Write a program that uses bitwise operations to: (1) generate and display all power-of-two numbers in the range +128 to -128, and (2) display an arbitrary user-input integer. Sample Session: Decimal Binary 128 0000 0000 0000 0000 0000 0000 1000 0000 64 0000 0000 0000 0000 0000 0000 0100 0000 32 0000 0000 0000 0000 0000 0000 0010 0000 16 0000 0000 0000 0000 0000 0000 0001 0000 8 0000 0000 0000 0000 0000 0000 0000 1000 4 0000 0000 0000 0000 0000 0000 0000 0100 2 0000 0000 0000 0000 0000 0000 0000 0010 1 0000 0000 0000 0000 0000 0000 0000 0001 0 0000 0000 0000 0000 0000 0000 0000 0000

C++ for Engineers and Scientists
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ISBN:9781133187844
Author:Bronson, Gary J.
Publisher:Bronson, Gary J.
Chapter5: Repetition Statements
Section5.5: A Closer Look: Loop Programming Techniques
Problem 12E: (Program) Write a program that tests the effectiveness of the rand() library function. Start by...
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This assignment introduces bitwise-manipulation operators (&, |, <<, >>, >>>, ^, and ~),
which are not discussed elsewhere in the book. The bitwise-manipulation operators perform
simultaneous bit manipulations and enable programs to process large quantities of binary
information efficiently. This project is here because the solution uses a conditional operator.
The binary & and | operators can implement bitwise “and” and “or” operations on
corresponding bits in a pair of 32-bit int operands. This bit-manipulation capability enables
Java to efficiently process large quantities of raw binary information. We use this capability to
encrypt information sent over the Internet and to process graphical images. Suppose you have
a 32-bit pattern of 1’s and 0’s in an integer called mask. You can use mask to either set to 1
or reset to 0 any subset of the bits in another integer called data:
• data |= mask;1 drives to 1 all bits in data that correspond to 1 bits in mask.
• data &= mask; drives to 0 all bits in data that correspond to 0 bits in mask.
In particular, you can use a mask having only one 1 bit to see if that particular bit is 1 in data.
If (mask & data) != 0, that data bit is 1. If (mask & data) == 0, that data bit is
0.
The << shift-left operator shifts the bit pattern to the left by a number of bits indicated by the
operand to the right of the operator, and it shifts zero into the right end. Each left shift multiplies
the numerical value of the int operand by 2. The >> arithmetic shift-right operator shifts the
bit pattern to the right by a number of bits indicated by the operand to the right of the operator.
To preserve the sign, it shifts into the left end whatever was there before. Each arithmetic-right
shift divides the numerical value of the int operand by two. The >>> logical shift-right
operator shifts the bit pattern to the right by a number of bits indicated by the operand to the
right of the operator, and it shifts zero into the left end. The >>> operator is the logical opposite
of the << operator.
Java also includes a complement operator, ~. The complement operator simply reverses the
polarity of all bits in the following int operand. Each bit that is 1 becomes 0, and each bit
that is 0 it becomes 1. For example, if number is initially 0, ~number is -1, and vice versa.
 1 data |= mask; is the same as data = (mask | data); and data &= mask; is the same as
data = (mask & data);
Write a program that uses bitwise operations to: (1) generate and display all power-of-two numbers
in the range +128 to -128, and (2) display an arbitrary user-input integer.
Sample Session:
Decimal Binary
128 0000 0000 0000 0000 0000 0000 1000 0000
64 0000 0000 0000 0000 0000 0000 0100 0000
32 0000 0000 0000 0000 0000 0000 0010 0000
16 0000 0000 0000 0000 0000 0000 0001 0000
8 0000 0000 0000 0000 0000 0000 0000 1000
4 0000 0000 0000 0000 0000 0000 0000 0100
2 0000 0000 0000 0000 0000 0000 0000 0010
1 0000 0000 0000 0000 0000 0000 0000 0001
0 0000 0000 0000 0000 0000 0000 0000 0000
-1 1111 1111 1111 1111 1111 1111 1111 1111
-2 1111 1111 1111 1111 1111 1111 1111 1110
-4 1111 1111 1111 1111 1111 1111 1111 1100
-8 1111 1111 1111 1111 1111 1111 1111 1000
-16 1111 1111 1111 1111 1111 1111 1111 0000
-32 1111 1111 1111 1111 1111 1111 1110 0000
-64 1111 1111 1111 1111 1111 1111 1100 0000
-128 1111 1111 1111 1111 1111 1111 1000 0000
Enter any integer: 127
127 0000 0000 0000 0000 0000 0000 0111 1111
Implement your solution using one class. In your class, provide two methods – main and
display.
Your main method should:
• Declare int number = 128;
• Include a while loop that loops while number is >= -128.
• Call the display method, which prints the value of its passed-in number
parameter.
• If number is greater than zero, use the >>= operator to do one arithmetic-right shift.
• If number equals zero, use the ~ operator to complement it.
• If number is less than zero, use the <<= operator to do one left shift.
• After the while loop, ask the user to input any number, and call the display
method to print that number.
Write the display(number) method like this:
• Receive a number parameter.
• Print number’s value and a tab.
• Assign to a local variable named mask the value 1 shifted left 31 times. This puts a 1
in bit 31 and zeros in all other bits.
• Use a for loop to step through all 32 bits, doing the following in each step:
o Use a conditional operator whose condition is (mask & number != 0) to
print either 1 or 0.
o After every fourth bit, print a single space to make the output readable.

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