Unreachable Java code

Unreachable Java code

When I try to complie this code, Lines 41-45 give me an "Unreachable code" statement. Same thing happens when I put in a few lines to handle exceptions. Is there something wrong that I am doing? This is a modified example code from the book SAMS Teach yourself Java in 24 hours. Using it as a refresher.

import java.util.*;  import java.util.concurrent.TimeUnit;  public class Clock {    public static void main(String[] arguments) {      Calendar now = Calendar.getInstance();      int hour = now.get(Calendar.HOUR_OF_DAY);      int minute = now.get(Calendar.MINUTE);      int month = now.get(Calendar.MONTH) + 1;      int day = now.get(Calendar.DAY_OF_MONTH);      int year = now.get(Calendar.YEAR);        //Display greeting      if (hour < 12){          System.out.println("Good Morning! \n");      }else if (hour < 17){          System.out.println("Good afternoon! \n");      } else {          System.out.println("Good evening! \n");      }      //Time message start      while(1 < 2){           try              {                  final String os = System.getProperty("os.name");                    if (os.contains("Windows"))                  {                      Runtime.getRuntime().exec("cls");                  }                  else                  {                      Runtime.getRuntime().exec("clear");                  }              }              catch (final Exception e)              {                  //  Handle any exceptions.              }          }  //Errors occur here          try {               TimeUnit.SECONDS.sleep(100);          }   catch (InterruptedException e) {              //Handle exception              }  //Errors end here              System.out.println("The time currently is:" + hour + ":" + minute);              System.out.println("Date: " + month + "/" + day + "/" + year);          }      }  

Answer by Emd4600 for Unreachable Java code

The code there is unreachable, as the while loop condition, 1 < 2, is always true, and so you are always in the while loop. To avoid it, you can:

• Change the while loop condition to something that can be false.
• Add a break statement somewhere in your while loop, to exit it.

Answer by manouti for Unreachable Java code

The while loop will never break since 1 < 2 is always true. Therefore, the part after the while loop will never be reached, hence the compiler error.

Answer by James Wierzba for Unreachable Java code

You have a infinite loop with no break statement.

Answer by optimus_Prime for Unreachable Java code

As your while loop is an infinite loop, so your code will never reach to your this block.

//Errors occur here      try {           TimeUnit.SECONDS.sleep(100);      }   catch (InterruptedException e) {          //Handle exception          }  //Errors end here          System.out.println("The time currently is:" + hour + ":" + minute);          System.out.println("Date: " + month + "/" + day + "/" + year);      }  

As your code

while(1 < 2){       try          {              final String os = System.getProperty("os.name");                if (os.contains("Windows"))              {                  Runtime.getRuntime().exec("cls");              }              else              {                  Runtime.getRuntime().exec("clear");              }          }          catch (final Exception e)          {              //  Handle any exceptions.          }      } // from here again start from loop staring point and looping this again and again  

so in compile time compiler is warning you that your second try block and rest statements are unreachable. you program will never able to execute this blocks. That's why it is saying unreachable.

Answer by lewtenantDan for Unreachable Java code

You have an infinite loop. 1 is always less than 2.

Suggestions:

1. Use while(true) and set breaks
2. Don't put your try/catch in a loop at all. It is only performing one check and the loop seems unnecessary.

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