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Learn about binary, octal, hexadecimal, and decimal number systems, and how to perform bitwise operations. Understand the weighted values for each position and how to convert between different number systems.
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Binary, Octal, Hexadecimal, and Decimal • Binary • Binary numbering system has only two possible values for each digit: 0 and 1. For example, binary number decimal number 0 0 1 1 10 2 11 3 100 4 101 5 110 6 1100 1010 202
Decimal Numbers • Decimal • The digits' weight increases by powers of 10. The weighted values for • each position is determined as follows: 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)
Binary, Octal, Hexadecimal, and Decimal • Binary • The digits' weight increases by powers of 2. The weighted values for each position is • determined as follows: 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 + 0 + 8 + 0 + 2 + 0 = 202 (decimal)
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. • write the number in its absolute value. • complement the binary number. • plus 1. • Example, represent –2 in binary two’s complement with 16 bits for short int. • Binary value of 2: 0b0000 0000 0000 0010 • Binary complement of 2: 0b1111 1111 1111 1101 • Plus 1: +1 • Binary two’s complement representation of -2: 0b1111 1111 1111 1110
Give binary two’s complement form of a negative number, find • the absolute value of the negative value as follows. • Complement the binary number. • Plus 1. • Example, find the decimal value of (0b1111 1111 1111 1110)2 in binary • two’s complement form with 16 bits. • Binary two’s complement (0b1111 1111 1111 1110)2 • Binary complement (0b0000 0000 0000 0001)2 • Plus 1 +1 • Absolute value: (0b0000 0000 0000 0010)2 = 210 • Negative value: -2
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: 0b0000 0000 0000 0010 (binary representation of 2) 0b1111 1111 11111110 (two’s complement representation of -2) 0b0000 0000 0000 0000 (2+(-2))
Example > short i, j > i = 0b0000000000000010 2 > j = 0b1111111111111110 -2 > i+j 0
Octal • 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: • 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 value1 010 000 111 101 110 100 011 equals to • octal value (12075643)8.
Octal to Binary Conversion • Convert the octal number to its 3-bit binary equivalent. • Combine all the 3-bit sections. • For example, the octal value45761023 equals to binary value • 100 101 111 110 001 000 010 011. • 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
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 Hexadecimal
The weighted values for each position is as follows: The conversion between binary value and hexadecimal value is similar to octal number,but using four-bit sections. The hexadecimal value 20A represents decimal value 522. 2*256 + 0*16 + 10*1 = 522
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 to16.
Bitwise Operators • There are six bitwise operators:
/* File: bitop.ch (run in Ch only) Use Ch features “%b” and 0b */ #include <stdio.h> int main() { char a = 0b10110100; char b = 0b11011001; char c; printf("a = 0b%8b\n", a); printf("b = 0b%8b\n", b); c = a & b; printf("a & b = 0b%8b\n", c); c = a | b; printf("a | b = 0b%8b\n", c); c = a ^ b; printf("a ^ b = 0b%8b\n", c); c = b << 1; printf("b << 1 = 0b%8b\n", c); c = a >> 1; printf("a >> 1 = 0b%8b\n", c); c = ~a; printf("~a = 0b%8b\n", c); return 0; } Output: a = 0b10110100 b = 0b11011001 a & b = 0b10010000 a | b = 0b11111101 a ^ b = 0b01101101 b << 1 = 0b10110010 a >> 1 = 0b11011010 ~a = 0b01001011
Logic Operators • There are four logic operators: • 1) ! --- logic NOT • 2) &&--- logic AND • 3) || --- inclusive OR • 4) ^^ --- exclusive OR (available in Ch only)
int %d • float %f • char %c • long float %lf • String %s etc • Month= 12 Expense= 111.1
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
#include <stdio.h> int main ( ) { int net_salary, full_salary; printf(“Welcome to UniMAP”); printf(“Enter salary”); scanf(“%d”, net_salary); //missing ‘&’ full_salary = net _salary+175; printf(“Net salary is %d”, full_salary); return 0; }
#include <stdio.h> int main ( ) { int net_salary, full_salary; printf(“Welcome to UniMAP”); printf(“Enter salary”); scanf(“%d”, &net_salary); printf(“Net salary is %d”, full_salary); full_salary = net _salary+175; return 0; }
