Number Systems and Bitwise Operation 1 Binary Octal

Number Systems and Bitwise Operation 1

Binary, Octal, Hexadecimal, and Decimal n Binary numbering system has only two possible values for each digit: 0 and 1. For example, binary number 0 1 10 11 100 101 1100 1010 decimal number 0 1 2 3 4 5 6 202 2

Decimal Numbers n Decimal The digits' weight increases by powers of 10. The weighted values for each position is determined as follows: 104 103 102 101 10000 100 10 1 For example, A decimal number 4261 can be thought of as follows. 4 * 1000 + 2 * 100 + 6 * 10 + 1 * 1 = 4000 + 200 + 60 + 1 = 4261 (decimal) 3

Binary, Octal, Hexadecimal, and Decimal n Binary The digits' weight increases by powers of 2. The weighted values for each position is determined as follows: 27 26 25 24 23 22 21 20 128 64 32 16 8 4 2 1 For example, binary 10 is decimal 2. the binary value 1100 1010 represents the decimal value 202. 1 * 128 + 1 * 64 + 0 * 32 + 0 * 16 + 1 * 8 + 0 * 4 + 1 * 2 + 0 * 1 = 128 + 64 + 0 + 8 + 0 + 2 + 0 = 202 (decimal) 4

Binary Two’s Complement The left-most bit is the sign bit. If it is 1, then the number is negative. Otherwise, it is positive. Give a negative value, represent it in binary two’s complement form as follows. 1. write the number in its absolute value. 2. complement the binary number. 3. plus 1. Example, represent – 2 in binary two’s complement with 16 bits for short int. n n Binary value of 2: 0 b 0000 0010 Binary complement of 2: 0 b 1111 1101 Plus 1: +1 Binary two’s complement representation of -2: 0 b 1111 1110 5

Give binary two’s complement form of a negative number, find the absolute value of the negative value as follows. 1. Complement the binary number. 2. Plus 1. Example, find the decimal value of (0 b 1111 1110) 2 in binary two’s complement form with 16 bits. n n n Binary two’s complement Binary complement Plus 1 Absolute value: Negative value: (0 b 1111 1110) 2 (0 b 0000 0001) 2 +1 (0 b 0000 0010) 2 = 210 -2 6

Subtraction of a value in the computer can be treated as addition of its two’s complement. For example, the subtraction of (2 -2) can be performed as 2+(-2) as follows: 0 b 0000 0010 (binary representation of 2) 0 b 11111110 (two’s complement representation of -2) 0 b 0000 (2+(-2)) 7

n Example > short i, j > i = 0 b 000000010 2 > j = 0 b 111111110 -2 > i+j 0 8

Octal n The octal system is based on the binary system with a 3 -bit boundary. The octal number system uses base 8 includes 0 through 7. The weighted values for each position is as follows: 1. 83 82 81 80 512 64 8 1 Binary to Octal Conversion • • Break the binary number into 3 -bit sections from the least significant bit (LSB) to the most significant bit (MSB). Convert the 3 -bit binary number to its octal equivalent. For example, the binary value 1 010 000 111 101 110 100 011 equals to octal value (12075643)8. 9

Octal to Binary Conversion 2. • • Convert the octal number to its 3 -bit binary equivalent. Combine all the 3 -bit sections. For example, the octal value 45761023 equals to binary value 100 101 110 001 000 011. 3. Octal to Decimal Conversion To convert octal number to decimal number, multiply the value in each position by its octal weight and add each value together. For example, the octal value (167)8 represents decimal value 119. 1*64 + 6*8 + 7*1 = 119 10

n Hexadecimal Similar to octal, the hexadecimal system is also based on the binary system but using 4 -bit boundary. The hexadecimal number system uses base 16 including the digits 0 through 9 and the letters A, B, C, D, E, and F. The letters A through F represent the decimal numbers 10 through 15. For the decimal values from 0 to 15, the corresponding hexadecimal values are listed below. Decimal 15 14 13 12 11 10 9 Hexadecima F l E D C B A 9 8 7 6 5 4 3 2 1 0 11

The weighted values for each position is as follows: 163 162 161 160 4096 256 16 1 The conversion between binary value and hexadecimal value is similar to octal number, but using four-bit sections. The hexadecimal value 20 A represents decimal value 522. 2*256 + 0*16 + 10*1 = 522 12

Following table provides all the information you need to convert from one type number into any other type number for the decimal values from 0 to 16. Binary Octal Decimal Hex 0000 00 1001 11 09 09 0001 01 1010 12 10 A 0010 02 02 02 1011 13 11 B 0011 03 03 03 1100 14 12 C 0100 04 04 04 1101 15 13 D 0101 05 05 05 1110 16 14 E 0110 06 06 06 1111 17 15 F 0111 07 07 07 10000 20 16 10 1000 10 08 08 13

Bitwise Operators n There are six bitwise operators: Operator Name Description & bitwise AND The bit is set to 1 if the corresponding bits in the two operands are both 1. | bitwise OR The bit is set to 1 if at least one of the corresponding bits in the two operands is 1. ^ bitwise exclusive OR The bit is set to 1 if exactly one of the corresponding bits in the two operands is 1. << left shift Shift the bits of the first operand left by the number of bits specified by the second operand; fill from right with 0 bits. >> right shift Shift the bits of the first operand right by the number of bits specified by the second operand; filling from the left is implementation dependent. ~ One’s complement Set all 0 bits to 1, and all 1 bits to 0. 14

Example: a 1 0 1 0 0 b 1 1 0 0 1 a&b 1 0 0 0 0 a|b 1 1 1 0 1 a^b 0 1 1 0 1 b << 1 1 0 0 1 0 a >> 1 1 1 0 1 0 ~a 0 1 0 1 1 15

/* File: bitop. ch (run in Ch only) Use Ch features “%b” and 0 b */ #include <stdio. h> Output: a = b = a & b = a | b = a ^ b = b << 1 = a >> 1 = ~a = 0 b 10110100 0 b 11011001 0 b 10010000 0 b 11111101 0 b 01101101 0 b 10110010 0 b 11011010 0 b 01001011 int main() { char a = 0 b 10110100; char b = 0 b 11011001; char c; printf("a = printf("b = c = a & b; printf("a & b = c = a | b; printf("a | b = c = a ^ b; printf("a ^ b = c = b << 1; printf("b << 1 = c = a >> 1; printf("a >> 1 = c = ~a; printf("~a = 0 b%8 bn", a); 0 b%8 bn", b); 0 b%8 bn", c); 0 b%8 bn", c); return 0; } 16

Logic Operators n There are four logic operators: 1) ! --- logic NOT 2) && --- logic AND 3) || --- inclusive OR 4) ^^ --- exclusive OR (available in Ch only) a b !a a && b a || b a ^^ b 0 0 1 1 0 1 1 1 0 0 0 1 1 0 17

Formatted Output with printf 18

int %d float %f char %c long float %lf String %s etc Month= 12 Expense= 111. 1 19

Formatted Output with printf-cont 20

Formatted input with scanf 21

Formatted input with scanf-cont 22

Program debugging Syntax error Mistakes caused by violating “grammar” of C C compiler can easily diagnose during compilation Run-time error Called semantic error or smart error Violation of rules during program execution C compiler cannot recognize during compilation Logic error Most difficult error to recognize and correct Program compiled and executed successfully but answer wrong 23

Program debugging-syntax error snapshot 24

printf(“Welcome to KUKUM”) //missing ‘; ’ the end of line

Program debugging-run time error snapshot 26

scanf(“%d”, set_salary); //missing ‘&’

#include <stdio. h> int main ( ) { int net_salary, full_salary; printf(“Welcome to Uni. MAP”); printf(“Enter salary”); scanf(“%d”, net_salary); //missing ‘&’ full_salary = net _salary+175; printf(“Net salary is %d”, full_salary); return 0; }

Program debugging-logic error snapshot 29

full_salary expression is after “Net salary”

#include <stdio. h> int main ( ) { int net_salary, full_salary; printf(“Welcome to Uni. MAP”); printf(“Enter salary”); scanf(“%d”, &net_salary); printf(“Net salary is %d”, full_salary); full_salary = net _salary+175; return 0; }

End Q & A! 32