What is the most effective way to create BigInteger instance from int value?

Question

I have a method (in 3rd-party library) with BigInteger parameter:

public void setValue (BigInteger value) { ... }

I don't need 'all its power', I only need to work with integers. So, how can I pass integers to this method? My solution is to get string value from int value and then create BigInteger from string:

int i = 123;
setValue (new BigInteger ("" + i));

Are there any other (recommended) ways to do that?

score 52 · Accepted Answer · answered Apr 15 '10 at 14:16

52

BigInteger.valueOf(i);

answered Apr 15 '10 at 14:16

Michael Borgwardt

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Cool :) Didn't noticed that method myself. – Roman Apr 15 '10 at 14:21
I lost by a 6 seconds.. (I just made up that number). – Rosdi Kasim Apr 16 '10 at 04:14

score 8 · Answer 2 · edited Apr 15 '10 at 14:20

8

Use the static method BigInteger.valueOf(long number). int values will be promoted to long automatically.

edited Apr 15 '10 at 14:20

Joey

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answered Apr 15 '10 at 14:16

Bozhidar Batsov

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score 4 · Answer 3 · answered Apr 15 '10 at 14:17

4

You can use this static method: BigInteger.valueOf(long val)

answered Apr 15 '10 at 14:17

Petar Minchev

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score 3 · Answer 4 · edited Aug 26 '15 at 18:32

3

setValue(BigInteger.valueOf(123L));

edited Aug 26 '15 at 18:32

Toby Speight

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answered Apr 15 '10 at 14:17

Paul Croarkin

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What is the most effective way to create BigInteger instance from int value?

4 Answers4