我的 while 循环中的条件似乎不正确？

Question

尝试编写一个程序，投掷 3 次骰子，一旦连续获得 3 个 6，它就会打印出尝试了多少次。代码末尾有问题，|| 之间的差异而&&，好像是相反的，看看...

package javaapplication12;
import java.util.*;

public class JavaApplication12 {
public static void main(String[] args) {
  Random rand = new Random();
  // this should try to cast a die 3 times and keep doing that until we get 3 6s in a row and counter 
  // how many tries it takes to get that.
  int array[] = new int[3]; // creating 3 places for castin our die 3 times in a row
  int counter1 = 0; // creating a counter to track each time we try to cast 3 die ( 3 cast = 1 try)
  do{

      counter1++; 
      System.out.println("try " + counter1); // prints out counter in the beginning of every try.

      for (int counter = 0; counter < array.length; counter++ ){
          array[counter]=(rand.nextInt(6)+1);  // a loop that fills the array with random numbers from 1-6
      }
      for(int x: array)
          {System.out.println(x);} // this is for our own to check if we have 3 6s in a row,
                                   // prints out all the numbers in out array
  }

  //so this is the I'veusing part, i've written 3 scenarios that can be written for the condtion in our
  // do- while loop...

  while (array[0]+array[1]+array[2] != 18); // this works just fine.

  while (array[0] !=6 || array[1] != 6 || array[2] != 6); // imo this should not work but surprisingly it does

  while (array[0] !=6 && array[1] != 6 && array[2] != 6); // this should work, but it doesnt.


  } 
}

Answer 1

你的第一个条件是：总和不是 18，所以它确实应该可以正常工作。

第二个条件是：至少一个die不是6个，所以应该也能正常工作。

你的最后一个条件：没有骰子是6，所以即使一个骰子掷出6，它也会坏掉，所以它不起作用是有道理的。

更新： AND (&&) 表示参数的两边都必须为真，输出才为真， OR (||) 意味着参数的一侧需要为真才能使结果为真。

您需要重温boolean algebra的基本操作。

Answer 2

它应该是这样工作的。这个：

do{
}while(array[0] !=6 && array[1] != 6 && array[2] != 6)

意味着只要条件为真，循环就会继续。换句话说——如果任一部分被评估为假，那么循环将停止。所以如果 array[0]==6 那么循环就会中断。

Answer 3

 while (array[0] !=6 || array[1] != 6 || array[2] != 6);

这很好用，因为它意味着："while one of the three attempts is not 6"

所以退出条件是："all the die are 6"

while (array[0] !=6 && array[1] != 6 && array[2] != 6);

这不起作用，因为它意味着："while neither of the three attempts is 6"

所以退出条件是："at least one die is 6"

 while (array[0]+array[1]+array[2] != 18);

这很好用，因为它意味着："the die result sum is 18"

所以退出条件是："all the die are 6"（因为这是求和 18 的唯一方法）

Answer 4

我相信您的困惑来自 De Morgan's laws of negation。基本上，当您对一组条件进行否定时，您应该将 && 更改为 ||，反之亦然，当您对单个条件进行否定时。

或者简单地说：

!(A && B) == !A || !B
!(A || B) == !A && !B

在你的情况下，你想这样做：

!(array[0] == 6 && array[1] == 6 && array[2] == 6)

又名。 "while it's not true that the first is 6 AND the second is 6 AND the third is six"

由于上面的规律，为了把!带进去，需要把&&换成||，导致

!(array[0] == 6) || !(array[1] == 6) || !(array[2] == 6)

简化为

array[0] != 6 || array[1] != 6 || array[2] != 6

Answer 5

一般：

do{
    something....
}while(some condition is true);

如果要用'&&'运算符表示

do{
    something....
}
while (!(array[0] == 6 && array[1] == 6 && array[2] == 6));

意思是

  do { 
     something...
  }while(the expression every element in the array is equal to 6 is false);

我的 while 循环中的条件似乎不正确？

the condition in my while loop doesn't seem right?

java

die