java为什么这一行是charg=（char）（br.read（））；被跳过？

Question

在这个程序中，每当我运行程序时，char g = (char)(br.read())行就会被跳过。在输入年龄后，如果我将性别输入为m，我将得到错误java。lang.NumberFormatException

import java.io.*;
public class Tax
{
    public static void main()throws IOException
    {
        InputStreamReader isr  = new InputStreamReader(System.in);
        BufferedReader br = new BufferedReader(isr);
        System.out.println("Enter age");
        int age = Integer.parseInt(br.readLine());
        System.out.println("Enter gender as m or f");
        char g = (char)(br.read());
        System.out.println("Enter Taxable Income");
        int ti = Integer.parseInt(br.readLine());
        double in;
        if (age > 65 || g == 'f')
        System.out.println("Wrong Category");
        else
        {
            if (ti <= 160000)
            in = 0;
            else if(ti > 160000 && ti<=500000)
            in = (ti - 160000) * (10.0/100.0);
            else if(ti > 500000 && ti<=800000)
            in = (ti - 50000) * (20.0/100.0) + 34000;
            else
            in = (ti - 800000) * (30.0/100.0) + 94000;
            System.out.println("Income Tax = " + in);
        }
    }
}

但是，如果程序按如下方式修改，即如果char g = (char)(br.read());被String g = br.readLine()替换，则它可以工作

import java.io.*;
public class Tax
{
    public static void main()throws IOException
    {
        InputStreamReader isr  = new InputStreamReader(System.in);
        BufferedReader br = new BufferedReader(isr);
        System.out.println("Enter age");
        int age = Integer.parseInt(br.readLine());
        System.out.println("Enter gender as m or f");
        String g = br.readLine();
        System.out.println("Enter Taxable Income");
        int ti = Integer.parseInt(br.readLine());
        double in;
        if (age > 65 || g.equalsIgnoreCase("f"))
        System.out.println("Wrong Category");
        else
        {
            if (ti <= 160000)
            in = 0;
            else if(ti > 160000 && ti<=500000)
            in = (ti - 160000) * (10.0/100.0);
            else if(ti > 500000 && ti<=800000)
            in = (ti - 50000) * (20.0/100.0) + 34000;
            else
            in = (ti - 800000) * (30.0/100.0) + 94000;
            System.out.println("Income Tax = " + in);
        }
    }
}

Answer 1

你只读取了一个字节，这意味着如果你写了

m 22

如果在数字前有空格，并且键入

m
22

这意味着m之后的行的其余部分为空

我就是这样写的

Scanner in  = new Scanner(System.in);
System.out.println("Enter age");
int age = in.nextInt(); in.nextLine();
System.out.println("Enter gender as m or f");
String mf = in.next(); in.nextLine();
System.out.println("Enter Taxable Income");
int ti = in.nextInt(); in.nextLine();

Answer 2

尝试使用扫描仪，如中所示

     char g;
     Scanner scan = new Scanner (System.in);
     g = scan.nextChar();`

Answer 3

问题是，您只从读卡器中读取了一个字符，而行分隔符部分仍有待读取

因此，如果BufferedReader逻辑上包含：

30\r\nm\r\n20

然后：

• 第一次调用readLine()将消耗"30\r\n"
• 调用read()将消耗"m"而离开\r\n20
• 第二次调用readLine()将消耗"\r\n"，返回一个空字符串

如果您实际上键入“30”，点击return，然后键入“m20”并点击return，它将起作用-但显然这不是您想要的：）

要么使用扫描仪，要么只使用readLine()并对照“m”和“f”检查行的内容