This question already has answers here:
How does Java handle integer underflows and overflows and how would you check for it?
(12 answers)
Do Java arrays have a maximum size?
(10 answers)
Closed last month.
I was trying to create a int[] type array in Java with a user input based length. Below is the code for the same :
int[] change = new int[max+1];
When the value of max = 2147483647, the compiler shows the below error code :
type here
java.lang.NegativeArraySizeException: -2147483648
I know that the upper limit for int in Java is -2147483648 to 2147483647 (-231 to 231-1), but why is the size of array going negative with a addition (is it looping back to the lower limit?).
The problem remains the same with long[] type array. So, does Java not support array size above 231-1 ?
This question already has answers here:
Initialize a long in Java
(4 answers)
Closed 8 years ago.
When declaring the number 0xFFFFFFFF (4294967295) as a long, java will set the variable to the number -1.
public static final long Bits32 = 0xFFFFFFFF; //Bits32 = -1
I assume that java initially converts the hexadecimal to an integer and then converts this to the long variable.
The obvious work around would to set Bits32 to the number 4294967295 instead. However this doesn't seem like a neat solution to me.
Dose anyone know how I would be able to declare a long to this number without having to manually convert the hexadecimal?
Cheers,
Chris.
Define as a Hex Long:
long Bits32 = 0xFFFFFFFFL; //Bits32 = -1
System.out.println(Bits32);
System.out.println(Long.toHexString(Bits32));
Note the L at the end of the 0xFFFFFFFF
The output is
4294967295
ffffffff
This question already has answers here:
How to handle numbers bigger than machine representative number?
(3 answers)
Closed 8 years ago.
I want to store a very big number of range 15477096172227810860 approximate. This number is store in a String and when i am converting it into Long its giving me NumberFormatException. I think long does not sport that much lengthy value. Please give me some solution.
Thanks
Java BigInteger or BigDecimal can be used for large numbers. It will also be easy to declare from a String.
BigInteger num = new BigInteger("15477096172227810860");
Note that it's not a "basic datatype" (primitive), so you will have to call methods to get the value or "modify" it. (It's immutable fyi). As mentioned by other answers, the length of the number is limited by the memory available.
A BigInteger can store an arbitrarily large number (usually the memory available sets the limit).
Best part (and only?) is to use BigInteger
Though, it depends on what you wnt to do with that number
This question already has answers here:
Is there an upper bound to BigInteger? [duplicate]
(3 answers)
What does BigInteger having no limit mean?
(4 answers)
Closed 9 years ago.
I was multiplying very two huge BigIntegervalues in a program. It failed. What are the limits of BigInteger and BigDecimal ?
You won't get NumberFormatException multiplying large numbers. If the number produced is too large, you will get a cryptic NegativeArraySizeException as the size of the array overflows.
You are more likely to get an out of memory error.
The limit is 32 * 2^32-1 bits for BigInteger or about 2^(4 billion).
You can get a NumberFormatException if you
create a BigInteger from an empty byte[]
use a signum < -1 or > +1
try to parse a number in base >36 or < 2
have a string with illegal digits.
When you get an exception you should also look at the message and the stack trace as this usually gives you the real cause.
there shouldn't be a limit, except for memory, but maybe there is, according to the implementation of the class (for example, some fields there might be int or long).
This question already has answers here:
How does Java handle integer underflows and overflows and how would you check for it?
(12 answers)
Closed 9 years ago.
I am using the Long primitive type which increments by 1 whenever my 'generateNumber'method called. What happens if Long reaches to his maximum limit? will throw any exception or will reset to minimum value?
here is my sample code:
class LongTest {
private static long increment;
public static long generateNumber(){
++increment;
return increment;
}
}
Long.MAX_VALUE is 9,223,372,036,854,775,807.
If you were executing your function once per nanosecond, it would still take over 292 years to encounter this situation according to this source.
When that happens, it'll just wrap around to Long.MIN_VALUE, or -9,223,372,036,854,775,808 as others have said.
It will overflow and wrap around to Long.MIN_VALUE.
Its not too likely though. Even if you increment 1,000,000 times per second it will take about 300,000 years to overflow.
Ranges from -9,223,372,036,854,775,808 to +9,223,372,036,854,775,807.
It will start from -9,223,372,036,854,775,808
Long.MIN_VALUE.
Exceding the maximum value of a long doesnt throw an exception, instead it cicles back. If you do this:
Long.MAX_VALUE + 1
you will notice that the result is the equivalent to Long.MIN_VALUE.
From here: java number exceeds long.max_value - how to detect?