How can I pass number greater than 999 999 999 as parameter in java

Question

Can I pass number greater than 999 999 999 as parameter in java.

When I do like this it gives compiler error the literal 999 999 999 9 of type is out of range

 passNumber(9999999999);

 public String passNumber(long number){
     if(number > 999999999)
         throw new BigNumberException("Number too large")
 }

See [this](http://stackoverflow.com/questions/8924896/java-long-number-too-large-error) and [this](http://stackoverflow.com/questions/769963/javas-l-number-long-specification-question) questions. — Lion, Jul 15 '12 at 09:27

score 9 · Accepted Answer · answered Jul 15 '12 at 09:21

9

It's because 9,999,999,999 is considered as an int by the compiler and is larger than Integer.MAX_VALUE (2,147,483,647).

You can use a long: 9999999999L.

answered Jul 15 '12 at 09:21

assylias

321,522
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660
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JB Nizet · Answer 2 · 2012-07-15T09:33:58.153

7

Use 9999999999L to tell the compiler it's a long literal, and not an int literal.

edited Jul 15 '12 at 09:33

answered Jul 15 '12 at 09:21

JB Nizet

678,734
91
1,224
1,255

score 4 · Answer 3 · answered Jul 15 '12 at 09:22

4

That's because 9 billion is out of integer range (signed int up to 2,147,483,647 and unsigned int up to 4,294,967,295).

Take a look here to learn more.

answered Jul 15 '12 at 09:22

Steven Lu

41,389
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How can I pass number greater than 999 999 999 as parameter in java

3 Answers3