Java mistaking primitive data types?

Question

Problem: I can't store the number '600851475143'. I realize this number is bigger than what an int can hold and is smaller than the maximum long value. However, my program isn't registering the variable "number" as a long, it is registering it as a int. Can someone shed some light onto this problem?

** - Line of the problem

public class Problem3{
//What is the largest prime factor of the number 600851475143
public static void main(String[] args){
  ***long number = 600851475143 , total = 0;
    for(long x = (number-1)/2; x>1; x--)
      if(number%x == 0 && isPrime(x)) total += x;
    System.out.println(total);
}
private static boolean isPrime(long determine){
  for(long x = determine/2 - 1; x>1; x--)
    if(determine%x ==0) return false;
  return true;
}

}

SOLUTION: As Jim said below, in order to of type long, one has to put a "L" or "l" at the end of the number. "An integer literal is of type long if it ends with the letter L or l; otherwise it is of type int. It is recommended that you use the upper case letter L because the lower case letter l is hard to distinguish from the digit 1." - From Oracle site on Primitive types.

A little more info: Java's L number (long) specification

I see 2147483647 in your code, not 600851475143. Not that that should cause number to be an int and not a long... — BoltClock
– BoltClock, Commented May 11, 2012 at 23:46
Sorry, I was trying to solve my issue. The "2147483647" is the max number that can be stored as an int. I realized that any number above this number was causing an error. I edited my original post to avoid potential confusion for people in the future. — user1390463
– user1390463, Commented May 12, 2012 at 3:52

Jim Garrison · Accepted Answer · 2012-05-11 23:48:49Z

12

Long literals need to be expressed with a trailing "L", as in 600851475143L

answered May 11, 2012 at 23:48

Jim Garrison

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1 Comment

user1390463 Over a year ago

Thank you Jim. This helped a lot.

Tiago Peczenyj · Accepted Answer · 2012-05-11 23:47:49Z

4

Put a small 'L' in this literal value:

 600851475143L

The reason is:

An integer literal is of type long if it ends with the letter L or l; otherwise it is of type int. It is recommended that you use the upper case letter L because the lower case letter l is hard to distinguish from the digit 1.

http://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html

You can use this notation too, to be more clear:

 600_851_475_143L

answered May 11, 2012 at 23:47

Tiago Peczenyj

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3 Comments

user207421 Over a year ago

Doesn't have to be a small 'l', it can be a big L too.

Brigham Over a year ago

I recommend against using lowercase L, since it often looks like a 1 (one). Instead, use uppercase L.

Tiago Peczenyj Over a year ago

Yes, sometimes is hard to distinguish l from 1. Thanks

user207421 · Accepted Answer · 2012-05-11 23:48:45Z

4

"my program isn't registering the variable "number" as a long, it is registering it as a int".

That's not correct. You declared it as a long. It's a long.

What you must be getting is something quite different: a compile error on the constant 600851475143. Try 600851475143L. I suggest that if you read the compiler error message more carefully you would have seen that.

answered May 11, 2012 at 23:48

user207421

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2 Comments

user1390463 Over a year ago

Thank you, I did further research after. My compiler only said "Error: integer number too large: 6008514751143".

user207421 Over a year ago

@user1390463 Exactly as I said. The compiler has told you exactly what is wrong. Your first mistake was to mangle what the message told you into "isn't registering the variable "number" as a long".

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