我收到了一段过于复杂且效率极低的代码。我的任务是：
重新考虑此代码，使其效率更高。
do
{
if(noOfQuest<1)
{
System.out.println("FILE_ERROR: Please ensure line \"noOfEmails\" is more than 0");
}
if(noOfQuest==1)
{
if(iCount==0)
{
System.out.println("Count is on ["+iCount+"]");
System.out.println(prefinalMess);
System.out.println(questionArr[0]);
ansArray[0]=ans.next();
System.out.println("Answer 1 where question is ["+noOfQuest+"]" +ansArray[0]);
if(IsAnswer1Blank(ansArray)==true)
{
System.out.println("Please ensure you enter a response");
continue;
}
System.out.println(finalMess);
}
}
if(noOfQuest==2)
{
if(iCount==0)
{
System.out.println("Count is on ["+iCount+"]");
System.out.println(questionArr[0]);
ansArray[0]=ans.next();
if(IsAnswer1Blank(ansArray)==true)
{
System.out.println("Please ensure that you enter a response for question 1");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==1)
{
System.out.println("Count is on ["+iCount+"]");
System.out.println(prefinalMess);
System.out.println(questionArr[1]);
ansArray[1]=ans.next();
if(IsAnswer2Blank(ansArray)==true)
{
System.out.println("Please ensure that you enter a response for question 2");
continue;
}
iCount++;
System.out.println(finalMess);
}
}
if(noOfQuest==3)
{
if(iCount==0)
{
System.out.println("Count is on [" +iCount+"]");
System.out.println(questionArr[0]);
ansArray[0]=ans.next();
if(IsAnswer1Blank(ansArray)==true)
{
System.out.println("Please ensure that you enter a response for question 1");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==1)
{
System.out.println(questionArr[1]);
ansArray[1]=ans.next();
if(IsAnswer2Blank(ansArray)==true)
{
System.out.println("Please ensure that you have entered a response for question 2");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==2)
{
System.out.println("Count is on ["+iCount+"]");
System.out.println(prefinalMess);
System.out.println(questionArr[2]);
ansArray[2]=ans.next();
if(IsAnswer3Blank(ansArray)==true)
{
System.out.println("Please ensure that you have entered a response for question 3");
continue;
}
iCount++;
System.out.println(finalMess);
}
}
if(noOfQuest==4)
{
if(iCount==0)
{
System.out.println("Count is on [" +iCount+"]");
System.out.println(questionArr[0]);
ansArray[0]=ans.next();
if(IsAnswer1Blank(ansArray)==true)
{
System.out.println("Please ensure that you enter a response for question 1");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==1)
{
System.out.println("Count is on [" +iCount+"]");
System.out.println(questionArr[1]);
ansArray[1]=ans.next();
if(IsAnswer2Blank(ansArray)==true)
{
System.out.println("Please ensure that you have entered a response for question 2");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==2)
{
System.out.println("Count is on ["+iCount+"]");
System.out.println(questionArr[2]);
ansArray[2]=ans.next();
if(IsAnswer3Blank(ansArray)==true)
{
System.out.println("Please ensure that you have entered a response for question 3");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==3)
{
System.out.println("Count is on ["+iCount+"]");
System.out.println(prefinalMess);
System.out.println(questionArr[3]);
ansArray[3]=ans.next();
if(IsAnswer4Blank(ansArray)==true)
{
System.out.println("Please ensure that you have entered a response for question 3");
continue;
}
iCount++;
System.out.println(finalMess);
}
}
if(noOfQuest==5)
{
if(iCount==0)
{
System.out.println("Count is on [" +iCount+"]");
System.out.println(questionArr[0]);
ansArray[0]=ans.next();
if(IsAnswer1Blank(ansArray)==true)
{
System.out.println("Please ensure that you enter a response for question 1");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==1)
{
System.out.println("Count is on [" +iCount+"]");
System.out.println(questionArr[1]);
ansArray[1]=ans.next();
if(IsAnswer2Blank(ansArray)==true)
{
System.out.println("Please ensure that you have entered a response for question 2");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==2)
{
System.out.println("Count is on ["+iCount+"]");
System.out.println(questionArr[2]);
ansArray[2]=ans.next();
if(IsAnswer3Blank(ansArray)==true)
{
System.out.println("Please ensure that you have entered a response for question 3");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==3)
{
System.out.println("Count is on ["+iCount+"]");
System.out.println(questionArr[3]);
ansArray[3]=ans.next();
if(IsAnswer4Blank(ansArray)==true)
{
System.out.println("Please ensure that you have entered a response for question 4");
continue;
}
iCount++;
System.out.println(finalMess);
}
if(iCount==4)
{
System.out.println("Count is on ["+iCount+"]");
System.out.println(prefinalMess);
System.out.println(questionArr[4]);
ansArray[4]=ans.next();
if(IsAnswer5Blank(ansArray)==true)
{
System.out.println("Please ensure that you have entered a response for question 5");
continue;
}
}
iCount++;
System.out.println(finalMess);
}
if(iCount==0)
{
if(noOfQuest==0)
{
System.out.println("FILE_ERROR: Please ensure line \"noOfQuest\" in file is between 1 and 5." + "\n" +
"File located: C:/Exercise 7 - Emails/configuration.properties");
}
iCount++;
if(noOfQuest>5)
{
System.out.println("FILE_ERROR: Please ensure line \"noOfQuest\" in file is between 1 and 5." + "\n" +
"File located: C:/Exercise 7 - Emails/configuration.properties");
}
iCount++;
}
if(iCount==1)
{
break;
}
break;
}
while(true);
当变量“noOfQuest”设置为“0”时，应用程序按预期关闭。但是，当变量设置为大于5时，应用程序将挂起。 我该如何解决这个问题？
我已经编辑了这个问题，包括我最终使用的答案，下面是Lawrence Aiello建议的答案（接受答案）。
if(iCount==0)
{
if(noOfQuest==0 || noOfQuest>5)
{
System.out.println("FILE_ERROR: Please ensure line \"noOfQuest\" in file is greater than 0." + "\n" +
"File located: C:/Exercise 7 - Emails/configuration.properties");
System.exit(0);
}
}
重构为for-nested do-while循环以检查所有条件
答案 0 :(得分：1)
因为你的程序处于无限循环中，只有当noOfQuest为0时才会退出。因此，如果它永远不会达到这个值，循环将永远继续，这导致你正在谈论的悬挂
在那里放置一个if语句，为noOfQuest处理大于5的值。
答案 1 :(得分：0)
如果系统地检查不同的离散值（1,2,3，...），请使用switch case语句。还要添加默认路径，以免卡住。为您节省了大量代码。
答案 2 :(得分：0)
我遇到的一个简单的解决方案是以下代码：
if(iCount==0)
{
if(noOfQuest==0 || noOfQuest>5)
{
System.out.println("FILE_ERROR: Please ensure line \"noOfQuest\" in file is greater than 0." + "\n" +
"File located: C:/Exercise 7 - Emails/configuration.properties");
System.exit(0);
}
}
此部分检查iCount是否为0（始终在运行时）并检查问题的数量是0还是大于5.如果是，它将打印消息并关闭应用程序，使用“System.exit（ 0）”。此命令将关闭当前的JVM会话。