Java：如何在用户需要时使用标记值退出程序
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时间：2020-11-02 12:50:07 来源:igfitidea点击:
Java: How to use a sentinel value to quit a program when the user wants
提问by JcireR
The only thing i am missing is to use a sentinel value, like zero, to quit the loop when the user wants, even without enter any gussing.
我唯一缺少的是使用一个哨兵值，如零，在用户需要时退出循环，即使没有输入任何 gussing。
import java.util.Scanner;
import java.util.Random;
public class RandomGuessing {
//-----------------------------------------------------------------------------------------------
//This application prompt to the user to guess a number. The user can still playing until
//guess the number or want to quit
//-----------------------------------------------------------------------------------------------
public static void main(String[] args) {
Scanner scan = new Scanner (System.in);
Random rand = new Random();
int randNum = rand.nextInt(100) + 1;
System.out.println(randNum);
System.out.println("\t\tHi-Lo Game with Numbers\t\t\n\t Guess a number between 1 and 100!!!\n");
String ans;
int attemptsCount = 0;
do {
System.out.print("Guess the number: ");
int input = scan.nextInt();
while(input != randNum){
attemptsCount++;
if(input < randNum){
System.out.println();
System.out.print("low guessing\nEnter new number: ");
input = scan.nextInt();
}
else{
System.out.println();
System.out.print("high guessing\nEnter new number: ");
input = scan.nextInt();
}
}
System.out.println();
System.out.println("Congrats!!! You guess right\nAttempts: "+attemptsCount);
System.out.println();
System.out.print("You want to play again (yes/no): ");
ans = scan.next();
randNum = rand.nextInt(100) + 1; //generate new random number between same above range,
//if the user wants to keep playing.
}while (ans.equalsIgnoreCase("yes"));
}
}
回答by Bohemian
Here's a simple approach:
这是一个简单的方法：
// inside do loop
System.out.print("Guess the number (-1 to quit): ");
int input = scan.nextInt();
if (input == -1) {
break;
}
回答by roeygol
Try using this code for exiting totaly:
尝试使用此代码退出 totaly：
System.exit(0);
