CSC231 - Assembly · mith College C omputer Science Dominique Thiébaut [email protected] CSC231...

mith College

Computer Science

Dominique Thiébaut [email protected]

CSC231 - AssemblyWeek #8

D. Thiebaut, Computer Science, Smith College

https://xkcd.com/571/

Can't Sleep


Logic Design


Image from http://www.willegal.net/appleii/images/motherboard.jpg

Apple II Motherboard

http://www.willegal.net/appleii/images/motherboard.jpg


Nasa Computer(Apollo)


A Bit of History

• Claude Shannon • 21 years old • 1937 • MIT Master's Thesis • All arithmetic operations

on binary numbers can be performed usingboolean logic operators

https://en.wikipedia.org/wiki/Claude_Shannon

https://en.wikipedia.org/wiki/Claude_Shannon


https://www.youtube.com/watch?v=xTQDIiSWK_k




Designing a 2-bitAdder with Logic


ALU

CU

Processor


ALU

yixi

sumi

carryi

Compute add x, y


yx

z

…

x0 y0x1 y1x2 y2xn yn x3 y3

z0z1z2z3zn


x0y0

z0

0

+ 0 ____

= 0 0

0+ 1____

= 0 1

1+ 0____

= 0 1

1+ 1____

= 1 0

carry


x0 y0 Carry z0

0 0 0 0

0 1 0 1

1 0 0 1

1 1 1 0


x0 y0 Carry z0

0 0 0 0

0 1 0 1

1 0 0 1

1 1 1 0

Carry = x0 and y0

z0 = x0 xor y0


x0

z0

XORAND

y0

carry


Moral of the Story: Addition is performed

by logic operations using natural binary numbers…

(unsigned arithmetic)


NEGATIVE NUMBERS


How can we represent signed binary numbers when

all we have are bits (0/1)?

Whichever system we use should work

with the binary adder in the ALU…


Binary Hex Unsigned Decimal

0000 0 00001 1 10010 2 20011 3 30100 4 40101 5 50110 6 60111 7 71000 8 81001 9 91010 A 101011 B 111100 C 121101 D 131110 E 141111 F 15

4-bit Nybble



0 000 0 00 001 1 10 010 2 20 011 3 30 100 4 40 101 5 50 110 6 60 111 7 71 000 8 81 001 9 91 010 A 10 1 011 B 111 100 C 121 101 D 131 110 E 141 111 F 15

4-bit NybbleSign Bit



0 000 0 00 001 1 10 010 2 20 011 3 30 100 4 40 101 5 50 110 6 60 111 7 71 000 8 81 001 9 91 010 A 10 1 011 B 111 100 C 121 101 D 131 110 E 141 111 F 15

PositiveNumbers

NegativeNumbers

4-bit Nybble


Signed MagnitudeNumber System


Signed Magnitude Rule: To find the opposite of a number, just flip

its MSB

0 101 (+5)

1 101(-5)

and vice versa…



Signed Magnitude

0 000 0 0 +00 001 1 1 +10 010 2 2 +20 011 3 3 +30 100 4 4 +40 101 5 5 +50 110 6 6 +60 111 7 7 +71 000 8 8 -01 001 9 9 -11 010 A 10 -2 1 011 B 11 -31 100 C 12 -41 101 D 13 -51 110 E 14 -61 111 F 15 -7

4-bit Nybble


Does this System work With the ALU Adder?


Signed Magnitud

e0 000 0 0 +00 001 1 1 +10 010 2 2 +20 011 3 3 +30 100 4 4 +40 101 5 5 +50 110 6 6 +60 111 7 7 +71 000 8 8 -01 001 9 9 -11 010 A 10 -2 1 011 B 11 -31 100 C 12 -41 101 D 13 -51 110 E 14 -61 111 F 15 -7

3 + -3——- = 0

4 + -1——- = 3


1's ComplementNumber System


1's Complement Rule: To find the opposite of a number,

just flip all its bits

0 101 (+5)

1 010(-5)

and vice versa…



1'sComplement

0 000 0 0 +00 001 1 1 +10 010 2 2 +20 011 3 3 +30 100 4 4 +40 101 5 5 +50 110 6 6 +60 111 7 7 +71 000 8 8 -71 001 9 9 -61 010 A 10 -5 1 011 B 11 -41 100 C 12 -31 101 D 13 -21 110 E 14 -11 111 F 15 -0

4-bit Nybble


Does this System work With the ALU Adder?


1'sComplement

0 000 0 0 +00 001 1 1 +10 010 2 2 +20 011 3 3 +30 100 4 4 +40 101 5 5 +50 110 6 6 +60 111 7 7 +71 000 8 8 -71 001 9 9 -61 010 A 10 -5 1 011 B 11 -41 100 C 12 -31 101 D 13 -21 110 E 14 -11 111 F 15 -0

3 + -3——- = 0

4 + -1——- = 3

5 + -3——- = 2


2's ComplementNumber System


2's Complement Rule: To find the opposite of a number,

just flip all its bits, and add 1

0 101(+5)

1 011(-5)

and vice versa…

1 010 + 1

0 100 + 1



2'sComplement

0 000 0 0 +00 001 1 1 +10 010 2 2 +20 011 3 3 +30 100 4 4 +40 101 5 5 +50 110 6 6 +60 111 7 7 +71 000 8 8 -81 001 9 9 -71 010 A 10 -6 1 011 B 11 -51 100 C 12 -41 101 D 13 -31 110 E 14 -21 111 F 15 -1

4-bit Nybble


Does this System workWith the ALU Adder?


2'sComplement

0 000 0 0 +00 001 1 1 +10 010 2 2 +20 011 3 3 +30 100 4 4 +40 101 5 5 +50 110 6 6 +60 111 7 7 +71 000 8 8 -81 001 9 9 -71 010 A 10 -6 1 011 B 11 -51 100 C 12 -41 101 D 13 -31 110 E 14 -31 111 F 15 -1

3 + -3——- = 0

4 + -1 ——- = 3

5 + -3——- = 2


InterestingProperty

• What is the binary representation of -1 as a byte?

• What is the binary representation of -1 as a word?

• What is the binary representation of -1 as a dword?


int x = 0x7fffffff - 5;

for ( int i=0; i<10; i++ )System.out.println( x++ );

getcopy Loop0x7fffffff.java


What did you just learn about Java ints?


negneg op8 neg op16 neg op32 op: mem, reg

alpha db 1beta dw 4x dd 0xF06

neg byte[alpha] mov ax,1234 neg ax

neg dword[x]

neg oprnd

To get the

2's complement

of an int


Range of 2's Comp't. intsBinary Hex Unsigned

Decimal0000 0 00001 1 10010 2 20011 3 30100 4 40101 5 50110 6 60111 7 71000 8 81001 9 91010 A 101011 B 111100 C 121101 D 131110 E 141111 F 15


Type Minimum Maximum # Bytes

unsigned byte 0 255 1

signed byte -128 127 1

unsigned short 0 65,535 2

signed short -32,768 32,767 2

unsigned int 0 4,294,967,295 4

signed int -2,147,483,648 2,147,483,647 4

signed long

−9,223,372,036,854,775,808

9,223,372,036,854,775,807 8

We Stopped Here Last Time

http://www.genengnews.com/media/images/genhighlight/Mar12_2014_8296129_StopSigns_GeronTrialHalt2322361121.jpg

(and finished first exercise on next slide)

http://www.genengnews.com/media/images/genhighlight/Mar12_2014_8296129_StopSigns_GeronTrialHalt2322361121.jpg


The 13th Floorhttps://www.youtube.com/watch?v=dtYdZkPmFoU

https://www.youtube.com/watch?v=dtYdZkPmFoU


The LOOP Instruction


looploop label

x dd 1sum dd 0 mov ecx, 10addUp: mov eax, dword[x] add dword[sum], eax inc dword[x] loop addUp ;ecx<—ecx-1 ;if ecx!=0, ; goto addUp

loop label


looploop label


loop label

Label


looploop label ;ecx <— ecx-1 ;if ecx!= 0, ; goto label ;else continue


loop label

Label

Labels _start: mov eax, 4

mov ecx, 10 for1: … … loop for1 for2: … …

loop for2

• Start with a letter (or a dot)

• End with a colon (when declared)

• Represent an address in the code section

• Must be unique in program


Tracinga Code Example


mov ecx, 3 mov eax, 1for: call _printDec inc eax loop for ;ecx<—ecx-1 ;if ecx!=0, ; goto for

eax

ecx



?eax

3ecx



1eax

3ecx



1eax

3ecx

1



1 2eax

3ecx

1



1 2eax

3 2ecx

1



1 2eax

3 2ecx

12



1 2 3eax

3 2ecx

12



1 2 3eax

3 2 1ecx

12



1 2 3eax

3 2 1ecx

123



1 2 3 4eax

3 2 1ecx

123



1 2 3 4eax

3 2 1 0ecx

123


mov ecx, 3 mov eax, 1for: call _printDec inc eax loop for ;ecx<—ecx-1 ???? ;if ecx!=0, ; goto for

1 2 3 4eax

3 2 1 0ecx

123


Example 1Sum of 1..10


; computes sum(1,2, …10)x dd 1sum dd 0 mov ecx, 10 mov eax, dword[x]addUP: add dword[sum], eax inc eax loop addUp ;ecx<—ecx-1 ;if ecx!=0, ; goto addUp mov eax, dword[sum] call _printDec


; computes sum(1,2, …10)x dd 1sum dd 0 mov ecx, 10 addUP: add dword[sum], ecx loop addUp ;ecx<—ecx-1 ;if ecx!=0, ; goto addUp mov eax, dword[sum] call _printDec

Do we needeax?


Example 2Fibonaccis

http://i.dailymail.co.uk/i/pix/2016/07/17/13/1A32203E000005DC-3694326-image-m-17_1468760305397.jpg

http://i.dailymail.co.uk/i/pix/2016/07/17/13/1A32203E000005DC-3694326-image-m-17_1468760305397.jpg


_start: mov eax, 1 ; fibn mov ebx, 1 ; fibn-1 call _printDec call _println

mov ecx, 10-1 ; we printed 1, 9 more to go for: mov edx, ebx mov ebx, eax add eax, edx call _printDec call _println loop for

getcopy fib.asm

; print the first 10 Fibonacci terms


Looping Through

Arrays


Looping Through

Arrays

LOOP INSTRUCTION

INDIRECTADDRESSINGMODE


Indirect AddressingMode

The addressing mode refers to the way the operand of an instruction is generated. We already know register mode, immediate mode, and direct mode.


Tracing One Example

of Indirect Addressing (Base Addressing)


12

33

78

56

3E

F0

3

1

Memory

0x110450x110460x110470x110480x110490x1104A0x1104B0x1104C

???ebx?al

section .data A db 1,3,0xF0,0x3E,0x56 B db 0x78,0x33,0x12

section .text

_start: mov al, 'z' mov ebx, A mov byte[ebx], 0

mov ebx, B mov byte[ebx], al


12

33

78

56

3E

F0

3

1

Memory

0x110450x110460x110470x110480x110490x1104A0x1104B0x1104C


section .text



???ebx'z'al


12

33

78

56

3E

F0

3

1

Memory

0x110450x110460x110470x110480x110490x1104A0x1104B0x1104C


section .text



11045ebx'z'al


12

33

78

56

3E

F0

3

1 0

Memory

0x110450x110460x110470x110480x110490x1104A0x1104B0x1104C


section .text



11045ebx'z'al


12

33

78

56

3E

F0

3

1 0

Memory

0x110450x110460x110470x110480x110490x1104A0x1104B0x1104C


section .text



1104Aebx'z'al


12

33

78 'z'

56

3E

F0

3

1 0

Memory

0x110450x110460x110470x110480x110490x1104A0x1104B0x1104C


section .text



1104Aebx'z'al


Example 2:Setting an Array

to All 0s


; Array Table contains 10 wordsTable dw 1,2,3,4,5,6 dw 7,8,9,10 mov ecx, ____ ;# of elements mov ebx, ____ ;address of ;Tableclear: mov word[ebx], ____;value to store add ebx,____ ;make ebx point ;to next word loop clear ;ecx<—ecx-1 ;if ecx!=0, ; goto clear


ExercisesProblem #1: Store the first 10 Fibonacci terms in an array of ints (32 bits)

Problem #2: Given a DNA sequence of 1,000,000 characters stored in an array of bytes, and all characters in uppercase, transform it into its lowercase equivalent. The characters are A, C, G, T and N.


We stopped here last time…

CSC231 - Assembly · mith College C omputer Science Dominique Thiébaut [email protected] CSC231...

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