我正在尝试输出它,以便它在 num1、num2、num3、num4 或 num5 = 999 时打印,并停止扫描仪输入继续,因为它与我的代码不同。希望这是有道理的,如果您能提供帮助,我们将不胜感激。
package a5q4;
import java.util.Scanner;
/**
*
* @author Drew
*/
public class A5Q4 {
/**
* @param args the command line arguments
*/
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
int num1;
int num2;
int num3;
int num4;
int num5;
System.out.println("Enter different numbers between 1 and 100");
num1 = input.nextInt();
if (num1 == 999) {
num2 = 0;
num3 = 0;
num4 = 0;
num5 = 0;
}
else {
num2 = input.nextInt();
}
if (num2 == 999) {
num3 = 0;
num4 = 0;
num5 = 0;
}
else {
num3 = input.nextInt();
}
if (num3 == 999) {
num4 = 0;
num5 = 0;
}
else {
num4 = input.nextInt();
}
if (num4 == 999) {
num5 = 0;
}
else {
num5 = input.nextInt();
}
最佳答案
首先,您可以在一行上声明并初始化多个int
。如果我理解您的问题,您想测试给定的 num
值是否不是 999
如果不是,则阅读 nextInt
。您需要嵌套 if
来使用单独的 num
值来执行此操作。比如,
int num1 = 0, num2 = 0, num3 = 0, num4 = 0, num5 = 0;
System.out.println("Enter different numbers between 1 and 100");
num1 = input.nextInt();
if (num1 != 999) {
num2 = input.nextInt();
if (num2 != 999) {
num3 = input.nextInt();
if (num3 != 999) {
num4 = input.nextInt();
if (num4 != 999) {
num5 = input.nextInt();
}
}
}
}
如果没有 boolean
否定 (!
),它必须位于嵌套 else
(s) 中,
System.out.println("Enter different numbers between 1 and 100");
num1 = input.nextInt();
if (num1 == 999) {
} else {
num2 = input.nextInt();
if (num2 == 999) {
} else {
num3 = input.nextInt();
if (num3 == 999) {
} else {
num4 = input.nextInt();
if (num4 == 999) {
} else {
num5 = input.nextInt();
}
}
}
}
关于Java if else 语句有帮助吗?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/33747599/