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This chapter is from the book

This chapter is from the book

Bitwise Operators (<<, >>, |, &, ^, ~)

An additional set of operators that is common to virtually all programming languages is the set of operators for manipulating values in their binary formats: the bit operators.

Shift Operators (<<, >>, <<=, >>=)

Sometimes you want to shift the binary value of a number to the right or left. In executing a left shift, all bits in a number’s binary representation are shifted to the left by the number of locations specified by the operand on the right of the shift operator. Zeroes are then used to backfill the locations on the right side of the binary number. A right-shift operator does almost the same thing in the opposite direction. However, if the number is a negative value of a signed type, the values used to backfill the left side of the binary number are ones and not zeroes. The shift operators are >> and <<, the right-shift and left-shift operators, respectively. In addition, there are combined shift and assignment operators, <<= and >>=.

Consider the following example. Suppose you had the int value -7, which would have a binary representation of 1111 1111 1111 1111 1111 1111 1111 1001. In Listing 3.37, you right-shift the binary representation of the number –7 by two locations.

Listing 3.37. Using the Right-Shift Operator

int x;
x = (-7 >> 2); // 11111111111111111111111111111001 becomes
               // 11111111111111111111111111111110
// Write out "x is -2."
System.Console.WriteLine("x = {0}.", x);

Output 3.17 shows the results of Listing 3.37.

Output 3.17.

x = -2.

Because of the right shift, the value of the bit in the rightmost location has “dropped off” the edge and the negative bit indicator on the left shifts by two locations to be replaced with 1s. The result is -2.

Although legend has it that x << 2 is faster than x * 4, do not use bit shift operators for multiplication or division. This might have been true in certain C compilers in the 1970s, but modern compilers and modern microprocessors are perfectly capable of optimizing arithmetic. Using shifting for multiplication or division is confusing and frequently leads to errors when code maintainers forget that the shift operators are lower precedence than the arithmetic operators.

Bitwise Operators (&, |, ^)

In some instances, you might need to perform logical operations, such as AND, OR, and XOR, on a bit-by-bit basis for two operands. You do this via the &, |, and ^ operators, respectively.

Listing 3.38 demonstrates how to use these bitwise operators. The results of Listing 3.38 appear in Output 3.18.

Listing 3.38. Using Bitwise Operators

byte and, or, xor;
and = 12 & 7;   // and = 4
or = 12 | 7;    // or = 15
xor = 12 ^ 7;   // xor = 11
    "and = {0} \nor = {1}\nxor = {2}",
    and, or, xor);

Output 3.18.

and = 4
or = 15
xor = 11

In Listing 3.38, the value 7 is the mask; it is used to expose or eliminate specific bits within the first operand using the particular operator expression. Note that, unlike the AND (&&) operator, the & operator always evaluates both sides even if the left portion is false. Similarly, the | version of the OR operator is not “short-circuiting.” It always evaluates both operands even if the left operand is true. The bit versions of the AND and OR operators, therefore, are not short-circuiting.

In order to convert a number to its binary representation, you need to iterate across each bit in a number. Listing 3.39 is an example of a program that converts an integer to a string of its binary representation. The results of Listing 3.39 appear in Output 3.19.

Listing 3.39. Getting a String Representation of a Binary Display

public class BinaryConverter
  public static void Main()
      const int size = 64;
      ulong value;
      char bit;

      System.Console.Write ("Enter an integer: ");
      // Use long.Parse() so as to support negative numbers
      // Assumes unchecked assignment to ulong.
      value = (ulong)long.Parse(System.Console.ReadLine());

      // Set initial mask to 100....
      ulong mask = 1UL << size - 1;
      for (int count = 0; count < size; count++)
          bit = ((mask & value) != 0) ? '1': '0';
          // Shift mask one location over to the right
          mask >>= 1;

Output 3.19.

Enter an integer: 42

Notice that within each iteration of the for loop (discussed later in this chapter), you use the right-shift assignment operator to create a mask corresponding to each bit position in value. By using the & bit operator to mask a particular bit, you can determine whether the bit is set. If the mask test produces a nonzero result, you write 1 to the console; otherwise, 0 is written. In this way, you create output describing the binary value of an unsigned long.

Note also that the parentheses in (mask & value) != 0 are necessary because inequality is higher precedence than the AND operator; without the explicit parentheses this would be equivalent to mask & (value != 0), which does not make any sense; the left side of the & is a ulong and the right side is a bool.

Bitwise Compound Assignment Operators (&=, |=, ^=)

Not surprisingly, you can combine these bitwise operators with assignment operators as follows: &=, |=, and ^=. As a result, you could take a variable, OR it with a number, and assign the result back to the original variable, which Listing 3.40 demonstrates.

Listing 3.40. Using Logical Assignment Operators

byte and = 12, or = 12, xor = 12;
and &= 7;   // and = 4
or  |= 7;   // or = 15
xor ^= 7;   // xor = 11
    "and = {0} \nor = {1}\nxor = {2}",
    and, or, xor);

The results of Listing 3.40 appear in Output 3.20.

Output 3.20.

and = 4
or = 15
xor = 11

Combining a bitmap with a mask using something like fields &= mask clears the bits in fields that are not set in the mask. The opposite, fields &= ~mask, clears out the bits in fields that are set in mask.

Bitwise Complement Operator (~)

The bitwise complement operator takes the complement of each bit in the operand, where the operand can be an int, uint, long, or ulong. ~1, therefore, returns the value with binary notation 1111 1111 1111 1111 1111 1111 1111 1110, and ~(1<<31) returns the number with binary notation 0111 1111 1111 1111 1111 1111 1111 1111.

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