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java - 确保我的 RNG 不会根据用户输入字符串崩溃

转载 作者:行者123 更新时间:2023-12-02 10:24:12 25 4
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我制作了一个随机数生成器,如果有人在没有数字的情况下使用它,它会崩溃如何确保用户输入是否是字符串,它不会崩溃,而是说“无效参数”?

static int random; 
static String string;
static Scanner scanner = new Scanner(System.in);
public static void main(String args[]) {


System.out.println(getRandomNum());
int guessResult =1;
int randomGuess= 0;
while(guessResult != -123) {

System.out.print("Guess a number between 0 and 50");

randomGuess = scanner.nextInt();
guessResult = checkGuess(randomGuess);

}

System.out.println("Yes the random number is " + randomGuess);

}

public static int getRandomNum() {


random = (int)(Math.random() * 51);
return random;
}

public static int checkGuess(int guess) {

if(guess == random) {

return -123;

}else {
System.out.println("please retry " +guess + " was not correct");
return guess;


}

}

}

最佳答案

您可以将输入扫描包装到 try-catch block 中:

try{
randomGuess = scanner.nextInt();
guessResult = checkGuess(randomGuess);
}
catch(Exception ex){
System.out.println("Invalid argument");
}

关于java - 确保我的 RNG 不会根据用户输入字符串崩溃,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/54098573/

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