所以我正在尝试使用 try-catch 和异常制作用户输入计算器。程序不断重复,无法获得数字的实际输入,还包括语句Wrong input. Please try again
。
有关如何解决这个问题的想法吗?
import java.util.*;
public class Calculator
{
public static void main(String[] args) {
int x=1;
do {
try {
System.out.println("Menu");
System.out.println("1-Addition");
System.out.println("2-Subtraction");
System.out.println("3-Multiplication");
System.out.println("4-Divison");
System.out.println("5-Modulos");
System.out.println("6-Exit");
System.out.println("Choose option: ");
Scanner scan = new Scanner(System.in);
int choice = scan.nextInt();
switch (choice) {
case 1:
System.out.print("Input two numbers:");
String dimension = scan.nextLine();
String[] parts = dimension.split(" ");
int a = Integer.parseInt(parts[0]);
int b = Integer.parseInt(parts[1]);
int c=a+b;
System.out.println("Sum = " +c);
break;
case 2:
System.out.print("Input two numbers:");
String dif = scan.nextLine();
String[] difference = dif.split(" ");
int num1 = Integer.parseInt(difference[0]);
int num2 = Integer.parseInt(difference[1]);
int d=num1-num2;
System.out.println("Difference = " +d);
break;
case 3:
System.out.print("Input two numbers:");
String multi = scan.nextLine();
String[] product = multi.split(" ");
int num3 = Integer.parseInt(product[0]);
int num4 = Integer.parseInt(product[1]);
int p=num3*num4;
System.out.println("Product = " +p);
break;
case 4:
System.out.print("Input two numbers:");
String div = scan.nextLine();
String[] quotient = div.split(" ");
int num5 = Integer.parseInt(quotient[0]);
int num6 = Integer.parseInt(quotient[1]);
int q=num5/num6;
System.out.println("Quotient = " +q);
break;
case 5:
System.out.print("Input two numbers:");
String mod = scan.nextLine();
String[] modulo = mod.split(" ");
int num7 = Integer.parseInt(modulo[0]);
int num8 = Integer.parseInt(modulo[1]);
int m=num7%num8;
System.out.println("Modulos = " +m);
break;
case 6:
System.out.println("Now exiting program...");
break;
}
} catch (Exception e) {
System.out.println("Wrong input. Try again.");
}
} while (x==1);
}
}
答案 0 :(得分:3)
你需要这样做
int choice = Integer.parseInt(scan.nextLine());
因为您正在使用readline()
String dimension = scan.nextLine();
当您使用\n
输入选项时,和nextInt()
已经显示在该流中,因为nextint()
永远不会通过按\n
来读取您留下的Enter
}按钮。
答案 1 :(得分:2)
你应该添加:
scan.skip("\n");
之后
int choice = scan.nextInt();
为什么呢?因为nextInt()从输入读取下一个整数。并且它在输入的末尾留下新的行字符。就像:
1<enter>
之后,您调用scan.nextLine()
,它会读取下一个新行字符的所有内容。实际上缓冲区中有1个新行字符。在这里你的nextLine()读取一个空字符串。
要解决该问题,您必须使用scan.skip("\n")
跳过缓冲的换行符。
编辑:
此外,您的代码不允许您退出,因为您没有更改x
变量。尝试在System.out.println("Now exiting program...");
之后更改它。它会工作;)
答案 2 :(得分:1)
这将帮助您确定根本原因:
} catch (Exception e) {
System.out.println("Wrong input. Try again.");
e.printStackTrace();
}