Appendix. Bit Operators

Size: px

Start display at page:

Download "Appendix. Bit Operators"

Rachel Strickland
6 years ago
Views:

Appendix C Bit Operations C++ operates with data entities, such as character, integer, and double-precision constants and variables, that can be stored as 1 or more bytes.

1 Appendix C Bit Operations C++ operates with data entities, such as character, integer, and double-precision constants and variables, that can be stored as 1 or more bytes. In addition, C++ provides for manipulating bits of character and integer constants and variables. Generally, these bit manipulations are used in computer science and engineering applications and typically aren t required in commercial applications. The operators used to perform bit manipulations are called bit operators and are listed in Table C.1. Table C.1 Operator & ^ ~ << >> Bit Operators Description Bit-by-bit AND Bit-by-bit inclusive OR Bit-by-bit exclusive OR Bit-by-bit ones complement Left shift Right shift

2 C-2 Bit Operations All the operators listed in Table C.1, except ~, are binary operators, requiring two operands. Each operand is treated as a binary number consisting of a series of 1s and 0s. The bits in each operand are then compared on a bit-by-bit basis, and the result is determined based on the selected operation. The AND Operator The AND operator, &, causes a bit-by-bit AND comparison between its two operands. The result of each bit-by-bit comparison is a 1 only when both bits being compared are 1s; otherwise, the result of the AND operation is a 0. For example, the following two 8-bit numbers are to be ANDed: 1ƒ0ƒ1ƒ1ƒ0ƒ0ƒ1ƒ1 1ƒ1ƒ0ƒ1ƒ0ƒ1ƒ0ƒ1 ƒ ƒ ƒ ƒ ƒ ƒ ƒ To perform an AND operation, each bit in one operand is compared with the bit occupying the same position in the other operand. The following sample AND operation illustrates the correspondence between bits for these two operands: ƒƒ1ƒ0ƒ1ƒ1ƒ0ƒ0ƒ1ƒ1 &ƒ1ƒ1ƒ0ƒ1ƒ0ƒ1ƒ0ƒ1 ƒƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒƒ1ƒ0ƒ0ƒ1ƒ0ƒ0ƒ0ƒ1 As shown, when both bits being compared are 1s, the result is a 1; otherwise, the result is a 0. The result of each comparison is, of course, independent of any other bit comparison. Program C.1 shows the use of an AND operation. In this program, the variable op1 is initialized to the octal value 325, which is the octal representation of the binary number , and the variable op2 is initialized to the octal value 263, which is the octal representation of the binary number They re the same two binary numbers used in the preceding example. Program C.1 #includeƒ<iostream> usingƒnamespaceƒstd; intƒmain() ƒƒintƒop1ƒ=ƒ0325,ƒop2ƒ=ƒ0263; ƒƒintƒop3ƒ=ƒop1ƒ&ƒop2; ƒƒcoutƒ<<ƒoctƒ<<ƒop1ƒ<<ƒ ƒandedƒwithƒ <<ƒop2ƒ<<ƒ ƒisƒ ƒ<<ƒop3ƒ<<ƒendl; ƒƒreturnƒ0;

3 Appendix C C-3 Program C.1 produces the following output: 325ƒANDedƒwithƒ263ƒisƒ221 The result of ANDing the octal numbers 325 and 263 is the octal number 221. The binary equivalent of 221 is the binary number , which is the result of the AND operation shown previously. AND operations are useful for masking, or eliminating, selected bits from an operand. The reason is that ANDing any bit (1 or 0) with a 0 forces the resulting bit to be a 0, and ANDing any bit (1 or 0) with a 1 leaves the original bit unchanged. For example, say the variable op1 has the bit pattern xƒxƒxƒxƒxƒxƒxƒx, in which each x can be 1 or 0, independent of any other x in the number. The result of ANDing this binary number with the binary number is as follows: ƒƒƒop1ƒ=ƒƒƒxƒxƒxƒxƒxƒxƒxƒx ƒƒƒop2ƒ=ƒƒƒ0ƒ0ƒ0ƒ0ƒ1ƒ1ƒ1ƒ1 ƒƒƒƒƒƒƒƒƒƒƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ- Resultƒ=ƒƒƒ0ƒ0ƒ0ƒ0ƒxƒxƒxƒx As you can see in this example, the 0s in op2 mask the bits in op1, and the 1s in op2 filter, or pass, the bits in op1 through with no change in their values. In this example, the variable op2 is called a mask. By choosing the mask appropriately, any bit in an operand can be selected, or filtered, out of an operand for inspection. For example, ANDing the variable op1 with the mask forces all bits of the result to be 0, except the third bit. The third bit of the result is a copy of the third bit of op1. Therefore, if the result of the AND operation is 0, the third bit of op1 must have been 0, and if the result of the AND is a non-zero number, the third bit must have been 1. Program C.2 uses this masking property to convert lowercase letters into their uppercase form, assuming the letters are stored with the ASCII code. The algorithm for converting letters is based on binary codes for lowercase and uppercase letters in ASCII being the same except for bit five, which is a 1 for lowercase letters and a 0 for uppercase letters. 1 For example, the binary code for the letter a is (hex 61), and the binary code for the letter A is (hex 41). Similarly, the binary code for the letter z is (hex 7A), and the binary code for the letter Z is (hex 5A). (See Appendix B in the book for the hexadecimal values of uppercase and lowercase letters.) Therefore, a lowercase letter can be converted into its uppercase form by forcing the fifth bit to 0. Program C.2 does this by masking the letter s code with the binary value , which has the hexadecimal value DF. 1 This algorithm assumes the conventional numbering scheme, starting with bit 0 as the rightmost bit, is used. With this convention, the rightmost bit (bit 0) is referred to as the least significant bit (LSB), and the leftmost bit is referred to as the most significant bit (MSB). In this example, the MSB is bit 7.

4 C-4 Bit Operations Program C.2 #includeƒ<iostream> usingƒnamespaceƒstd; constƒintƒto_upƒ=ƒ0xdf; constƒintƒmaxƒ=ƒ80; intƒmain() ƒƒƒcharƒword[max];ƒƒƒƒƒƒƒ//ƒenoughƒstorageƒforƒaƒcompleteƒline ƒƒƒvoidƒupper(charƒ*);ƒƒƒ//ƒfunctionƒprototype ƒƒƒcoutƒ<<ƒ Enterƒaƒstringƒofƒbothƒupperƒandƒlowercaseƒletters:\n ; ƒƒƒcin.getline(word,max,'\n'); ƒƒƒcoutƒ<<ƒ \ntheƒstringƒofƒlettersƒjustƒenteredƒis:\n ƒƒƒƒƒƒƒƒ<<ƒwordƒ<<ƒendl; ƒƒƒupper(word); ƒƒƒcoutƒ<<ƒ \nthisƒstringƒinƒuppercaseƒlettersƒis:\n ƒƒƒƒƒƒƒƒ<<ƒwordƒ<<ƒendl; ƒƒƒreturnƒ0; voidƒupper(charƒ*word) ƒƒwhileƒ(*wordƒ!=ƒ'\0') ƒƒƒƒ*word++ƒ&=ƒto_up; ƒƒreturn; A sample run of Program C.2 follows: Enterƒaƒstringƒofƒbothƒupperƒandƒlowercaseƒletters: abcdefghijklmnopqrstuvwxyz Theƒstringƒofƒlettersƒjustƒenteredƒis: abcdefghijklmnopqrstuvwxyz Thisƒstringƒinƒuppercaseƒlettersƒis: ABCDEFGHIJKLMNOPQRSTUVWXYZ Notice that lowercase letters are converted to uppercase form, and uppercase letters are unaltered. This result happens because bit five of all uppercase letters is 0, so forcing this bit to 0 with the mask has no effect. Only when bit five is 1, as for lowercase letters, is the input character altered.

5 Appendix C C-5 The Inclusive OR Operator The inclusive OR operator,, performs a bit-by-bit comparison of its two operands in a similar fashion as the bit-by-bit AND. The result of the OR comparison, however, is determined by the following rule: The result of the comparison is 1 if either bit being compared is a 1; otherwise, the result is a 0. You can see this rule in the following sample OR operation. As with all bit operations, the result of each comparison is independent of any other comparison. ƒƒ1ƒ0ƒ1ƒ1ƒ0ƒ0ƒ1ƒ1 ƒ1ƒ1ƒ0ƒ1ƒ0ƒ1ƒ0ƒ1 ƒƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒƒ1ƒ1ƒ1ƒ1ƒ0ƒ1ƒ1ƒ1 Program C.3 shows the use of an OR operation, using the octal values of the operands in the preceding example. Program C.3 #includeƒ<iostream> usingƒnamespaceƒstd; intƒmain() ƒƒintƒop1ƒ=ƒ0325,ƒop2ƒ=ƒ0263; ƒƒintƒop3ƒ=ƒop1ƒ ƒop2; ƒƒcoutƒ<<ƒoctƒ<<ƒop1ƒ<<ƒ ƒoredƒwithƒ ƒ<<ƒop2ƒ<<ƒ ƒisƒ ƒ<<ƒop3ƒ<<ƒendl; ƒƒreturnƒ0; Program C.3 produces the following output: 325ƒORedƒwithƒ263ƒisƒ367 The result of ORing the octal numbers 325 and 263 is the octal number 367. The binary equivalent of 367 is , which is the result of the OR operation shown previously. Inclusive OR operations are useful for forcing selected bits to take on a 1 value or for passing through other bit values unchanged. The reason is that ORing any bit (1 or 0) with a 1 forces the resulting bit to be a 1, and ORing any bit (1 or 0) with a 0 leaves the original bit unchanged. For example, the variable op1 has the bit pattern xƒxƒxƒxƒxƒxƒxƒx; each x can be 1 or 0, independent of any other x in the number. The result of ORing this binary number with the binary number is as follows: ƒƒƒop1ƒ=ƒƒƒxƒxƒxƒxƒxƒxƒxƒx ƒƒƒop2ƒ=ƒƒƒ1ƒ1ƒ1ƒ1ƒ0ƒ0ƒ0ƒ0 ƒƒƒƒƒƒƒƒƒƒƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ- Resultƒ=ƒƒƒ1ƒ1ƒ1ƒ1ƒxƒxƒxƒx

6 C-6 Bit Operations As you can see in this example, the 1s in op2 force the resulting bits to 1, and the 0s in op2 filter, or pass, the bits in op1 through with no change in their values. By using an OR operation, you can produce a masking operation similar to an AND operation, except the masking bits are set to 1s rather than 0s. Another way of looking at it is to say that ORing with a 0 has the same effect as ANDing with a 1. Program C.4 uses this masking property to convert uppercase letters in a word into their lowercase form by masking the letter s code with the binary value , which has the hexadecimal value 20. Program C.4 #includeƒ<iostream>ƒ usingƒnamespaceƒstd; constƒintƒmaxƒ=ƒ80; constƒintƒto_lowƒ=ƒ0x20; intƒmain() ƒƒcharƒword[max];ƒƒƒƒƒƒ//ƒenoughƒstorageƒforƒaƒcompleteƒline ƒƒvoidƒlowerƒ(charƒ*);ƒ//ƒfunctionƒprototype ƒƒcoutƒ<<ƒ Enterƒaƒstringƒofƒbothƒupperƒandƒlowercaseƒletters:\n ; ƒƒcin.getline(word,max,'\n'); ƒƒcoutƒ<<ƒ \ntheƒstringƒofƒlettersƒjustƒenteredƒis:\n ƒƒƒƒƒƒƒ<<ƒwordƒ<<ƒendl; ƒƒlower(word); ƒƒcoutƒ<<ƒ \nthisƒstringƒinƒlowercaseƒlettersƒis:\n ƒƒƒƒƒƒƒ<<ƒwordƒ<<ƒendl; ƒƒreturnƒ0; voidƒlower(charƒ*word) ƒƒwhileƒ(*wordƒ!=ƒ'\0') ƒƒƒƒ*word++ƒ =ƒto_low;ƒ ƒƒreturn;

7 Appendix C C-7 A sample run of Program C.4 follows. Enterƒaƒstringƒofƒbothƒupperƒandƒlowercaseƒletters: abcdefghijklmnopqrstuvwxyz Theƒstringƒofƒlettersƒjustƒenteredƒis: abcdefghijklmnopqrstuvwxyz Thisƒstringƒinƒlowercaseƒlettersƒis: abcdefghijklmnopqrstuvwxyz Notice that uppercase letters are converted to lowercase form, and lowercase letters are unaltered. This result happens because bit five of all lowercase letters is 1, so forcing this bit to 1 with the mask has no effect. Only when bit five is 0, as for uppercase letters, is the input character altered. The Exclusive OR Operator The exclusive OR operator, ^, performs a bit-by-bit comparison of its two operands. The result of the comparison is determined by the following rule: The result of the comparison is 1 if one, and only one, of the bits being compared is a 1; otherwise, the result is 0. The following example of an exclusive OR operation shows that when both bits being compared are the same value (both 1 or both 0), the result is 0. Only when both bits have different values (one bit a 1 and the other a 0) is the result 1. Again, each pair or bit comparison is independent of any other bit comparison. ƒƒ1ƒ0ƒ1ƒ1ƒ0ƒ0ƒ1ƒ1 ^ƒ1ƒ1ƒ0ƒ1ƒ0ƒ1ƒ0ƒ1 ƒƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒƒ0ƒ1ƒ1ƒ0ƒ0ƒ1ƒ1ƒ0 An exclusive OR operation can be used to create the opposite value, or complement, of any bit in a variable. The reason is that exclusive ORing any bit (1 or 0) with 1 forces the resulting bit to be the opposite value of its original state, and exclusive ORing any bit (1 or 0) with 0 leaves the original bit unchanged. For example, the variable op1 has the bit pattern xƒxƒxƒxƒxƒxƒxƒx; each x can be 1 or 0, independent of any other x in the number. Using the notation that x is the complement (opposite) value of x, the result of exclusive ORing this binary number with the binary number is as follows: ƒƒƒop1ƒ=ƒƒƒxƒxƒxƒxƒxƒxƒxƒx ƒƒƒop2ƒ=ƒƒƒ0ƒ1ƒ0ƒ1ƒ0ƒ1ƒ0ƒ1 ƒƒƒƒƒƒƒƒƒƒƒ ƒ ƒ ƒ ƒ ƒ ƒ ƒƒƒƒƒƒƒƒƒƒƒƒƒƒ_ƒƒƒ_ƒƒƒ_ƒƒƒ_ Resultƒ=ƒƒƒxƒxƒxƒxƒxƒxƒxƒx As you can see in this example, the 1s in op2 force the resulting bits to be the complement of their original bit values, and the 0s in op2 filter, or pass, the bits in op1 through with no change in their values.

8 C-8 Bit Operations Many encryption methods use the exclusive OR operation to code data by exclusive ORing each character in the string with a mask value. The choice of the mask value, which is referred to as the encryption key, is arbitrary, and any key value can be used. Program C.5 uses an encryption key of 52 to code a user-entered message. Program C.5 #includeƒ<iostream> usingƒnamespaceƒstd; constƒintƒmaxƒ=ƒ80; intƒmain() ƒƒcharƒmessage[max];ƒƒƒƒƒ//ƒenoughƒstorageƒforƒaƒcompleteƒline ƒƒvoidƒencrypt(charƒ*);ƒƒ//ƒfunctionƒprototype ƒƒcoutƒ<<ƒ \nenterƒaƒsentence:\n ; ƒƒcin.getline(message,max,'\n'); ƒƒcoutƒ<<ƒ \ntheƒsentenceƒjustƒenteredƒis:\n ƒƒƒƒƒƒƒ<<ƒmessageƒ<<ƒendl; ƒƒencrypt(message); ƒƒcoutƒ<<ƒ \ntheƒencryptedƒversionƒofƒthisƒsentenceƒis:\n ƒƒƒƒƒƒƒ<<ƒmessageƒ<<ƒendl; ƒƒreturnƒ0; voidƒencrypt(charƒ*message) ƒƒwhileƒ(*messageƒ!=ƒ'\0') ƒƒƒƒƒ*message++ƒ^=ƒ52; ƒƒreturn; Following is a sample run of Program C.5: Enterƒaƒsentence: Good morning Theƒsentenceƒjustƒenteredƒis: Goodƒmorning Theƒencryptedƒversionƒofƒthisƒsentenceƒis: s[[p Y[FZ]ZS

9 Appendix C C-9 Decoding an encrypted message requires exclusive ORing the coded message with the original encryption key. The Complement Operator The complement operator, ~, is a unary operator that changes each 1 bit in its operand to 0 and each 0 bit to 1. For example, if the variable op1 contains the binary number , ~op1 replaces this binary number with the number The complement operator is used to force any bit in an operand to 0, independent of the actual number of bits used to store the number. For example, the statement op1ƒ=ƒop1ƒ&ƒ~07;ƒƒƒ//ƒ07ƒisƒanƒoctalƒnumber or its shorter form op1ƒ&=ƒ~07;ƒƒƒ//ƒ07ƒisƒanƒoctalƒnumber sets the last three bits of op1 to 0, regardless of how op1 is stored in the computer. Either statement can, of course, be replaced by ANDing the last three bits of op1 with 0s, if the number of bits used to store op1 is known. In a computer that uses 16 bits to store integers, the AND operation is op1ƒ=ƒop1ƒ&ƒ ;ƒƒƒ//ƒinƒoctal or op1ƒ=ƒop1ƒ&ƒ0xfff8;ƒƒƒ//ƒinƒhexadecimal For a computer that uses 32 bits to store integers, the preceding AND operation also sets the leftmost or higher order 16 bits to 0, which is an unintended result. The following is the correct statement for 32 bits: op1ƒ=ƒop1ƒ&ƒ ;ƒƒƒ//ƒinƒoctal or op1ƒ=ƒop1ƒ&ƒ0xfffffff8;ƒƒƒ//ƒinƒhexadecimal Using the complement operator in this situation frees the programmer from having to determine the operand s storage size and, more important, makes the program portable between machines using different integer storage sizes. Different-Sized Data Items When the bit operators &,, and ^ are used with operands of different sizes, the shorter operand is always increased in bit size to match the larger operand s size. Figure C.1 illustrates extending a 16-bit unsigned integer into a 32-bit number. As the figure shows, the extra bits are added to the left of the original number and filled with 0s. This is the equivalent of adding leading 0s to the number, which has no effect on the number s value.

10 C-10 Bit Operations 16 bits XXXXXXXXXXXXXXXX X can be 1 or s The original 16 bits XXXXXXXXXXXXXXXX 32 bits Figure C.1 Extending 16-bit unsigned data to 32 bits When extending signed numbers, the original leftmost bit is reproduced in the extra bits added to the number. As shown in Figure C.2, if the original leftmost bit is 0, corresponding to a positive number, 0 is placed in each of the additional bit positions. If the leftmost bit is 1, which corresponds to a negative number, 1 is placed in the additional bit positions. In either case, the resulting binary number has the same sign and magnitude as the original number. 16 bits A sign bit of s 0XXXXXXXXXXXXXXX The original 16 bits X can be 1 or XXXXXXXXXXXXXXX 16 bits A sign bit of 1 1XXXXXXXXXXXXXXX 16 1s The original 16 bits XXXXXXXXXXXXXXX Figure C.2 Extending 16-bit signed data to 32 bits The Shift Operators The left shift operator, <<, causes the bits in an operand to be shifted to the left by a given amount. For example, the statement op1ƒ=ƒop1ƒ<<ƒ4;

11 Appendix C C-11 causes the bits in op1 to be shifted four bit positions to the left, filling any vacated bits with a 0. Figure C.3 illustrates the effect of shifting the binary number to the left by four bit positions Each bit is shifted to the left by the designated number of places Vacated bit positions are filled with 0s Figure C.3 An example of a left shift For unsigned integers, each left shift corresponds to multiplication by 2. This is also true for signed numbers using twos complement representation, as long as the leftmost bit doesn t switch values. Because a change in the leftmost bit of a twos complement number represents a change in both the sign and magnitude represented by the bit, this shift doesn t represent a simple multiplication by 2. The right shift operator, >>, causes the bits in an operand to be shifted to the right by a given amount. For example, the statement op2ƒ=ƒop1ƒ>>ƒ3; causes the bits in op1 to be shifted to the right by three bit positions. Figure C.4a shows the right shift of the unsigned binary number by three bit positions. As shown, the three rightmost bits are shifted off the end and are lost.

12 C-12 Bit Operations Each bit is shifted to the right by the designated number of places Vacated bit positions are filled with 0s Figure C.4a An unsigned arithmetic right shift For unsigned numbers, the leftmost bit isn t used as a sign bit. For this type of number, the vacated leftmost bits are always filled with 0s, as shown in Figure C.4a. For signed numbers, what s filled in the vacated bits depends on the computer. Most computers reproduce the number s original sign bit. Figure C.4b illustrates the right shift of a negative binary number by four bit positions, and the sign bit is reproduced in the vacated bits. Figure C.4c illustrates the equivalent right shift of a positive signed binary number. The sign bit is a Each bit is shifted to the right by the designated number of places Vacated bit positions are filled with 1s Figure C.4b The right shift of a negative binary number

13 Appendix C C-13 The sign bit is a Each bit is shifted to the right by the designated number of places Vacated bit positions are filled with 0s Figure C.4c The right shift of a positive binary number The type of fill shown in Figures C.4b and C.4c, where the sign bit is reproduced in vacated bit positions, is called an arithmetic right shift. In an arithmetic right shift, each single shift to the right corresponds to a division by 2. Instead of reproducing the sign bit in right-shifted signed numbers, some computers fill the vacated bits with 0s automatically. This type of shift is called a logical shift. For positive signed numbers, in which the leftmost bit is 0, both arithmetic and logical right shifts produce the same result. The results of these two shifts are different only when negative numbers are involved.

Number System. Introduction. Decimal Numbers

Number System. Introduction. Decimal Numbers Number System Introduction Number systems provide the basis for all operations in information processing systems. In a number system the information is divided into a group of symbols; for example, 26