The constructor BigInteger(long) is not visible

Question

So I'm creating a factorial program using BigInteger class. But I keep getting the above error.

public static BigInteger fact(long n){
        BigInteger result = BigInteger.ONE;
        for(int i = 1; i <= n; ++i){
            result = result.multiply(new BigInteger(i));
        }
        return result;
}

I already found the fix which is just add an empty string with result.

result = result.multiply(new BigInteger(i + ""))

My question is, why do we have to add that empty string ?

Answer 1

As per oracle docs, BigInteger does not have any constructor that takes int as an argument

Secondly you should use BigInteger.valueOf(i); instead of new BigInteger(i + "")

Answer 2

It is showing the correct error. If you look at the BigInteger class in the documentation , then you can see that there is no constructor that accepts an int . So you can't create BigInteger object by using new and passing an int .

There is a constructor that accepts a string, and by adding an empty string, you are casting your int to string.

You can use the following code:

int myint = 5; // For example
BigInteger myBigInter = BigInteger.valueOf(myint);

Answer 3

When you add an empty string to your constructor parameter, the java compiler transform your parameter to string : 5 + "" -> "5". the resulting consequence is java will use the BigInterger constructor with String parameter. This is why your code works.

 /**
     * Translates the decimal String representation of a BigInteger into a
     * BigInteger.  The String representation consists of an optional minus
     * sign followed by a sequence of one or more decimal digits.  The
     * character-to-digit mapping is provided by {@code Character.digit}.
     * The String may not contain any extraneous characters (whitespace, for
     * example).
     *
     * @param val decimal String representation of BigInteger.
     * @throws NumberFormatException {@code val} is not a valid representation
     *         of a BigInteger.
     * @see    Character#digit
     */
    public BigInteger(String val) {
        this(val, 10);
    }

So to use a clean code use this :

new BigInteger(Integer.toString(i), 10)

Answer 4

123+"" is same as Integer.toString(123)+"" ie adding an empty string to an integer is converting that integer to string.

And as per docs BigInteger has a constructor that takes the decimal string of an integer as an argument.

And as mentioned by SpringLearner BigInteger doesn't have a constructor that takes just int as an argument.

Answer 5

BigInteger constructor with long is private because the library developer wants you to invoke BigInteger.valueOf(long l) . The reason why constructor is private is provided in javadoc of valueOf,

Returns a BigInteger whose value is equal to that of the specified {@code long}. This "static factory method" is provided in preference to a ({@code long}) constructor because it allows for reuse of frequently used BigIntegers.

Answer 6

BigInteger has no constructor that accepts int . By adding an empty string, i is converted to a String .

You could also do something like new BigInteger(String.valueOf(i))

Edit: Why do you have to use a String ? BigInteger is (usually) used if you need bigger values than Integer.MAX_VALUE or even Long.MAX_VALUE . So, if you want to create a BigInteger that holds a value greater than that, you need to pass it as an object/variable that can hold that value. And a String is able to hold it.