我試圖提示用戶輸入一個數字，然后詢問他們是否要輸入另一個數字。如果是，則他們重新啟動第一個 while 循環，如果不是，則他們退出以打印他們選擇的數字。但是似乎無論我做什么，我都無法擺脫第一個循環。它只是一遍又一遍地重復數字。我試著把 userInput = "y", break, and continue 放在 where valid= true; 的末尾 目前是，它們都產生相同的結果。陷入無限循環。ArrayList<Integer> inputs = new ArrayList<Integer>();System.out.println("Enter some numbers: "); String userInput = "y"; boolean valid = false; do { while (scnr.hasNextInt()) { inputs.add(scnr.nextInt()); System.out.println("Would you like to enter another?"); valid = true; } System.out.println(inputs); } while (!userInput.equalsIgnoreCase("n"));
2 回答
料青山看我應如是
TA貢獻1772條經驗 獲得超8個贊
問題是你沒有跳出內循環。在這里我將如何編寫相同的代碼：
import java.util.ArrayList;
import java.util.Scanner;
public class ReadNumbers {
public static void main(String[] args) {
ArrayList<Integer> inputs = new ArrayList<Integer>();
System.out.println("Enter some numbers: ");
try (Scanner scnr = new Scanner(System.in)) {
do {
inputs.add(scnr.nextInt());
System.out.println("Would you like to enter another y/n?");
} while (scnr.next().equalsIgnoreCase("y"));
}
System.out.println(inputs);
}
}
這會產生以下輸出：
Enter some numbers:
14
Would you like to enter another y/n?
y
15
Would you like to enter another y/n?
y
17
Would you like to enter another y/n?
y
44
Would you like to enter another y/n?
n
[14, 15, 17, 44]
讀取數字的另一種方法是從空格分隔的行中讀取幾個：
慕姐4208626
TA貢獻1852條經驗 獲得超7個贊
我認為你把這個復雜化了。您只需要一個循環，該循環一直運行到用戶輸入“n”為止。
在此之前，要求用戶輸入一個數字，讀入，然后詢問他們是否要繼續。適當地更新循環條件，你就完成了：
public static void main(String... args) {
ArrayList<Integer> inputs = new ArrayList<Integer>();
try (Scanner scnr = new Scanner(System.in)) {
boolean valid = true;
while (valid) {
System.out.println("Enter some numbers: ");
if (scnr.hasNextInt()) {
inputs.add(scnr.nextInt());
}
System.out.println("Would you like to enter another?");
String response = scnr.next();
valid = response.trim().equalsIgnoreCase("y");
}
}
System.out.println(inputs);
}
產生輸出
Enter some numbers:
1
Would you like to enter another?
y
Enter some numbers:
2
Would you like to enter another?
y
Enter some numbers:
3
Would you like to enter another?
n
[1, 2, 3]
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