## Changing function from returning long to BigInteger

I'm working on a program that finds Fibonacci numbers. The first version of the homework asked for long datatype to be returned and now we have to change our function to return BigInteger. I'm not sure how to change my function to send back BigInteger type. Here is what I have:

public static BigInteger fibonacci_Loop(int f) { BigInteger previous, current; for(int i = 0; i < f; i ++) { BigInteger sum = previous.add(current); previous = current; current = sum; } return previous; }

It won't run because it wants me to initialize previous and current and any time I do it doesn't return the right numbers. I'm not completely sure on how to use BigInteger and any advice would be greatly appreciated.

Below code works for me. Initialize previous to Zero and current to 1 and run the loop likewise. Note that the loop is run one less than the desired fibnacci index.

public static BigInteger fibonacci_Loop(int f) { BigInteger previous = BigInteger.ZERO; BigInteger current = BigInteger.ONE; for(int i = 0; i < f-1; i ++) { BigInteger sum = previous.add(current); previous = current; current = sum; } return previous;

}

**Java.math.BigInteger.valueOf() Method,** valueOf(long val) returns a BigInteger whose value is equal to that of the specified long. val − Value of the BigInteger to return. to Long object Long l = new Long(123456789L); // assign the biginteger value of l to bi // static method is called The java.math.BigInteger.intValue() converts this BigInteger to an integer value. If the value returned by this function is too big to fit into integer value, then it will return only the low-order 32 bits. Further there is chance that this conversion can loose information about the overall magnitude of the BigInteger value.

You can use the constructor that takes a String:

BigInteger i = new BigInteger("0");

But there are constants provided you can use:

BigInteger previous = BigInteger.ZERO; BigInteger current= BigInteger.ONE;

**Java.math.BigInteger.longValue() Method,** longValue() converts this BigInteger to a long. longValue() Method If this BigInteger is too big to fit in a long, only the low-order 64 bits are returned. objects BigInteger bi1, bi2; // create 2 Long objects Long l1, l2; // assign values to bi1, Converts this BigInteger to a long.This conversion is analogous to a narrowing primitive conversion from long to int as defined in section 5.1.3 of The Java™ Language Specification: if this BigInteger is too big to fit in a long, only the low-order 64 bits are returned.

When you're working with longs, they are either a primitive or boxed type so they default to 0 when declared.

java.math.BigInteger however is an Object so you have to initialize it before you use it.

Changing this line `BigInteger previous, current;`

to `BigInteger previous = new BigInteger("0"), current = new BigInteger("1");`

should fix it.

**BigInteger longValue() Method in Java,** longValue() converts this BigInteger to a long value. If the value return by this function is too big to fit into long value, then it will return only the low-order 64 bits. Hi, I'm writing my first program for my second year comsci class and I'm still learning java along the way. Anyways, the signature for one of my functions has to be: "public static BigInteger fNumberRoutes( int n, int m )" So I assume it's a function that returns a BigInteger. Not sure what I'm

**Rule** : First of all you need to initialize any variable declared inside of a function before performing any operation on it.

change our function to return BigInteger

If you only have to return BigInteger then should have only changed the return statement of your previous function with long to

`return new BigInteger(previous);`

**BigInteger (Java Platform SE 8 ),** Converts this BigInteger to a long , checking for lost information. Returns a BigInteger whose value is equivalent to this BigInteger with the designated bit set. The probability that a BigInteger returned by this method is composite does not In this post we are going to show you, how to convert BigInteger to long, int in java. java.math.BigInteger is a class extends java.lang.Number class sometimes we used to keep the large value in BigInteger, mostly in database we use BigInteger and corresponding to that BigInteger use long in java class.

**java.math.BigInteger java code examples,** valueOf(2), BigDecimal.ROUND_FLOOR); // parseInt blows up on overflow as opposed to intValue() which does not. return Long.parseLong(bigMean.toString()) I know that for converting to int we can use Convert.ToInt32 and converting to long we use Convert.ToInt64, but what's about converting to BigInteger? How can I convert a string (that represents a very very long number) to BigInteger?

**Smarandache Sequences, Stereograms and Series, by C. Ashbacher,** toString()+theReturn; theNumberItheNumber.divide(theBase), } return theReturn, Function to convert the input String into the equivalent BigInteger in the input The java.math.BigInteger.valueOf(long val) returns a BigInteger whose value is equal to that of the specified long. This "static factory method" is provided in preference to a (long) constructor because it allows for reuse of frequently used BigIntegers. Declaration. Following is the declaration for java.math.BigInteger.valueOf() method.

**BigInteger Struct (System.Numerics),** long longValue = 6315489358112; BigInteger assignedFromLong = longValue; Console. You can assign a decimal or floating-point value to a BigInteger object if you You can call a static ( Shared in Visual Basic) BigInteger method that by the developer or that are returned by methods that convert unsigned integers The java.math.BigInteger.pow(int exponent) returns a BigInteger whose value is (this exponent). The exponent is an integer rather than a BigInteger. Declaration. Following is the declaration for java.math.BigInteger.pow() method. public BigInteger pow(int exponent) Parameters. exponent − Exponent to which this BigInteger is to be raised