我的程序规格如下。1.所有四个数字都不同。2.千位数字是十位数的三倍。3.数字是奇数。4.数字的总和是27。我遗漏了整个程序的一些代码。它具有干净的编译器,但运行时会自动终止。我认为问题出在数据类型的转换中。
int randomNumber = rand.nextInt(9000) + 1000; String randomString; boolean found = true; while (found) { // converting to string to find position of digits and value randomString = String.valueOf(randomNumber); // converting to char to transfer back to in while knowing the position of the digits char position0a = randomString.charAt(0); char position1a = randomString.charAt(1); char position2a = randomString.charAt(2); char position3a = randomString.charAt(3); // coverted back to int int position0 = Character.getNumericValue(position0a); int position1 = Character.getNumericValue(position1a); int position2 = Character.getNumericValue(position2a); int position3 = Character.getNumericValue(position3a); int sumCheck = position0a + position1a + position2a + position3a; int digit30Check = 3 * position2; //checking addition to 27 String sumCheck27 = "27"; String sumCheck28 = String.valueOf(sumCheck); // checking all digits are different if (position0 != position1 && position0 != position2 && position0 != position3 && position1 != position2 && position1 != position3 && position2 != position3) { if (position3 != digit30Check) // check if the digit in the thousands place 3 * tens { if (sumCheck27.equals(sumCheck28)) // check if the sum is 27 { if (position0 != 1 && position0 != 3 && position0 != 5 && position0 != 7 && position0 != 9 && position1 != 1 && position1 != 3 && position1 != 5 && position1 != 7 && position1 != 9 && position2 != 2 && position2 != 3 && position2 != 5 && position2 != 7 && position2 != 9 && position3 != 3 && position3 != 3 && position3 != 5 && position3 != 7 && position3 != 9) { // checks for odd digits found = false; System.out.println(randomNumber); } else randomNumber = rand.nextInt(9000) + 1000; } else randomNumber = rand.nextInt(9000) + 1000; } else randomNumber = rand.nextInt(9000) + 1000; } else randomNumber = rand.nextInt(9000) + 1000; // end while }
boolean found = false; while (found)
仅此一项就确保了while循环将永远不会进入,因为它found是false。while循环中的任何内容都没有任何区别,因为它将永远不会执行。
found
你可能想写
while (!found)
除了此错误外,您的情况也过于复杂。这是您可以简化它们的方法:
if ((position0 == (3 * position2)) && // note that position0 is the "thousands place", not position3 ((position0+position1+position2+position3) == 27) && // sum of digits (position3 % 2 == 1) && // odd number (position0 != position1 && position0 != position2 && position0 != position3 && position1 != position2 && position1 != position3 && position2 != position3)) { // different digits found = true; }