This chapter is from the book
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.
Beginner Topic: Bits and Bytes
All values within a computer are represented in a binary format of 1s and 0s, called binary digits (bits). Bits are grouped together in sets of eight, called bytes. In a byte, each successive bit corresponds to a value of 2 raised to a power, starting from 20 on the right, to 27 on the left, as shown in Figure 3.1.
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Figure 3.1. Corresponding Placeholder Values
In many scenarios, particularly when dealing with low-level or system services, information is retrieved as binary data. In order to manipulate these devices and services, you need to perform manipulations of binary data.
As shown in Figure 3.2, each box corresponds to a value of 2 raised to the power shown. The value of the byte (8-bit number) is the sum of the powers of 2 of all of the eight bits that are set to 1.
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Figure 3.2. Calculating the Value of an Unsigned Byte
The binary translation just described is significantly different for signed numbers. Signed numbers (long, short, int) are represented using a “twos complement” notation. This is so that addition continues to work when adding a negative number to a positive number as though both were positive operands. With this notation, negative numbers behave differently than positive numbers. Negative numbers are identified by a 1 in the leftmost location. If the leftmost location contains a 1, you add the locations with 0s rather than the locations with 1s. Each location corresponds to the negative power of 2 value. Furthermore, from the result, it is also necessary to subtract 1. This is demonstrated in Figure 3.3.
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Figure 3.3. Calculating the Value of a Signed Byte
Therefore, 1111 1111 1111 1111 corresponds to –1 and 1111 1111 1111 1001 holds the value –7. 1000 0000 0000 0000 corresponds to the lowest negative value that a 16-bit integer can hold.
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.
Beginner Topic: Logical Operators Explained
If you have two numbers, as shown in Figure 3.4, the bitwise operations will compare the values of the locations beginning at the leftmost significant value and continuing right until the end. The value of “1” in a location is treated as “true,” and the value of “0” in a location is treated as “false.”
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Figure 3.4. The Numbers 12 and 7 Represented in Binary
Therefore, the bitwise AND of the two values in Figure 3.4 would be the bit-by-bit comparison of bits in the first operand (12) with the bits in the second operand (7), resulting in the binary value 000000100, which is 4. Alternatively, a bitwise OR of the two values would produce 00001111, the binary equivalent of 15. The XOR result would be 00001011, or decimal 11.
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
byteand, or, xor; and = 12 & 7; // and = 4 or = 12 | 7; // or = 15 xor = 12 ^ 7; // xor = 11 System.Console.WriteLine("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 classBinaryConverter {public static voidMain() {const intsize = 64;ulongvalue;charbit; 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....ulongmask = 1UL << size - 1;for(intcount = 0; count < size; count++) { bit = ((mask & value) != 0) ?'1':'0'; System.Console.Write(bit); // Shift mask one location over to the right mask >>= 1; } System.Console.WriteLine(); } }
Output 3.19.
Enter an integer: 42 0000000000000000000000000000000000000000000000000000000000101010
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
byteand = 12, or = 12, xor = 12; and&=7; // and = 4 or|=7; // or = 15 xor^=7; // xor = 11 System.Console.WriteLine("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.