java integer number too large

1. Error:Integer number too large:600851475143 stackoverflow.com

Hi I have written this code: but it will show this error for this line : obj.function(600851475143);

2. why 09 is a too large integer number? stackoverflow.com

They think it is:

3. integer number too large: coderanch.com

Im storing dates in a table using System.currentTimeMillis() format. To read them back I tried this Date currentDatetime = new Date(1133382920156); SimpleDateFormat formatter = new SimpleDateFormat(«hh:mm dd-MMMM-yyyy zz» ); String myDate = formatter.format(currentDatetime); Then display myDate. But im getting integer number too large: error: It works fine if I do this. Date currentDatetime = new Date(System.currentTimeMillis()); SimpleDateFormat formatter = new SimpleDateFormat(«hh:mm .

4. Integer too large? coderanch.com

I am currently getting the following error message: C:javaPoliceDatabase>javac TestApp.java .InvalidRecord.java:70: integer number too large: 0181 if ( (newDate 1220) ) ^ 1 error Could someone explain to me why 0181 is to large? The range as I understand it for an integer is: -2147,483648 to 2,147483,647 which surely means 0181 is fine?

5. integer number too large coderanch.com

6. integer number too large coderanch.com

Hi Experts, I was trying a program to convert the miliseconds we get after from date object (Post Jan 1970) to date object again. Here what I was trying and I am getting interger number too large error. I am using long datatype though. Please advice. import java.util.*; import java.text.*; public class MillisecondToDate < public static void main(String[] args)throws Exception < .

7. integer number too large forums.oracle.com

Does that mean if you declare a long and the number is still within an Integer value its actually an int and not a long? You’re abusing terminology here. Let me rephrase: «Does that mean that a numeric literal is of integer primitive type even if its value lies outside the range of the integer primitives?» Yes. You have to explicitly .

8. Integer number too large and :expected in switch forums.oracle.com

9. Integer number too large? — Mike Myatt forums.oracle.com

Just anticipating your next question: If you write this: long big = 100000000000; You will get a compile-time error message such as «integer number too large: 100000000000». You must suffix the literal value with L for long: long big = 100000000000L; If you want to dirve anyone else who looks at your code crazy, lowercase L does the same thing: long .

When the code above is run, it produces an error on the line obj.function(600851475143); . Why?

8 Answers 8

600851475143 cannot be represented as a 32-bit integer (type int ). It can be represented as a 64-bit integer (type long ). long literals in Java end with an «L»: 600851475143L

Append suffix L : 23423429L .

By default, java interpret all numeral literals as 32-bit integer values. If you want to explicitely specify that this is something bigger then 32-bit integer you should use suffix L for long values.

You need to use a long literal:

But I would expect that function to run out of memory (or time) .

The java compiler tries to interpret 600851475143 as a constant value of type int by default. This causes an error since 600851475143 can not be represented with an int.

To tell the compiler that you want the number interpretet as a long you have to add either l or L after it. Your number should then look like this 600851475143L .

Since some Fonts make it hard to distinguish «1» and lower case «l» from each other you should always use the upper case «L».

I try to put 03145 in as an integer, but the complier keeps on complaining «integer number too large». What can I do to fix it? I can’t chop off the zero in the beginning.

the program works fine if I input «89345», the whole program is too big to be copied in here. This is from the main program

BarCode zip = new BarCode(20138);

5 Answers

You should show the exact lines of code. I’m guessing you are missing a semi-colon or something that will cause the compiler to see it as a much larger number. The leading 0 will treat it as octal, but that just gives you even more digits to work with!

But, something tells me it isn’t the compiler that is complaining. If you cannot cut of the leading 0, perhaps you are talking about reading from a data file, so you are getting a runtime exception. In that case, show the lines of code where you read in the data and your data file. It may be running things together.

I’m not sure what’s wrong.

I use Eclipse IDE for java, and When print the int k where it’s value is 03145, it does not error. It prints 1637

It must be interpreted as a different Base number by java or something

Try this in your browser window:

Parse it into a string, and use a RegExp to check if it has leading 0s, and then just substring it.