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Java - Integer highestOneBit() method
Description
The Java Integer highestOneBit() method returns an int value with at most a single one-bit, in the position of the highest-order ("leftmost") one-bit in the specified int value. It returns zero if the specified value has no one-bits in its two's complement binary representation, that is, if it is equal to zero.
Declaration
Following is the declaration for java.lang.Integer.highestOneBit() method
public static int highestOneBit(int i)
Parameters
i − This is the int value.
Return Value
This method returns an int value with a single one-bit, in the position of the highest-order one-bit in the specified value, or zero if the specified value is itself equal to zero.
Exception
NA
Example 1
The following example shows the usage of Integer highestOneBit() method to get an int value with at most a single one-bit in the position of the highest-order. We've created a int variable and assigned it a positive int value. Then using toBinaryString() method, we're printing the binary format of the value. Using bitCount(), we're printing the one bits count and then value of highest one bit is printed using highestOneBit() method.
package com.tutorialspoint; public class IntegerDemo { public static void main(String[] args) { int i = 170; System.out.println("Number = " + i); /* returns the string representation of the unsigned integer value represented by the argument in binary (base 2) */ System.out.println("Binary = " + Integer.toBinaryString(i)); // returns the number of one-bits System.out.println("Number of one bits = " + Integer.bitCount(i)); /* returns an int value with at most a single one-bit, in the position of the highest-order ("leftmost") one-bit in the specified int value */ System.out.println("Highest one bit = " + Integer.highestOneBit(i)); } }
Output
Let us compile and run the above program, this will produce the following result −
Number = 170 Binary = 10101010 Number of one bits = 4 Highest one bit = 128
Example 2
The following example shows the usage of Integer highestOneBit() method to get an int value with at most a single one-bit in the position of the highest-order. We've created a int variable and assigned it a negative int value. Then using toBinaryString() method, we're printing the binary format of the value. Using bitCount(), we're printing the one bits count and then value of highest one bit is printed using highestOneBit() method.
package com.tutorialspoint; public class IntegerDemo { public static void main(String[] args) { int i = -170; System.out.println("Number = " + i); /* returns the string representation of the unsigned integer value represented by the argument in binary (base 2) */ System.out.println("Binary = " + Integer.toBinaryString(i)); // returns the number of one-bits System.out.println("Number of one bits = " + Integer.bitCount(i)); /* returns an int value with at most a single one-bit, in the position of the highest-order ("leftmost") one-bit in the specified int value */ System.out.println("Highest one bit = " + Integer.highestOneBit(i)); } }
Output
Let us compile and run the above program, this will produce the following result −
Number = -170 Binary = 11111111111111111111111101010110 Number of one bits = 28 Highest one bit = -2147483648
Example 3
The following example shows the usage of Integer highestOneBit() method to get an int value with at most a single one-bit in the position of the highest-order. We've created a int variable and assigned it a positive int value. Then using toBinaryString() method, we're printing the binary format of the value. Using bitCount(), we're printing the one bits count and then value of highest one bit is printed using highestOneBit() method.
package com.tutorialspoint; public class IntegerDemo { public static void main(String[] args) { int i = 0; System.out.println("Number = " + i); /* returns the string representation of the unsigned integer value represented by the argument in binary (base 2) */ System.out.println("Binary = " + Integer.toBinaryString(i)); // returns the number of one-bits System.out.println("Number of one bits = " + Integer.bitCount(i)); /* returns an int value with at most a single one-bit, in the position of the highest-order ("leftmost") one-bit in the specified int value */ System.out.println("Highest one bit = " + Integer.highestOneBit(i)); } }
Output
Let us compile and run the above program, this will produce the following result −
Number = 0 Binary = 0 Number of one bits = 0 Highest one bit = 0