machine code, number system, and c variables and operations

Machine Code, Number System,andand

C Variables and Operations

A. Sahu amd S. V .RaoDept of Comp. Sc. & Engg.Dept of Comp. Sc. & Engg.

Indian Institute of Technology Guwahati

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Outline• Compiler Phases for C Programming

–Compiler, Assembler, Linker, Loader–Source, Assembly, Object, Executable , y, j ,

• Number SystemBi O t l d H–Binary, Octal and Hex

• Flow Charts• C Programming: Variable, data type and operationsand operations

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Machine code andMachine code and CompilationCompilation

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Compiling program: Object file, bl f l f lassembly file, exe file

$gcc –S test.cG t bl l filGenerate assembly language file .s

$ gcc test.s$

Source, Assembly TXT$gcc –c test.c

Generate object file TXT

$gcc test.oGenerate executable file

Object, EXEBinary

$./a.outrun the executable

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Compilation flow

test.c C Source file

gcc test.cgcc ‐S test.c

t ttest.s

gcc ‐c test.sgcc ‐c test.c

ASM file

gcc test.sgcc c test.s

Test.o

gcc c test.c

Obj file

gcc test.s

gcc ‐c test.o

5a.out

Exe file

Human‐Readable Machine LanguageHuman Readable Machine Language• Computers like ones and zeros…

0001110010000110• Humans like symbols…

ADD R6 R2 R6; increment index reg

0001110010000110

• Assembler is a program that turns symbols intomachine instructions.

ADD R6,R2,R6; increment index reg.

– ISA‐specific: Close correspondence between symbols and instruction set

M i f d• Mnemonics for opcodes• Labels for memory locations

– additional operations for allocating storage and

7‐6

additional operations for allocating storage and initializing data

Object File Format• Object file contains

– Starting address (location where program must be loaded), followed by Machine instructions

• Example– Beginning of “count character” object file looks like this:

00110000000000000101010010100000

.ORIG x3000

AND R2, R2, #0

LD R3 PTR00100110000100011111000000100011

LD R3, PTR

TRAP x23

Assembly Language: Human Readable%

include 'system.inc'include system.inc section .data hello db 'Hello, World!',

0Ah hb tes eq $ hello0Ah hbytes equ $-hello section .text global _start _start:

push dword hbytespush dword hellopush dword hello push dword stdoutsys.writepush dword 0

$nasm -f elf hello.asmpush dword 0

sys.exit$ld -s -o hello hello.o$./hello

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Hello, World!

Language LevelsLanguage LevelsHigh Level Language A

S

Assembler Language

BST

SP

Assembler Language TRA

EE

Machine Language ATI

D

Micro ‐programming „Firmware“ ON

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Hardware

High Level to Micro CodeHigh Level to Micro Code• High Level language

– Formulating program for certain application areas

–Hardware independent• Assembler languages

–Machine oriented language–Programs orient on special hardware g pproperties

–More comfortable than machine code

10(e.g. by using symbolic notations)

High Level to Micro CodeHigh Level to Micro Code• Machine code:

–Set of commands directly executable via CPUvia CPU

–Commands in numeric code–Lowest semantic level–Generally 2 executing oportunities:Generally 2 executing oportunities:

• Interpretiv via micro codeDi l i i h d

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• Directly processing via hardware

High Level to Micro CodeHigh Level to Micro Code• Micro programming:

– Implementing of executing of machine commands (Control unit ‐ controller)

–Machine command executed/shown as sequence of micro code commandssequence of micro code commands

–Micro code commands:Si li t t lli• Simpliest process controlling

–Moving of data, Opening of grids, Tests

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Number System y(binary, octal, dec and

hexa decimal)

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Computer: Number SystemComputer: Number System • Computers like ones and zeros…

• Humans like symbols…0001110010000110

Humans like symbols…• We need to know: binary number system

Also Octal and Hex number system– Also Octal and Hex number system– Type conversions

7‐14

Famous Number SystemFamous Number System• Decimal System: 0 ‐9

–May evolves: because human have 10 finger • Roman Systemy

–May evolves to make easy to look and feel–Pre/Post Concept: (IV V & VI) is (5‐1 5 &–Pre/Post Concept: (IV, V & VI) is (5‐1, 5 & 5+1)

• Binary System Others (Oct Hex)• Binary System, Others (Oct, Hex)–One can cut an apple in to two

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Significant DigitsSignificant Digits

Binary: 11101101Binary: 11101101

Most significant digit Least significant digitMost significant digit Least significant digit

Decimal: 1063079Decimal: 1063079

Most significant digit Least significant digitMost significant digit Least significant digit

Decimal (base 10)Decimal (base 10)• Uses positional representation• Each digit corresponds to a power of 10 based on its position in the number10 based on its position in the number

• The powers of 10 increment from 0, 1, 2, etc. as you move right to left–1 479 = 1 * 103 + 4 * 102 + 7 * 101 + 9 *–1, 479 = 1 10 + 4 10 + 7 10 + 9 100

Binary (base 2)Binary (base 2) • Two digits: 0, 1• To make the binary numbers more readable, the digits are often put in groups of 4–1010 = 1 * 23 + 0 * 22 + 1 * 21 + 0 * 20 1010 1 2 + 0 2 + 1 2 + 0 2

= 8 + 2 = 10 10

–1100 1001 = 1 * 27 + 1 * 26 + 1 * 23 + 1 * 20 = 128 + 64 + 8 + 1= 128 + 64 + 8 + 1 = 201

How to Encode Numbers: Binary NumbersNumbers

• Working with binary numbersI b t h l t k f 10–In base ten, helps to know powers of 10• One, Ten, Hundred, Thousand, ...

–In base two, helps to know powers of 2• One, Two, Four, Eight, Sixteen, .., , , g , ,• Count up by powers of two

24 23 22 21 2029 28 27 26 25 24 23 22 21 2029 28 27 26 25a

16 8 4 2 1512 256 128 64 32

Important Property of Binary NumberImportant Property of Binary Number• Number of different number can be possible for a N bit binary numbera N bit binary number – 2N, for 2 bit number it is 4 (00, 01,10 and 11)

S ti t N bit bi b t• Summation two N bit binary number cannot exceed 2*2N

• 20+21+22+23+…+2N=2N+1‐1 example– 1+2+4+8=15=16‐1

Octal (base 8)Octal (base 8)• Shorter & easier to read than binary8 di it 0 1 2 3 4 5 6 7• 8 digits: 0, 1, 2, 3, 4, 5, 6, 7,

• Octal numbers to DecimalOctal numbers to Decimal1368 = 1 * 82 + 3 * 81 + 6 * 80 8

= 1 * 64 + 3 * 8 + 6 * 1= 94= 9410

Hexadecimal (base 16)Hexadecimal (base 16)• Shorter & easier to read than binary• 16 digits:

–0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A, B, C, D, E, F

• “0x” often precedes hexadecimal numbers

0x123 = 1 * 162 + 2 * 161 + 3 * 160 = 1 * 256 + 2 * 16 + 3 * 1 1 256 2 16 3 1= 256 + 32 + 3291= 291

CountingDec Binary Oct Hex Dec Binary Oct Hex

0 00000 0 0 8 01000 10 8

1 00001 1 1 9 01001 11 9

2 00010 2 2 10 01010 12 A

3 00011 3 3 11 01011 13 B

4 00100 4 4 12 01100 14 C

5 00101 5 5 13 01101 15 D

6 00110 6 6 14 01110 16 E

7 00111 7 7 15 01111 17 F

8 01000 10 8 16 10000 20 10

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Fractional NumberFractional Number • Point: Decimal Point, Binary Point, Hexadecimal point

• Decimal247.75 = 2x102+4x101+7x100 + 7x10‐1+5x10‐2

• BinaryBinary10.101= 1x21+0x20 + 1x2‐1+0x2‐2+1x2‐3

• Hexadecimal• Hexadecimal6A.7D=6x161+10x160 + 7x16‐1+Dx16‐2

Converting To and From DecimalConverting To and From Decimal

Successive Division

Weighted M lti li ti

Decimal100 1 2 3 4 5 6 7 8 9

Weighted Multiplication

Successive Division

Multiplication

Successive Di i i

Weighted M lti li ti

Octal80 1 2 3 4 5 6 7

Hexadecimal160 1 2 3 4 5 6 7 8 9 A B C D E F

Division Multiplication

BinaryBinary20 1

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Decimal↔ BinaryDecimal ↔ Binary

SuccessiveSuccessiveDivision

a) Divide the decimal number by 2; the remainder is the LSB of the binary) y ; ynumber.

b) If the quotation is zero, the conversion is complete. Otherwise repeat step (a) using the quotation as the decimal number. The new remainder is the ( ) g qnext most significant bit of the binary number.

Weighted

a) Multiply each bit of the binary number by its corresponding bit‐

Multiplication

weighting factor (i.e., Bit‐0→20=1; Bit‐1→21=2; Bit‐2→22=4; etc).

b) Sum up all of the products in step (a) to get the decimal number.

Decimal to Binary : Division Methody• Divide decimal number by 2 and insert remainder into new binary number.y– Continue dividing quotient by 2 until the quotient is 0.

• Example: Convert decimal number 12 to binary

12 div 2 = ( Quo=6 , Rem=0) LSB6 div 2 = (Quo=3, Rem=0)3 div 2 = (Quo=1,Rem=1)3 div 2 (Quo 1,Rem 1)1 div 2 = ( Quo=0, Rem=1) MSB

1210= 1 1 00 2

Decimal to Octal ConversionDecimal to Octal ConversionThe Process: Successive Division• Divide number by 8; R is the LSB of the octal number• Divide number by 8; R is the LSB of the octal number • While Q is 0

U i th Q th d i l b• Using the Q as the decimal number. • New remainder is MSB of the octal number.

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1

LSB 6 r 11 94 8 ←=

0

3 r 1 11 8 = 9410 = 1368

MSB 1 r 0 1 8 ←= 28

Decimal to Hexadecimal ConversionDecimal to Hexadecimal Conversion

The Process: Successive Division• Divide number by 16; R is the LSB of the hex

number • While Q is 0

• Using the Q as the decimal number. • New remainder is MSB of the hex number.

LSBE 5

9416

0

LSB E r 94 16 ←=

9410 = 5E16MSB 5 r

0 5 16 ←=

10 16

Substitution CodeSubstitution Code

Convert 1110 0110 10102 to hex using the 4‐bit substitution code :

E 6 A

1110 0110 1010

E 6 A

E6A16

Flow chart and ProblemFlow chart and Problem solvingsolving

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Computing in this courseComputing in this course • Given a Problem :

– In English description

• Solve using Computer– Design methods to solve the problem – Analyze the designed solution for correctnessy g– Design flow chart to for the design method to solve the problem

– Write Pseudocode/source code– Compile and rule the code : You may get some error p y g– Test the code (with some Input): you get some error

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Computing– In English description

• Solve using Computer

Computing

– Design methods to solve the problem – Analyze the designed solution for correctnessy g– Design flow chart to for the design method to solve the problem ProgrammingProgramming


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Problem Solving Example 1• Given a positive integer, is this number divisible by 3?y

• Method 1Di id h b b h d f h–Divide the number by three and test for the reminder• Repeated substation of 3 till you get the result less than 3

• Divide using usual method and get reminder

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Problem Solving Example 1• Given a positive integer, is this number divisible by 3?

• Method 2 – Sum all the digits of the number and test theSum all the digits of the number and test the divisibility of Sum by three, repeat the same.

– If sum is divisible by three then number is divisible yby three. How you got to know?

– Some one must have solved this problem and proved the correctness of the solution. • Solution will work for all the cases

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Problem Solving Example 2• Problem: Searching for your friend

– Your friend is near to square and you are exactly q y yat the square

– You don’t know which direction and how much distance from the square

– You need to find your friend Friend

• Any Solution?– How you will search him?How you will search him?– Any approachThink for 2 3 minutes you– Think for 2‐3 minutes

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you


– You don’t know which direction and how much distance from the square

• Solution?– K=1;– While(not found)

Friend

• For all directions – Go K meter and return to square

• K=k+1;K k 1;

• Analysis :Total distance covered – 4*2(1+2+ +X)=8*X*(X+1)/2=4X2+4X you– 4 2(1+2+..+X)=8 X (X+1)/2=4X +4X

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you


– You don’t know which direction and how much di t f thdistance from the square

• Better Solution? YESK 1– K=1;

– While(not found)• Go in Next directions

Friend

Go in Next directions – Go K meter and return to square

• K=k*2;

A l i di t d• Analysis: distance covered – At max (1+2+22+23+..+X+2X+4X+8X)

< 16*X < 4X2+4X (earlier solution) you< 16*X < 4X2+4X (earlier solution)

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you


Computing– In English description

• Solve using Computer

Computing

– Design methods to solve the problem – Analyze the designed solution for correctnessy g– Design flow chart to for the design method to solve the problem ProgrammingProgramming


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Thanks

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machine code, number system, and c variables and operations

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