我一直试图向用户展示一些更好的东西。我正在尝试设置如果他输入了错误的输入,那么他必须重新开始,直到他输入介于0到9000之间的正确数字为止。这也是一个乘法迭代和一个递归。但是我不确定他犯错时为什么我设置的消息没有显示。有任何想法为什么和改进想法,有任何代码示例吗?
import java.util.InputMismatchException;
import java.util.Scanner;
public class Multiplication {
public static int multIterative(int a, int b) {
int result = 0;
while (b > 0) {
result += a;
b--;
}
return result;
}
public static int multRecursive(int a, int b) {
if (a == 0 || b == 0) {
return 0;
}
return a + multRecursive(a, b - 1);
}
public static void main(String[] args) {
int a = 0;
int b = 0;
Scanner userInput = new Scanner(System.in);
do {
System.out.print("Please enter first Integer: ");
System.out.print("Please enter second Integer: ");
try {
a = userInput.nextInt();
b = userInput.nextInt();
} catch (InputMismatchException e) {
System.out.println("Must enter an integer!");
userInput.next();
} catch (StackOverflowError e) {
System.out.println("Thats too much");
userInput.next();
}
} while (a >= 9000 || b >= 9000);
System.out.println("The Multiplication Iteration would be: "
+ multIterative(a, b));
System.out.println("The Multiplication Recursion would be: "
+ multRecursive(a, b));
}
}
最佳答案
我建议改用Integer.parseInt(...)
并检查NumberFormatException
。
以下做时工作:
do
try {
System.out.print("Please enter first Integer: ");
a = Integer.parseInt(userInput.next());
System.out.print("Please enter second Integer: ");
b = Integer.parseInt(userInput.next());
if (a >= 9000 || b >= 9000)
throw new StackOverflowError();
break;
} catch (NumberFormatException e) {
System.out.println("Must enter an integer!");
} catch (StackOverflowError e) {
System.out.println("Thats too much");
}
while (true);
关于java - Java用户界面改进,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/28654410/