Java中的try-catch-inside-while循环存在问题
我是新来的,我正在学习Java。在我的一个程序中,我做了一个猜谜游戏。猜测游戏应该一直要求用户输入猜测,直到他们猜对数字为止 这是我的代码:Java中的try-catch-inside-while循环存在问题,java,Java,我是新来的,我正在学习Java。在我的一个程序中,我做了一个猜谜游戏。猜测游戏应该一直要求用户输入猜测,直到他们猜对数字为止 这是我的代码: import java.util.InputMismatchException; import java.util.Random; import java.util.Scanner; public class Main { public static void main(String[] args) { final int min
import java.util.InputMismatchException;
import java.util.Random;
import java.util.Scanner;
public class Main {
public static void main(String[] args) {
final int minValue = 1;
final int maxValue = 10;
final boolean displayHints = true; // Display whether the number is too high or too low when guessed incorrectly?
int tries = 1;
int guess = 0; // We need to give 'guess' a (temporary) value or else the 'while' loop will create an error
boolean error = false;
Random generator = new Random(); // Create scanner 'generator'
int random = generator.nextInt(maxValue) + minValue; // Define 'random' variable with a random value
if (random == guess) { // In case 'random' = 'guess'
guess = -852654;
}
Scanner input = new Scanner(System.in); // Create a scanner
System.out.println("Random number: " + random); // Hey, no cheating! (for debugging purposes)
System.out.println("Try to guess the magic number! (from " + minValue + " to " + maxValue + ")");
while (random != guess) {
do { // Supposed to ask the user to input a number until they enter a valid number. This is the part of the code that is not working.
System.out.println("\nInput your guess now!");
try {
guess = input.nextInt();
error = false;
} catch (InputMismatchException e) {
System.err.println("That's not a number!\n");
error = true;
continue;
}
} while (error);
if (guess == random) {
System.out.println("Correct!");
System.out.println("Number of tries: " + tries + ".");
input.close();
} else {
tries++;
if (displayHints) {
if (guess < random) {
System.out.println("Sorry, too low!");
} else if (guess > random) { // not strictly necessary
System.out.println("Sorry, too high!");
}
} else {
System.out.println("Sorry, that was not the right number");
}
}
}
}
}
代码的其余部分工作正常。您忘记使用错误的输入。尝试使用
catch
块中输入错误的行
} catch (InputMismatchException e) {
System.err.println("That's not a number!\n");
error = true;
String notANumber = input.nextLine(); // add
continue;
}
而且,println
已经在正在打印的字符串末尾添加了一个换行符,因此无需向正在打印的字符串添加额外的\n
字符
通过上述更改,下面是do while循环的输入/输出示例:
Input your guess now!
banana
That's not a number!
Input your guess now!
8
正如Targetman所解释的,您需要使用错误的输入,因为如果出现
inputmaschException
,则不会使用令牌
另一种解决方案是使用hasnetint()
,将您从try/catch
块中解救出来:
扫描仪实际上从未获得有效的输入,因此当您到达
guess=input.nextInt()时,它会反复抓取香蕉代码>
我的解决办法是将输入作为字符串读入,并将其解析为整数。然后您只需要捕获一个NumberFormatException
,而不是inputmaschException
我会这样做:
try {
guess = Integer.parseInt(input.next());
error = false;
} catch (NumberFormatException e) {
System.err.println("That's not a number!\n");
error = true;
}
正如许多人提到的,您需要使用错误的输入。我遇到了一个非常类似的问题,我在这里没有看到合适的回答,但我在别处找到了一个
试着把下面的线放在你的挡块末端
input.nextLine();
这将清除缓冲区并解决您的问题。最简单的方法就是更改
guess = input.nextInt();
到
这将解决问题,只需更改一小行代码。复制并尝试它
但我还是觉得你的代码看起来很乱。我会这样做
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
Random r = new Random ();
int x = r.nextInt(10);
int y = 0;
int counter=0;
do{
System.out.println("Guess a number between 0-10: ");
try{
y = Integer.valueOf(input.next());
}catch (Exception e){
System.out.println("That is not a number ");
continue;
}
counter ++;
if (counter>5){
System.out.println("So you still don't know how to guess quicker?");
}
if (y<x){
System.out.println("You gessed wrong, the number is higher");
}
else if (y>x){
System.out.println("You gessed wrong, the number is lower");
}
else if (y==x)
System.out.println("You gessed right, the number is: " + x);
}while(y!=x);
System.out.println("You guessed the number in: " + counter + " times");
if(counter <=4){
System.out.println("You found out how to guess the number quickly");
}
}
publicstaticvoidmain(字符串[]args){
扫描仪输入=新扫描仪(System.in);
Random r=新的Random();
int x=r.nextInt(10);
int y=0;
int计数器=0;
做{
System.out.println(“猜一个介于0-10之间的数字:”);
试一试{
y=整数.valueOf(input.next());
}捕获(例外e){
System.out.println(“这不是一个数字”);
继续;
}
计数器++;
如果(计数器>5){
System.out.println(“那么你还不知道如何更快地猜?”;
}
如果(yx){
System.out.println(“您选错了,数字更低”);
}
else如果(y==x)
System.out.println(“你选对了,数字是:”+x);
}而(y!=x);
System.out.println(“您猜到了:“+counter+”times”中的数字);
if(计数器)\n
将给出一个空行作为间隔。谢谢!但是,请注意,input.readLine();
应该是input.nextLine();
。这就是我最后做的:if(input.hasNextInt()){guess=input.nextInt();error=false;}else{System.err.println(“这不是一个数字!”);if(input.hasNextLine()){input.nextLine();}错误=true;}
guess = input.nextInt();
guess = Integer.valueOf(input.next());
public static void main(String[] args) {
Scanner input = new Scanner(System.in);
Random r = new Random ();
int x = r.nextInt(10);
int y = 0;
int counter=0;
do{
System.out.println("Guess a number between 0-10: ");
try{
y = Integer.valueOf(input.next());
}catch (Exception e){
System.out.println("That is not a number ");
continue;
}
counter ++;
if (counter>5){
System.out.println("So you still don't know how to guess quicker?");
}
if (y<x){
System.out.println("You gessed wrong, the number is higher");
}
else if (y>x){
System.out.println("You gessed wrong, the number is lower");
}
else if (y==x)
System.out.println("You gessed right, the number is: " + x);
}while(y!=x);
System.out.println("You guessed the number in: " + counter + " times");
if(counter <=4){
System.out.println("You found out how to guess the number quickly");
}
}