解析长到负数

Question

code:

public class Main{
    public static void main(String[] a){
        long t=24*1000*3600;
        System.out.println(t*25);
        System.out.println(24*1000*3600*25);
    }
}

This prints :

2160000000

-2134967296

Why?

---

Thanks for all the replies.

Is the only way to use L after the number?

I have tried the (long)24*1000*3600*25 but this is also negative.

Answer 1

You reached the max of the int type which is Integer.MAX_VALUE or 2^31-1. It wrapped because of this, thus showing you a negative number.

For an instant explanation of this, see this comic:

Answer 2

To explain it clearly,

System.out.println(24*1000*3600*25);

In the above statement are actually int literals. To make treat them as a long literal you need to suffix those with L .

System.out.println(24L*1000L*3600L*25L);

Caveat, a small l will suffice too, but that looks like capital I or 1 , sometimes. Capital I doesn't make much sense here, but reading that as 1 can really give hard time. Furthermore, Even sufficing a single value with L will make the result long .

Answer 3

In the first case you are printing a long but in the second, you are printing it as int .

And int has a range from: -2^31 to 2^31 - 1 which is just below what you are calculating (int max: 2147483647 you: 2160000000) so you overflow the int to the negative range.

You can force the second one to use long as well:

System.out.println(24L*1000*3600*25);

Answer 4

You should suffix the numbers with 'l'. Check the snippet below:

   public static void main(String[] a){
        long t=24*1000*3600;
        System.out.println(t*25);
        System.out.println(24l*1000l*3600l*25l);
    }

Answer 5

Integral literals are treated as type int by default. 24*1000*3600*25 is greater than Integer.MAX_VALUE so overflows and evaluates to -2134967296. You need to explicitly make one of them a long using the L suffix to get the right result:

System.out.println(24L*1000*3600*25);

Answer 6

If you want to do mathematical operations with large numerical values without over flowing, try the BigDecimal class.

Let's say I want to multiply

200,000,000 * 2,000,000,000,000,000,000L * 20,000,000

int testValue = 200000000;
System.out.println("After Standard Multiplication = " +
                                                       testValue * 
                                                       2000000000000000000L * 
                                                       20000000);

The value of the operation will be -4176287866323730432, which is incorrect.

By using the BigDecimal class you can eliminate the dropped bits and get the correct result.

int testValue = 200000000;        
System.out.println("After BigDecimal Multiplication = " +
                              decimalValue.multiply(
                              BigDecimal.valueOf(2000000000000000000L).multiply(
                              BigDecimal.valueOf(testValue))));

After using the BigDecimal, the multiplication returns the correct result which is

80000000000000000000000000000000000

Answer 7

(int)Long.valueOf("2345678901").longValue();