1

I have an assignment in which I need to create a function that tells you how many 1's are in the binary notation of a integer. I did this already by creating my own algorithm. The second step is to use java.math.BigInteger.bitCount() to accomplish the same thing. I looked this up in the Java API but can someone put this into English and explain how it's relevant to finding the number of 1's in the binary notation of an integer, and perhaps also an example. I tried googling but found nothing but the following definition.

public int bitCount()

Description: Returns the number of bits in the two's complement representation of this number that differ from its sign bit. This method is useful when implementing bit-vector style sets atop BigIntegers.

Improve this question
0

5 Answers 5

Reset to default
4

In the two's complement representation of a negative integer, the sign bit is 1, whereas in that of a non-negative integer, the sign bit is 0. So for a positive integer, bitCount() returns the number of bits that are not 0, i.e., that are 1.

Improve this answer
Sign up to request clarification or add additional context in comments.

Comments

1

Read the javadoc again, and then assume a positive number -- which means that the sign bit will be zero.

Once you understand that, think about what to do in the negative case.

Improve this answer

Comments

0 
jcomeau@intrepid:/tmp$ cat test.java; javac test.java; java test
import java.math.BigInteger;
public class test {
 public static void main(String[] args) {
  System.out.println("one bits: " + new BigInteger("0f0f0f0f0f0f0", 16).bitCount());
 }
}
one bits: 24

Modified code for new comment:

jcomeau@intrepid:/tmp$ cat test.java; javac test.java; java test 0xf0f0f0f0f0f0 0x200 200 1 0
import java.math.BigInteger;
public class test {
 public static void main(String[] args) {
  BigInteger number = null;
  for (String arg : args) {
   if (arg.startsWith("0x")) {
    number = new BigInteger(arg.substring(2), 16);
   } else {
    number = new BigInteger(arg);  // decimal by default
   }
   System.out.println("one bits in " + arg + ": " + number.bitCount());
  }
 }
}
one bits in 0xf0f0f0f0f0f0: 24
one bits in 0x200: 1
one bits in 200: 3
one bits in 1: 1
one bits in 0: 0

Obviously, from reading the javadoc you posted, this may or may not conform to what you or your professor expect for negative numbers. This will return 0 "1" bits for negative 1, for example, since all the bits are the same as the sign bit. But it should work for all positive values.

Improve this answer

3 Comments

This generally works, but when I try to get the one bits for an integer that ends in 0 like 10 or 200 for example, it's always gives me the answer as if it's only the first digit e.g. one bits for 200 equals the number of one bits for 2. Do you know why that might be?
200 hex only has one bit, or do you mean 200 decimal? if you used a radix of 16 as I did in the code sample above, you're inputting numbers as hex.
Thanks! Yes I was mistakenly using a radix of 16. That's what was causing the problem.
0

The most significant bit of a signed integer is the sign-bit. If the number is >= 0 then the sign bit is 0. If the number is < 0 then the sign-bit is 1.

So the function gives you the number of bits that different from the sign-bit. Makes the job pretty easy for positive numbers. The answer for negative numbers will also depend on the total numbers of bits being used to represent the number.

Improve this answer

Comments

0

Why don't you just use Integer's bitCount (not BigInteger's)? It returns the number of 1 bits, regardless of sign.

Improve this answer

Comments

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.