Question

这是我整个程序代码的ASCII表打印部分。我提示用户输入所需的角色和列，但我无法将用户输入连接到表的实际生产。我知道我的变量没有正确匹配，我应该如何改变我的变量，以便在表的实现中使用rs，re，cs，ce？我很困惑自己，所以我在这里寻求帮助。谢谢！

    else if(input == 3)
    {
        System.out.println("Table B.1 ASCII Character Set, p.741 of textbook");
        System.out.println("What is starting row (0-12): ");
            int rs = keyboard.nextInt();
        System.out.println("What is end row (0-12): ");
            int re = keyboard.nextInt();
        System.out.println("What is start col (0-9): ");
            int cs = keyboard.nextInt();
        System.out.println("What is end col (0-9)");
            int ce = keyboard.nextInt();

        if (rs<0 || rs>12)
        {
            System.out.println( "Must be int between (0-12)");
            rs = keyboard.nextInt();
            if (!keyboard.hasNextInt())
            {
            System.out.println("Must be int between (0-12)");
            keyboard.nextInt();
            }
        }
        if (re<0 || re>12)
        {
            System.out.println( "Must be int between (0-12)");
            re = keyboard.nextInt();
            if (!keyboard.hasNextInt())
            {
            System.out.println("Must be int between (0-12)");
            keyboard.nextInt();
            }
        }
        if (cs<0 || cs>9)
        {
            System.out.println( "Must be int between (0-9)");
            cs = keyboard.nextInt();
            if (!keyboard.hasNextInt())
            {
            System.out.println("Must be int between (0-9)");
            keyboard.nextInt();
            }
        }
        if (ce<0 || ce>9)
        {
            System.out.println( "Must be int between (0-9)");
            ce = keyboard.nextInt();
            if (!keyboard.hasNextInt())
            {
            System.out.println("Must be int between (0-9)");
            keyboard.nextInt();
            }
        }
        keyboard.next();

        char hex; 
        char ascii = 0*20; 
        int row = 2;
        int column;

        System.out.print("\n\n");
        System.out.print("                            ");
        System.out.println("ASCII Table");
        System.out.print("                            ");
        System.out.print("\n    ");

        for (hex = '0'; hex<= '9'; hex++)
          System.out.print("  " + hex);
        for (hex = 'A'; hex<= 'F'; hex++)
          System.out.print("  " + hex);

        System.out.println("\n");

        while (ascii < 0*80) 
        {
          System.out.print("  " + row);
          for (column = 0; column < 16; column++) 
          {
            System.out.print("  " + ascii);
            ascii++;
          }
          System.out.print("\n\n");
          row++;
        }
    }

Answer 1

这一行

keyboard.next();

在没有提示的情况下请求输入，并且在输入某些内容之前，ASCII表不会出现。你可以安全地删除它。

未正确指定底部的十六进制文字。 0*20是零的二十，而不是十六进制的文字。十六进制文字是零，后跟字母x，例如0x20。将您的两个十六进制文字分别从0*20和0*80更改为0x20和0x80。

通过这些更改，我得到了这个输出：

Table B.1 ASCII Character Set, p.741 of textbook
What is starting row (0-12): 
0
What is end row (0-12): 
12
What is start col (0-9): 
0
What is end col (0-9)
9


                            ASCII Table

      0  1  2  3  4  5  6  7  8  9  A  B  C  D  E  F

  2     !  "  #  $  %  &  '  (  )  *  +  ,  -  .  /

  3  0  1  2  3  4  5  6  7  8  9  :  ;  <  =  >  ?

  4  @  A  B  C  D  E  F  G  H  I  J  K  L  M  N  O

  5  P  Q  R  S  T  U  V  W  X  Y  Z  [  \  ]  ^  _

  6  `  a  b  c  d  e  f  g  h  i  j  k  l  m  n  o

  7  p  q  r  s  t  u  v  w  x  y  z  {  |  }  ~

打印ASCII表java

1 个答案: