If you disable the compiler’s primitive optimization, then integer literals will be BigInteger.
From: Paul King <pa...@asert.com.au> Sent: Monday, December 9, 2019 11:01 AM To: Groovy_Developers <dev@groovy.apache.org> Subject: Re: Groovy multiplication Meta note: This thread is related to the usage of Groovy and best belongs on the users mailing list where it will gain more eyeballs and also more likely to be found in the future by other users in the same scenario. The dev list is for aspects relating to the development of the language itself. But to answer the current question, Integer literals aren't made BigInteger by default but if either side of a multiplication is a BigInteger, the answer will be BigInteger, so you get the correct answer in the posed problem with e.g.: 86400G*1000*30 Cheers, Paul. On Tue, Dec 10, 2019 at 2:48 AM Angelo Schneider <angelo.schnei...@oomentor.de<mailto:angelo.schnei...@oomentor.de>> wrote: Would that not be handled by BigInteger? I assumed integer literals are treated as BigInter just like float/double literals are treated as BigDecimals. Greetings Angelo Am 09.12.2019 um 00:51 schrieb Edmond Kemokai <ekemo...@gmail.com<mailto:ekemo...@gmail.com>>: I'd read up on the MAX_VALUE doc but didn't see any mention of this behavior... Thanks guys!!
