NumberFormatException For input String "11111000000000000000"

Question

I attempted the following command in Scala and get a NumberFormatException, but I am not sure why. It may be something elementary, but I would certainly appreciate an extra set of eyes. Thank you so much!

"11111000000000000000".toLong

java.lang.NumberFormatException: For input string: "11111000000000000000" at java.lang.NumberFormatException.forInputString(NumberFormatException.java:65) at java.lang.Long.parseLong(Long.java:592) at java.lang.Long.parseLong(Long.java:631) at scala.collection.immutable.StringLike$class.toLong(StringLike.scala:276) at scala.collection.immutable.StringOps.toLong(StringOps.scala:30)
... 33 elided

Don't forget to select an answer if you find any of them helpful. — Cary Swoveland, Aug 13 '15 at 23:13

score 6 · Answer 1 · answered Aug 08 '15 at 16:51

The method

Long.parseLong(String s)

parses the string argument as a signed decimal long (uses radix of 10). 11111000000000000000, when treated as a base 10 number is larger than the maximum value of Long, which is why the java.lang.NumberFormatException is being thrown.

Chances are you're looking for

Long.parseLong("11111000000000000000", 2)

which treats the number as binary (base 2).

Otherwise you may want to check out java.math.BigInteger which can handle arbitrary precision integers.

score 3 · Answer 2 · answered Aug 08 '15 at 16:35

3

 9223372036854775807 // maximum value of Long
11111000000000000000 // your value

Notice something?

answered Aug 08 '15 at 16:35

Jörg W Mittag

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Can questions be answered with another question? – Vince Aug 08 '15 at 16:39
@VinceEmigh It is OK as long it is not main part of the answer, so I would say that in this case everything is fine. – Pshemo Aug 08 '15 at 16:44
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@VinceEmigh, evidently, the answer is "yes"? – The Archetypal Paul Aug 08 '15 at 19:09
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It seems to me that +AtomHeartFather found a better way to answer the question in a clear way. He even spotted a potential misunderstanding (radix) and addressed it. – Madoc Aug 08 '15 at 19:33

score 2 · Answer 3 · edited May 23 '17 at 10:26

2

long range for 64bit

–9,223,372,036,854,775,808 to

 9 ,223,372,036,854,775,807

or use BigInteger ex: Using the constructor

String val="111110000000000000000"

BigInteger(String val)

Translates the decimal String representation of a BigInteger into a BigInteger.

reference: BigInteger or not BigInteger?

String to BigInteger java

edited May 23 '17 at 10:26

Community

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answered Aug 08 '15 at 16:51

Piyush Mittal

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NumberFormatException For input String "11111000000000000000"

3 Answers3