Java compiler shows error that "integer is too large" [duplicate]

Question

I am writing a code in which I am trying to assign a value in long variable. But java compiler is showing error that too large integer number. I am trying to store 600851475143 in long type still.

class Sum {
    static public void main(String args[]){
        long num=600851475143;
    }
}

Note that it is standard practice to put public before static: public static void main(...) — assylias
– assylias, Commented Feb 16, 2013 at 19:06
When asking questions, remember to also post the real [compiler] error message and, better, search for the error before asking a question .. — user166390
– user166390, Commented Feb 16, 2013 at 19:11
stackoverflow.com/questions/3757763/… , stackoverflow.com/questions/8924896/… - from [java] integer too large search .. — user166390
– user166390, Commented Feb 16, 2013 at 19:13

PermGenError · Accepted Answer · 2013-02-16 19:05:25Z

9

append 'L' or 'l' at the end of the number to make it a long literal.you can use both lowercase(l)or uppercase(L), but uppercase(L) is recommend for readability.

 long num=600851475143L;

answered Feb 16, 2013 at 19:05

PermGenError

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Comments

Subhrajyoti Majumder · Accepted Answer · 2013-02-16 19:23:08Z

3

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.

Reference

So use this -

long num=600851475143l;

or better

long num=600851475143L;

edited Feb 16, 2013 at 19:23

answered Feb 16, 2013 at 19:06

Subhrajyoti Majumder

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3 Comments

jlordo Over a year ago

I prefer a capital L, because l looks to similar to 1.

Subhrajyoti Majumder Over a year ago

Actually it is recommended. :)

jlordo Over a year ago

I know. Wrote the comment long before you mentioned it in the answer ;)

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