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javastringbiginteger
Twinone
Twinone

Reputation: 3029

How do I convert a String to a BigInteger?

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)?

Upvotes: 100

Views: 255006

Answers (6)

Uku Loskit
Uku Loskit

Reputation: 42050

Using the constructor

public BigInteger(String val)

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

Javadoc

Upvotes: 166

Aaron
Aaron

Reputation: 1

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;
}

Upvotes: 0

Darshit Chokshi
Darshit Chokshi

Reputation: 589

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));
}

Upvotes: 10

Omar Mohammed
Omar Mohammed

Reputation: 21

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
}

Upvotes: 2

User9123
User9123

Reputation: 101

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.

Upvotes: 10

Vito Gentile
Vito Gentile

Reputation: 14426

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));

Upvotes: 30

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