Bit Operations Horton pp. 70 - 76

Why we need to work with bits Sometimes one bit is enough to store your data: say the gender of the student (e.g. 0 for men, 1 for women). We don’t have a 1-bit variable type in C++, so for gender, you will have to use a unsigned char or char type variable. But if you need 8 such 1-bit variables, say to record if the a student was present in the class when attendance was taken, then you can actually combine all into one char variable. class student { … private: unsigned char attendance; //now I can fit 8 bits into this, we will see how } In C++, there is special container class, called bitset, that is designed to store bits. But we will not see and use it. And you are not allowed to use it in this course since you have to learn bit manipulation via more basic mechanisms.

Packing bits “Packing” 8 1-bit variables into 1 char variable is easy. Say you know that the student were present in the first 3 class and not in the last 5. The variable attendance can be set as: unsigned char attendance = 7; 1 where the least significant 3 bits represent the first 3 attendances (just a choice, it could be the other way around as well). But in addition to being able to set a variable’s value, we need to be able to handle each bit separately, for which we need bit operators.

Hexadecimal literals and output In C++ a hexadecimal literal (number) is represented by preceeding 0x before the literal (0x is not part of the numeric value). E.g. 0xb4a2 // equivalent to 11x163 + 4x162 + 10x16 + 2 = 46242 (decimal) char ch = 0x7c; // ch contains 124 (7x16+ 12) When you want to display a numeric value in hexadecimal, use hex in cout before the value to be displayed. This affects all future integer outputs to be displayed in hexadecimal, so if you want to switch back to decimal, use dec later. short nums = -200; cout << "hexadecimal: " << hex << nums << " decimal: " << dec << nums << endl; Output: hexadecimal: ff38 decimal: -200 unsigned int numi = 0xfd04e0e4; cout << "hexadecimal: " << hex << numi << " decimal: " << dec << numi << endl; Output: hexadecimal: fd04e0e4 decimal: 4244955364

Bit Operators & Bitwise and | Bitwise or ^ Bitwise exclusive or ~ Complement (unary operator, takes only one operand) << shift left >> shift right Do not confuse && with & || with |

Bitwise Operations: AND Take the AND of the two numbers, bit by bit. a b a & b 1 char x,y,z; x = 0xd5; y = 0x6c; z = x&y; x 1 y 1 z 1

Bitwise AND unsigned char c1, c2, c3; c1 = 0x45; c2 = 0x71;

Bitwise OR Take the OR of the two numbers, bit by bit. unsigned char c1, c2, c3; c1 = 0x45; c2 = 0x71; c3 = c1 | c2; c1 : 0100 0101 c2 : 0111 0001 c3 : 0111 0101 (=0x75 = 7*16+5 = 11710 ) a b a | b 1

Bitwise XOR Take the Exclusive OR of the two numbers, bit by bit. unsigned char c1, c2, c3; c1 = 0x45; c2 = 0x71; c3 = c1 ^ c2; c1 : 0100 0101 c2 : 0111 0001 c3 : 0011 0100 (=0x34 = 3*16+4 = 5210 ) a b a ^ b 1

Bitwise Complement Complement operation (~) converts bits with value 0 to 1 and bits with value 1 to 0. This is unary operator (takes only one operand) unsigned char b1 = 0xc1; // 1100 0001 unsigned char b4 b4 = ~b1; // 0011 1110 a ~a 1

What do these two statements do? char x = 0xA5; if ( x & 0x08 ) //what does this mean? x = x | 0x02; //what happened to x? These are two of the most important bit operations! We will see more later, but basically you can check a particular bit and you can set a particular bit to 1 with these two operations.

Logic Operators versus Bitwise Logic Ops. The operators && , || and ! are not bitwise logic operators! The result of ! , && and || operations is an integral data type with the value 0 (every bit is a zero) or 1 (Least Significant bit is 1, all the others are 0).

Shift Operators how many times it is shifted Shift operators move bits left or right, discarding bits from one side and filling the other side with 0s (filling part has an exception that we will see in coming slides). << means shift left >> means shift right Do not confuse with stream operators how many times it is shifted

Bit operations: left shift Suppose we want to shift the bits of a number N, k bits to the left denoted N << k drop leftmost k bits append k zeros to the right Ex: y = x << 1; Another example: unsigned char c2 = 0x9C; //10011100 c = c2 << 2 //01110000 x 1 y 1

Bit Shifting as Multiplication Shift left by one position means multiplication by 2 (provided that the result is in the range) - Works as multiplication for both unsigned & 2’s complement numbers - May overflow. What is the effect of shifting a number left by 3? 1 1 1 = 19 1 1 1 1 1 1 1 = -3 1 1 1 = 38 1 1 1 1 1 1 = -6

Bit operations: right shift As opposed to left shift, the right shift works differently for signed and unsigned numbers. Suppose we want to shift N by k bits to the right (denoted N >> k): For unsigned numbers (i.e. unsigned char, unsigned int, etc.): drop rightmost k bits append k zeros to the left For signed numbers (i.e. char, int, etc.): drop the rightmost k bits append the sign bit k times to the left

right shift operator:examples Signed (all your variables are signed unless you specify “unsigned” specifically): positives: char c = 8; //0000 1000 c = c >> 2 //0000 0010 = 210 negatives: char c = - 8; //1111 1000 in 2s comp. c = c >> 2 //1111 1110 which is -210 Called arithmetic shift Unsigned: unsigned char d = 0xf8; //1111 1000 (24810) d = d >> 2 //0011 1110 which is 6210 Called logical shift

Bit Shifting as Division: summary 1 = 7 = 3 = 8 = 4 Always rounds down! Logical shift right for one bit divides by 2 for unsigned: 1 = 7 = 3 = -7 = -4 Always rounds down! CAUTION: Regular division of negative integers rounds down magnitude-wise. E.g. -7/2 is -3 Arithmetic shift right for one bit divides by 2 for signed numbers

Bit Shifting as Multiplication & Division Why useful? Simpler, thus faster, than general multiplication & division Can shift multiple positions at once: Multiplies or divides by corresponding power of 2. a << 5 (multiply by 25) a >> 5 (divide by 25) Remember, when you need to use only positive numbers or when you need to manipulate bits, use unsigned numbers.

Self-Quiz What are the resulting values of x,y and z? char x,y,z; x = 0x33; x = (x << 3) | 0x0F; y = (x >> 1) & 0x0F; z = x && y;

Example for setting, testing, and clearing the bits of bytes const unsigned char IO_ERROR = 0x01; //LSB (right-most bit) is 1, others are 0 const unsigned char CHANNEL_DOWN = 0x10; // 00010000 in binary Unsigned char flags = 0; //if ever CHANNEL_DOWN event happens, set its corresponding bit in the flags variable flags = flags | CHANNEL_DOWN; // set the 5th bit . . . //to check what errors may have happened... if ((flags & IO_ERROR ) != 0) // check the 1st bit cout << "I/O error flag is set"; else if ((flags & CHANNEL_DOWN) != 0) // check the 5th bit cout << "Channel down error flag is set"; flags = flags & ~IO_ERROR; // clear the ERROR flag This is also called masking

Example 2: Attendance unsigned char attendance = 0x00; //let the LSB be for quiz 1, the next one is for quiz 2, . . . //0 means unattended, 1 means attended //there are total of 8 quizzes; initally all unattended unsigned char mask=0x01; //Draft code showing how attendance array may be filled //if attended to quiz 1 attendance = attendance | mask; //if attended quiz 3 mask = mask << 2; // mask becomes 0000 0100 . . . //to display attendance information mask=0x01; for (int i=1; i <= 8; i++) { if (attendance & mask) cout << "Student attended quiz number " << i << endl; else cout << "Student missed quiz number " << i << endl; mask = mask << 1; }