输入一系列字节！ 非int类型将终止输入！
注意事项：
- 1.在主函数中注释掉sc.close()与上述错误完全相同。 - 2.将扫描器移动到主函数之上并将其声明为全局静态变量，结果与上述失败的结果完全相同。
第三次尝试：
现在我将两个扫描器都关闭了，这导致在主函数中激活了NoSuchElementException。看一下：
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();
}

在这个尝试中，至少我知道什么导致了NoSuchElementException的出现，我认为是因为关闭一个扫描器将会关闭整个代码的输入流。（如果我说错了，请纠正我！）
第三次尝试的输出结果是：
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，非常感谢您的帮助和投入时间和精力。

现在，我有一个关于这些行的小问题：

 if(sc.hasNextInt()) choice=sc.nextInt();
            else {
                sc.next();
                continue;
            }
            error = false;
  • 那么让我看看，如果我理解正确的话，这些行作为一种预防措施发挥作用，如果我错了，请纠正我！以防止异常弹出。

那么是否只写以下内容就足够了，抛弃try-catch块，因为NoSuchElementException没有机会出现，而且通过else块处理并防止了InputMismatchException：

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

如果我想通过try-catch块来处理这个错误，你认为这样写是否干净、免疫异常：（放弃NumberFormatException）。

- 那么展示你回答中的Handle variant，应该是这样的对吧？

                while (error){
                try {
                    choice=sc.nextInt();
                    error = false;                
                } catch (InputMismatchException /*| NumberFormatException*/ e) {
                    error = false;
                    //System.out.println("invalid type or format!");    
                    sc.next();
                    continue;
                }
            }
嘿，在你的第三次尝试中，你的输入是：-1，-1，e...然后你得到了所有包括“没有这样”的消息吗？或者你的输入是：-1，-1，e，X？（其中X是其他输入） - Villat
感谢您的帮助，但是当我向输入方法提供输入时，我明确地使用了try和catch来处理“InputMismatchException”，但这不是问题所在。这只会终止输入方法，我的主要问题是在第三次尝试关闭输入流时，“NoSuchElementException”捕获在主函数中以某种方式被激活了？ - StudentAccount4
我正在尝试理解您在第三次尝试中输入的字符，它们是-1，-1，e...还是-1，-1，e，X（其中X是另一个字符）？ - Villat
先生，输入的类型是字节型，它只接受数字以保存在字节中。如果输入中包含任何字母（例如 e 或 X），将导致输入终止。因此，扫描器只接受数字，其他任何内容都会导致输入被终止。 - StudentAccount4
只要输入流中不是整数，程序就会终止并以此为基础运行。因此，例如输入-1、1、3、x（其中x为& $§""!"%/)=），无论其符号或字母如何，都将终止输入。只要您的输入是数字，输入就不会被终止。 - StudentAccount4
1个回答
1
我对你的代码进行了一些更改（并删除了注释以使其更易读）。基本上，现在我只使用一个Scanner，并且在出现sc.nextInt()之前不会进入选项。
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();
}
谢谢你！你能再看一下帖子吗？我添加了一些想和你讨论的观点 ;) - StudentAccount4
嘿@studentaccount4，最好处理所有的异常，这样可以控制代码。如果我的回答有帮助，请接受并投票 :) - Villat
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