該代碼基本上會生成一個 Pincode 并要求用戶撰寫它。然后顯示輸入是否正確。此外,如果用戶在一兩次試驗后得到了正確的答案,它應該顯示“你成功地在(一兩次)試驗中猜到了你的密碼”
package assign4;
import java.util.*;
public class assign4 {
public static void main (String[]args) {
Scanner sc = new Scanner(System.in);
int min = 999;
int max = 10000;
int generatedPin = (int) (Math.random()*(max-min 1) min);
System.out.println("Please enter the pincode sent to you " generatedPin);
int input = sc.nextInt();
if (input == generatedPin) {
System.out.println("you succeeded to guess your pincode in one trial");
}
for(int i = 0; input != generatedPin && i<3; i ) {
if (i==0) {
System.out.println("wrong pin! two trial left!");
input = sc.nextInt();
} else
為什么不if運行以下條件?
if (i==1) {
System.out.println("wrong pin! one trial left!");
input = sc.nextInt();
if (input == generatedPin) {
System.out.println("you succeeded to guess your pincode in 1 trial");
}
}
else
if (i==2)
System.out.println("wrong pin! we will lock your account");
}
}}
uj5u.com熱心網友回復:
正如在您設定的回圈條件中一樣,如果 input == generatedPin 回圈將立即中斷,因此“if”成為無法訪問的代碼
uj5u.com熱心網友回復:
假設我正確理解了要求,以下代碼將滿足您的需求:
public class assign4 {
public static void main(String[] args) {
Scanner sc = new Scanner(System.in);
int min = 999;
int max = 10000;
int generatedPin = (int) (Math.random() * (max - min 1) min);
System.out.println("Please enter the pincode sent to you " generatedPin);
int input = sc.nextInt();
if (input == generatedPin) {
System.out.println("you succeeded to guess your pincode in one trial");
} else {
System.out.println("wrong pin! two trial left!");
for (int i = 0; i < 2; i ) {
if (i == 0) {
input = sc.nextInt();
if (input == generatedPin) {
System.out.println("you succeeded to guess your pincode in two trials");
break;
} else {
System.out.println("wrong pin! one trial left!");
}
} else {
input = sc.nextInt();
if (input == generatedPin) {
System.out.println("you succeeded to guess your pincode in three trials");
} else {
System.out.println("wrong pin! we will lock your account");
}
}
}
}
}
}
就像我的評論一樣,您的代碼難以閱讀,而我則不然。甚至能夠使其編譯。
基本上你需要的是一個運行兩次的 for 回圈,以便用戶總共有 3 次嘗試來猜測密碼。如果在第一個他猜到它你需要存在 for 回圈。如果他再次失敗,for 回圈將運行最后一次。
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