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mamruoc

asked on

Compare long

Hi!

seems like I can't do:

long a= 9876543435351;
long b =1232134343344;

if (a > b) {
 // do something
}

found that I had to use compareTo for long, but how?

regards
Avatar of CEHJ
CEHJ
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No - you have to use compareTo for Long, not long

if (a.longValue() > b.longValue()) {
}
(Where 'a' is defined as Long). The alternative is

if (a.compareTo(b) > 0) {
}
... and if they *are* defined as long, then your posted code will work
Avatar of mamruoc
mamruoc

ASKER

Error                  Cannot invoke longValue() on the primitive type long

hm....
a > b is correct for all simple number-carrying variables
compareTo( ) and equals( ) are for reference variables, when comparing (part of) the object's contents.
;JOOP!
Avatar of mamruoc

ASKER

I'm doing:

long time = 0 ;
Date dateObject = new Date(); /* Create dateobject to be able to use linux timestamp */

if((dateObject.getTime() < time)) {

}
>>seems like I can't do:

>>long a= 9876543435351;
>>long b =1232134343344;

Right, you can't since both values are too large to fit in a long


>>I 'm doing:

>>long time = 0 ;
>>Date dateObject = new Date(); /* Create dateobject to be able to use linux timestamp */

>>if((dateObject.getTime() < time)) {

>>}

That's OK. But I think you mean

if((dateObject.getTime() > time))
Avatar of mamruoc

ASKER

I'm doing:

long time = 0 ;
Date dateObject = new Date(); /* Create dateobject to be able to use linux timestamp */

if((dateObject.getTime() < time)) {

}

I'm trying to make a timeout function using linuxtimestamp....

time will be fill with timestamp and added, let's say 2000, for 2 sec timeout....
zzynx, you are confusing int and long: long can contain as much as 2^63
;JOOP!
>>I'm doing:

That's fine

Please post the code that's failing
>> Error               Cannot invoke longValue() on the primitive type long
longValue() must be invoked on a Long (capital L) not on a long (small l)
>> zzynx, you are confusing int and long: long can contain as much as 2^63
Did you try to compile

>>long a= 9876543435351;
>>long b =1232134343344;

It gives me
integer number too large: 9876543435351
integer number too large: 1232134343344
ASKER CERTIFIED SOLUTION
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CEHJ
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I see: that should've been

long a= 9876543435351L;
long b =1232134343344L;

(remark the L at the end)
Avatar of mamruoc

ASKER

public void run() {
      if (DEBUG) System.out.println("UDP AcceptThread accepting messages...");
          try {
            acceptSocket = new DatagramSocket(UDP_IN_PORT);
            byte[] buffer = new byte[defaultBufferLength];
            int counter=0;
            while (true) {
                  incoming = new DatagramPacket(buffer, buffer.length);
                  Date dateObject = new Date(); /* Create dateobject to be able to use linux timestamp */
                  try {
                        if(recording || !playing || acceptingNewStream) { //Save to file, if system busy playback or sending.
                              if(time == 0) { // timeout == 0 for initiate a new stream
                                                      in_baos = new ByteArrayOutputStream();
                                                      time = 1;
                                                }
                                                if((dateObject.getTime() < time) || (time == 1)) {      //time < timeout || timeout == 1
                                                      acceptSocket.receive(incoming); /* Accepts incoming udp stream */
                                                      try {
                                                            in_baos.write(incoming.getData());
                                                            if(incoming.getLength() > 0) {
                                                                  if(time == 1) {
                                                                        System.out.print("Accepting new stream");
                                                                  }
                                                                  time = (dateObject.getTime() + (timeOut*1000));
                                                                  acceptingNewStream = true;
                                                                  System.out.print(".");
                                                            }
                                                      }
                                                      catch (IOException e) {
                                                            System.out.println( "Writing to stream: " + e.getMessage());
                                                      }
                                                }
                                                else {
                                                      System.out.println("Something went wrong: ");
                                                }
                                          }
                                          else { // System not busy
                                                ///TODO: Fix playing when not recording or playback.
                                          }
                                    }
                                    catch (IOException e) {
                                          System.err.println(e);
                                    }
                              } // end while
                        }  // end try
                        catch (SocketException se) {
                              System.err.println(se);
                        }  // end catch
                  } // end run

I got a ticker thread running beside too, to timeout the whole system 'cause incoming = new DatagramPacket(buffer, buffer.length); seems to hang if there is no new incoming....
>>long a= 9876543435351;

should be

long a= 9876543435351L;
Try:

long a = 9876543435351l;
long b = 1232134343344l;

if (a > b) {
 // do something
}

(Notice the "l" at the end of the long numbers)
Said that :°)
>>long a= 9876543435351;
>>long b =1232134343344;

>Right, you can't since both values are too large to fit in a long

Nope they are not. They can easily fit in a long.
Yes, I know.
cf. previous comment.
As the title suggests, this thread is long.
Shouldn't this

        time = (dateObject.getTime() + (timeOut*1000));

be

        time = ((new Date()).getTime() + (timeOut*1000));
Avatar of mamruoc

ASKER

Neede acutally long,

but the suggestion about timout function was just what I needed.. :D

thanks alot everybody
Oh, no longer interested. OK.
Avatar of mamruoc

ASKER

Hi,

no hard feelings.. It's still intresting....

But CEHJ gave me a solution I needed....

:S
8-)
>>no hard feelings
No offence meant.