How do I convert a String to a BigInteger?

Question

I'm trying to read some really big numbers from standard input and add them together.

However, to add to BigInteger, I need to use BigInteger.valueOf(long);:

private BigInteger sum = BigInteger.valueOf(0);

private void sum(String newNumber) {
    // BigInteger is immutable, reassign the variable:
    sum = sum.add(BigInteger.valueOf(Long.parseLong(newNumber)));
}

That works fine, but as the BigInteger.valueOf() only takes a long, I cannot add numbers greater than long's max value (9223372036854775807).

Whenever I try to add 9223372036854775808 or more, I get a NumberFormatException (which is completely expected).

Is there something like BigInteger.parseBigInteger(String)?

score 162 · Accepted Answer · edited Apr 26 '23 at 10:22

162

Using the constructor

public BigInteger(String val)

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

Javadoc

edited Apr 26 '23 at 10:22

Joachim Sauer

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answered Mar 30 '13 at 10:03

Uku Loskit

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I tried the same but faced issue coz i missed to import java.math.BigInteger – Arun Aug 19 '18 at 13:50
1

ok, from integer: valueOf, from String, constructor, this API miss some coherence? – pdem Aug 26 '21 at 13:58

score 30 · Answer 2 · edited Jun 20 '20 at 09:12

30

According to the documentation:

BigInteger(String val)

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

It means that you can use a String to initialize a BigInteger object, as shown in the following snippet:

sum = sum.add(new BigInteger(newNumber));

edited Jun 20 '20 at 09:12

Community

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answered Mar 30 '13 at 10:07

Vito Gentile

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score 10 · Answer 3 · edited Jul 18 '19 at 13:44

10

BigInteger has a constructor where you can pass string as an argument.

try below,

private void sum(String newNumber) {
    // BigInteger is immutable, reassign the variable:
    this.sum = this.sum.add(new BigInteger(newNumber));
}

edited Jul 18 '19 at 13:44

fl0w

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answered Mar 30 '13 at 10:12

Darshit Chokshi

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score 9 · Answer 4 · edited Jul 01 '16 at 14:49

9

Instead of using valueOf(long) and parse(), you can directly use the BigInteger constructor that takes a string argument:

BigInteger numBig = new BigInteger("8599825996872482982482982252524684268426846846846846849848418418414141841841984219848941984218942894298421984286289228927948728929829");

That should give you the desired value.

edited Jul 01 '16 at 14:49

Rudy Velthuis

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answered Jul 01 '16 at 11:55

User9123

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score 2 · Answer 5 · edited Oct 01 '17 at 02:56

2

For a loop where you want to convert an array of strings to an array of bigIntegers do this:

String[] unsorted = new String[n]; //array of Strings
BigInteger[] series = new BigInteger[n]; //array of BigIntegers

for(int i=0; i<n; i++){
    series[i] = new BigInteger(unsorted[i]); //convert String to bigInteger
}

edited Oct 01 '17 at 02:56

answered Oct 01 '17 at 02:47

Omar Mohammed

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score 0 · Answer 6 · edited Apr 15 '20 at 15:38

If you may want to convert plaintext (not just numbers) to a BigInteger you will run into an exception, if you just try to: new BigInteger("not a Number")

In this case you could do it like this way:

public  BigInteger stringToBigInteger(String string){
    byte[] asciiCharacters = string.getBytes(StandardCharsets.US_ASCII);
    StringBuilder asciiString = new StringBuilder();
    for(byte asciiCharacter:asciiCharacters){
        asciiString.append(Byte.toString(asciiCharacter));
    }
    BigInteger bigInteger = new BigInteger(asciiString.toString());
    return bigInteger;
}

How do I convert a String to a BigInteger?

6 Answers6

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