Why does the constructor of java.util.Date act differently on parameters between int and long?

Question

Here's the scenario:

If I pass an int to Date as below, the output would be Thu Jan 22 20:56:36 CST 1970:

System.out.println(new Date(1393430400*1000));

When I pass a long, the output is just what I intend for: Thu Feb 27 00:00:00 CST 2014

System.out.println(new Date((long)1393430400*1000));

I've checked the method which the new Date(long l) invoked. It's just the same one:

public Date(long date) {
    fastTime = date;
}

Could anyone give me some explanations on this problem? Thanks a lot!

Also you might want to use `123L` for long `123`, not a cast — , Feb 27 '14 at 06:28

score 4 · Accepted Answer · answered Feb 27 '14 at 06:26

4

   System.out.println(new Date(1393430400*1000));

in this 1393430400 as int will overflow. Some different value will get multiply by 1000 and for that you will get the Date.But when you do

    System.out.println(new Date((long)1393430400*1000));

1393430400 this will be casted to long and the multiplication by 1000 will be long and no overflow.

answered Feb 27 '14 at 06:26

Rahul

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Thanks, I didn't realize it. – Judking Feb 27 '14 at 06:31

score 4 · Answer 2 · answered Feb 27 '14 at 06:26

4

System.out.println(new Date(1393430400*1000));

When you do that in crossed the integer range and overflowing happening there. So that is returning standard unix time(Thu Jan 22 20:56:36 CST 1970).

That is not the exact value it's getting parsed.

Where as the line

System.out.println(new Date((long)1393430400*1000));

is clearly indicating/conerting to the long value and the proper parsed date you are getting.

answered Feb 27 '14 at 06:26

Suresh Atta

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@Judking it will generate an compile time error. which JDK version you use ? – Harmeet Singh Taara Feb 27 '14 at 06:47

score 3 · Answer 3 · answered Feb 27 '14 at 06:31

Try printing the int value on its own:

System.out.println(1393430400 * 1000);

The output is 1860996096, which corresponds to the date you are seeing.

This occurs because Java's int is a signed 32-bit integer, meaning that the maximum value holdable is 2^31 - 1. When exceeding that value in calculations, integer overflow occurs, leading to values like the one you see above. For date creation, always use longs rather than ints, as longs are 64-bit and can store up to 2^63 - 1.

Why does the constructor of java.util.Date act differently on parameters between int and long?

3 Answers3