Java 为什么我的程序要求符号2次而不是1次(忽略第一个符号)?
这是一个简单的刽子手游戏:Java 为什么我的程序要求符号2次而不是1次(忽略第一个符号)?,java,user-input,Java,User Input,这是一个简单的刽子手游戏: public static void main(String[] args) { String[] words = {"writer", "that", "program"}; int wordNumber = (int) (Math.random() * words.length); System.out.print("Enter a letter in word "); for (int i = 0; i < words[w
public static void main(String[] args) {
String[] words = {"writer", "that", "program"};
int wordNumber = (int) (Math.random() * words.length);
System.out.print("Enter a letter in word ");
for (int i = 0; i < words[wordNumber].length(); i++)
System.out.print('*');
System.out.print(" > ");
Scanner input = new Scanner(System.in);
char letter;
do {
letter = input.nextLine().charAt(0);
boolean asterisksInWord = false;
String[] discoveredElements = new String[words[wordNumber].length()];
int countOfTries = 0;
int arrayCount = 0;
int asteriskCount = 0;
System.out.print("Enter a letter in word ");
do {
asterisksInWord = false;
boolean contain;
if (asteriskCount != 1) {
asteriskCount = 0;
for (char item : words[wordNumber].toCharArray()) {
contain = Arrays.asList(discoveredElements).contains(String.valueOf(item));
if (contain) {
System.out.print(item);
} else if (item == letter) {
System.out.print(item);
discoveredElements[arrayCount] = String.valueOf(item);
arrayCount++;
} else {
System.out.print('*');
asterisksInWord = true;
asteriskCount++;
}
}
}
if (asterisksInWord) {
System.out.print(" > ");
letter = input.nextLine().charAt(0);
if (asteriskCount != 1)
System.out.print("Enter a letter in word ");
} else
System.out.println("The word is " + words[wordNumber] +
" You missed " + (countOfTries + 2 - words[wordNumber].length()) + " time(s)");
countOfTries++;
} while (asterisksInWord);
System.out.print("Do you want to guess another word? Enter y or n >");
} while (input.nextLine().charAt(0) == 'y');
}
我的问题是,为什么它在询问“您想猜另一个单词吗?”时会忽略第一个符号“y”
我试着用同样的do-while条件创建一些测试程序,但是他们没有像那个程序那样要求2次字符 程序要求您输入两次,因为您已经这样编程了。打印后-
“您想猜另一个单词吗?输入y或n>”
-您需要在while条件中以及Do
之后的第一条语句中输入input.nextLine()
,因此它要求输入两次
也许您可以将问题在单词中输入字母移动到内部do
循环,而不是在if
条件if(asterisksInWord){
中执行该操作
另外,根据您当前的逻辑,它不会忽略您的第一个符号,即决定是否退出循环的真正符号,下一个输入实际上是您的第一个猜测。我建议您使用chatch=(char)System.in.read();
Enter a letter in word **** > t
Enter a letter in word t**t > h
Enter a letter in word th*t > a
The word is that You missed 0 time(s)
Do you want to guess another word? Enter y or n >y
y
Enter a letter in word **** >