Converting long to int in Java

Question

I'm using System.currentTimeMillis() for getting seconds since the epoch. This is an example.

 long enable_beacon_timestamp = System.currentTimeMillis()/1000;
 println(enable_beacon_timestamp);
 println(int(enable_beacon_timestamp));      
 enable_beacon(int(enable_beacon_timestamp));

And the output gives:

 >>1424876956
 >>1424876928

So the problem is that there is a mismatch in cast value. What I want is to get the first output the same as the integer.

Can you provide some background why this happen?.

That code should not compile. There's no `int` method in Java. — Luiggi Mendoza, Feb 25 '15 at 15:22
Sometimes, values in long will exceed the limit permited in the integers values. Lookup http://docs.oracle.com/javase/7/docs/api/java/math/BigInteger.html in order be able to cast correctly your long. — Kevin Avignon, Feb 25 '15 at 15:24
@KevinAvignon His number example is not exceeding the `Integer.MAX_VALUE` of 2,147,483,647, so it should be fine if he changed it to what I said above. — HelloWorld, Feb 25 '15 at 15:26
@Aradmey That worked! thanks!. Didn't know about that difference. — progloverfan, Feb 25 '15 at 15:28
Still be aware of Kevin Aivgnon's comment about the fact that long can exceed the maximum value of an `Integer` variable. — HelloWorld, Feb 25 '15 at 15:31

score 1 · Accepted Answer · answered Feb 25 '15 at 15:28

1

Your cast syntax is incorrect. You need also be aware that longs can be much bigger that the max value for int.

int y;
if ( enable_beacon_timestamp > (long)Integer.MAX_VALUE ) {
    // long is too big to convert, throw an exception or something useful
}
else {
    y = (int)enable_beacon_timestamp;
}

Try something like this perhaps...

answered Feb 25 '15 at 15:28

Dragan

455
4
12

Good answer but you should also test for underflow (albeit not required in this case). – Bathsheba Feb 25 '15 at 15:43
Good remark, definitely worth mentioning for reference readers tho System.currentTimeMillis() makes it obsolete to do so in this case. – Dragan Feb 25 '15 at 15:46

score 0 · Answer 2 · answered Feb 25 '15 at 15:38

0

you could write something such as

try{

   int castedValue = ( enable_beacon_timestamp < (long)Integer.MAX_VALUE )?(int)enable_beacon_timestamp : -1;
   }
catch (Exception e)
{
   System.out.println(e.getMessage());
}

answered Feb 25 '15 at 15:38

Kevin Avignon

2,853
3
19
40

score 0 · Answer 3 · answered May 18 '21 at 17:55

Java SE 8

To avoid doing the calculations yourself, you can use TimeUnit#convert.
To avoid getting undesirable result due to overflow, you can use Math.toIntExact which throws an exception if the value overflows an int.

Demo:

import java.util.concurrent.TimeUnit;

public class Main {
    public static void main(String[] args) {
        long seconds = TimeUnit.SECONDS.convert(System.currentTimeMillis(), TimeUnit.MILLISECONDS);
        System.out.println(seconds);

        try {
            int secondInt = Math.toIntExact(seconds);
            System.out.println(secondInt);
            // ...
        } catch (ArithmeticException e) {
            System.out.println("Encountered error while casting.");
        }
    }
}

Output:

1621358400
1621358400

Converting long to int in Java

3 Answers3

Java SE 8