Pengantar Organisasi Komputer

1
IKI10230Pengantar Organisasi KomputerKuliah no.
03 Sistem Bilangan
Sumber1. Paul Carter, PC Assembly Language2.
Hamacher. Computer Organization, ed-53. Materi
kuliah CS61C/2000 CS152/1997, UCB
25 Februari 2004 L. Yohanes Stefanus
(yohanes_at_cs.ui.ac.id)Bobby Nazief
(nazief_at_cs.ui.ac.id) bahan kuliah
http//www.cs.ui.ac.id/kuliah/POK/
2

• Penggunaan Tools
• GCC, NASM, EDEBUG32

3
C Compiler GCC

• Fungsi
• menerjemahkan program dalam bahasa C ke bahasa
  mesin
• Kompilasi
• gcc OPSI ltnama-filegt
• c menghasilkan .o (object-file)
• S menghasilkan .s (assembly-file) format ?
  NASM
• g (default) menghasilkan informasi untuk debug
• Cara 1
• gcc c myprog.c ? menghasilkan myprog.o
• gcc o myprog.exe myprog.o ltother-object-filesgt
• Cara 2
• gcc o myprog.exe myprog.c ltother-object-filesgt
• Contoh hello.c ? hello.exe
• Cara 1
• gcc c hello.c ? hello.o
• gcc o hello.exe hello.o
• Cara 2
• gcc o hello.exe hello.c

4
hello.c

• int main()
• printf("hello world\n")

5
Assembler NASM

• Fungsi
• menerjemahkan program dalam bahasa rakitan ke
  bahasa mesin
• Perakitan
• nasm OPSI ltnama-filegt
• f OBJ-TYPE
• f coff menghasilkan format COFF (.o)?
  digunakan oleh DJGPP
• f obj (default) menghasilkan format OMF (.obj)
• g (default) menghasilkan informasi untuk debug
• nasm f coff myprog.asm ? myprog.o
• gcc o myprog.exe myprog.o ltother-object-filesgt
• ? memungkinkan pengintegrasian dengan object-file
  yang bersumber dari bahasa C
• Contoh hello.asm ? hello.exe
• nasm f coff hello.asm ? hello.o
• gcc o hello.exe driver.c hello.o

6
hello.asm

• extern _printf
• segment .data
• the_str db "hello world", 10, 0
• segment .bss
• segment .text
• global _asm_main
• _asm_main
• enter 0,0
• push dword the_str
• call _printf printf(hello world\n)
• pop eax
• leave
• ret

7
driver.c

• int asm_main( void )
• int main()
• int ret_status
• ret_status asm_main()
• return ret_status

8
Debuger EDEBUG32

• Fungsi men-debug program (.exe)
• eksekusi program secara single-step (per
  instruksi)
• eksekusi program sampai dengan instruksi tertentu
  (breakpoint)
• evaluasi register memori
• Persyaratan
• sebelumnya program harus di-kompilasi/rakit
  dengan g
• Cara Menjalankan
• edebug32 ltfile-exegt

9
EDEBUG32 Commands

• go ltvgt g go, stop at ltvgt
• cont c continue execution
• step s step through current instruction
• next n step to next instruction
• list l u list instructions (takes addr, count)
• dump d dump memory (takes addr, count)
• print p print value of expression (takes expr)
• break b set breakpoint (takes which, addr)
• status breakpoint status
• regs r print registers
• set set register/memory
• npx display 80387 contents
• where display list of active functions
• whereis find a symbol/location (takes wildcard
  or value)
• cls clear screen
• help h,? print help
• quit q quit

10
tugas0a.asm (1/2)

• segment .data
• data1 db 11h
• data2 dw 2222h
• data3 dd 33333333h
• datatmp times 9 db 0ffh
• data4 times 16 db 0
• segment .bss
• stacks resd 1
• segment .text
• global _asm_main
• _asm_main
• enter 0,0
• pusha
• mov eax,10 decimal number, value 10
• mov ebx,10b binary number, value 2
• mov ecx,10h hexadecimal number, value
  16

11
tugas0a.asm (2/2)

• xor eax,eax
• xor ebx,ebx
• xor ecx,ecx
• xor edx,edx
• xor esi,esi
• xor edi,edi
• mov esi,data1 esi points to data1
• mov al,esi load 1 byte
• mov bx,esi load 1 word (2 bytes)
• mov ecx,esi load 1 double-word (2 words
  4 bytes)
• mov edx,data1
• mov esi,data2
• mov edi,data3
• mov data4,dl store 1 byte
• mov data4,dx store 1 word
• mov data4,edx store 1 double-word

12
Tugas 0 (1/4)

• Merakit tugas0a.asm
• nasm f coff tugas0a.asm
• nasm f coff asm_io.asm
• gcc o tugas0a.exe driver.c tugas0a.o asm_io.o
• Debug tugas0a.exe
• edebug32 tugas0a.exe
• gtgt
• gtgt g asm_main
• eax00000000 ebx000002a5 ecx00000000
  edx0000033f
• esi00000054 edi00010b50 ebp00090b30
  UP IE PL NZ AC PE NC
• ds01f7 es01f7 fs01e7 gs0207
  ssesp01f700090b14 cs01ef
• _asm_main()
• 000015e0 c8000000 enter 0,0
• gtgt l
• _asm_main()
• 000015e0 c8000000 enter 0,0
• 000015e4 60 pusha
• 000015e5 b80a000000 mov eax,0xa
• 000015ea bb02000000 mov ebx,0x2

13
Tugas 0 (2/4)

• Debug tugas0a.exe (lanjutan ...)
• gtgt s
• 000015e4 60 pusha
• gtgt r
• eax00000000 ebx000002a5 ecx00000000
  edx0000033f
• esi00000054 edi00010b50 ebp00090b10
  UP IE PL NZ AC PE NC
• ds01f7 es01f7 fs01e7 gs0207
  ssesp01f700090b10 cs01ef
• 000015e4 60 pusha
• gtgt s
• 000015e5 b80a000000 mov eax,0xa
• gtgt s
• 000015ea bb02000000 mov ebx,0x2
• gtgt r
• eax0000000a ebx000002a5 ecx00000000
  edx0000033f
• esi00000054 edi00010b50 ebp00090b10
  UP IE PL NZ AC PE NC
• ds01f7 es01f7 fs01e7 gs0207
  ssesp01f700090af0 cs01ef
• 000015ea bb02000000 mov ebx,0x2

14
Tugas 0 (3/4)

• Debug tugas0a.exe (lanjutan ...)
• gtgt s
• gtgt ...
• eax00000011 ebx00002211 ecx33222211
  edx33222211
• esi33332222 edi33333333 ebp00090b10
  UP IE PL ZR PE NC
• ds01f7 es01f7 fs01e7 gs0207
  ssesp01f700090af0 cs01ef
• 00001620 8815e0c80000 mov
  __what_size_app_thinks_it_is26,dl dl11
• 
  si2222
• ? mov data4,dl store 1 byte
• gtgt d __what_size_app_thinks_it_is26
• 0x0000c8e0 0x00000000 0x00000000 0x00000000
  0x00000000
• gtgt s
• 00001626 668915e0c80000 mov
  __what_size_app_think_thinks_it_is26,dx ...
• gtgt d 0xc8e0
• 0x0000c8e0 0x00000011 0x00000000 0x00000000
  0x00000000

15
Tugas 0 (4/4)

• Merakit tugas0b.asm
• nasm f coff tugas0b.asm
• gcc o tugas0b.exe driver.c tugas0b.o asm_io.o
• Eksekusi tugas0b.exe
• bandingkan hasil print-out dengan evaluasi isi
  register memori dengan menggunakan EDEBUG32!
• Laporan
• tugas0a ? rangkuman hasil pengamatan register
  memori
• hasil perbandingan hasil pengamatan dengan
  print-out tugas0b

16
tugas0b.asm (1/3)

• include "asm_io.inc"
• segment .data
• same as tugas0a.asm
• segment .bss
• same as tugas0a.asm
• segment .text
• global _asm_main
• _asm_main
• enter 0,0
• pusha
• dump_regs 1 initial condition
• mov eax,10
• mov ebx,10b
• mov ecx,10h
• mov edx,eax

17
tugas0b.asm (2/3)

• xor eax,eax
• xor ebx,ebx
• xor ecx,ecx
• xor edx,edx
• xor esi,esi
• xor edi,edi
• dump_regs 3 eax, ebx, ecx, edx, esi, edi
  should be 0
• dump_mem 1,data1,0 initial condition of
  data1
• dump_mem 2,data2,0
  data2
• dump_mem 3,data3,0
  data3
• mov esi,data1
• mov al,esi
• mov bx,esi
• mov ecx,esi
• mov edx,data1
• mov esi,data2
• mov edi,data3
• dump_regs 4 watch changes in eax, ebx, ecx,
  edx, esi, edi

18
tugas0b.asm (3/3)

• dump_mem 4,data4,0 initial condition of
  data4
• mov data4,dl
• dump_mem 5,data4,0 watch changes in data4
• mov data4,dx
• dump_mem 6,data4,0 watch changes in data4
• mov data4,edx
• dump_mem 7,data4,0 watch changes in data4
• popa
• mov eax, 0
• leave
• ret

19

• REVIEW
• Stored Program Computer

20
Review Bit dapat mepresentasikan apa saja !!!

• Bits dapat merepresentasikan apapun!
• Karakter? Latin
• 26 huruf gt 5 bits
• Huruf besar/kecil tanda lain gt 7 bits, berapa
  simbol huruf?
• Karakter, bahasa lain gt 16 (unicode)
• Logical values?
• 0 -gt False, 1 gt True
• Warna ? Berapa banyak warna gt berapa bits?
• Alamat? (berapa karakter alfabet ..)
• .. Tapi N bits ? hanya dapat merepresentasikan
  2N sesuatu

21
Review Bit ? Instruksi

• Instruksi (Operasi) dapat direpresentasikan oleh
  bit.
• Contoh
• 0 gt tepuk tangan
• 1 gt bersiul
• Eksekusi Instruksi
• 1. 0
• 2. 1
• 3. 1
• 4. 0
• 5. 0
• 6. 1
• 7. 0

22
Review Kumpulan bit disimpan di memori
Alamat

• Memori adalah tempat menyimpan kumpulan bit
  (instruksi/data)
• Suatu word adalah sejumlah bit data tetap,
  (mis. 16, atau 32 bit) pada satu lokasi di memori
• Byte-addressable memory menyimpan data multi-byte
  pada lokasi memori yang berurutan
• Alamat menunjuk ke lokasi word (byte-1)
  disimpan.
• Alamat dapat direpresen-tasikan oleh bit
• Alamat juga sebagai bilangan (yang dapat
  dimanipulasikan)

00000
data
01110
101101100110
11111 2k - 1
23
Review Stored-program Computer

• operasi yang dilakukan oleh komputer ditentukan
  oleh instruksi data yang tersimpan di memori

0846
IP
1686
IP
0061
IP
0061
0017
0017
0078
0078

• komputer dapat diprogram untuk memenuhi kebutuhan
  pengguna dengan jalan mengisi memori dengan
  instruksi data yang sesuai

24

• Representasi Data

25
Bilangan Biner

• Simbol 0,1
• Harga/Nilai suatu bilangan biner
• 1011010 1x26 0x25 1x24 1x23 0x22
  1x21 0x20 64 16
  8 2
  90
• Penulisan 1011010b
• Konversi Desimal ? Biner
• 90 / 2 45 sisa 0
• 45 / 2 22 sisa 1
• 22 / 2 11 sisa 0
• 11 / 2 5 sisa 1
• 5 / 2 2 sisa 1
• 2 / 2 1 sisa 0
• 1 / 2 0 sisa 1

26
Bilangan Heksa-Desimal

• Simbol 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F
• Harga/Nilai suatu bilangan heksa-desimal
• 5A 5x161 10x160
• 80 10 90
• Penulisan 5Ah atau 0x5A
• Konversi Desimal ? Heksa-desimal
• 90 / 16 5 sisa 10 (A)
• 5 / 16 0 sisa 5
• Konversi Heksa-desimal ? Biner
• 5A 101 1010
• Konversi Biner ? Heksa-desimal
• 1011010 101 1010
• 5 A 5A

27
Tabel Bilangan
28
Pengelompokkan Bit

• Bit String
• INTEL MIPS
• 4 bit nibble nibble
• 8 bit byte byte
• 16 bit word half-word
• 32 bit double-word word
• 64 bit quad-word double-word
• Alamat lokasi memori
• umumnya dinyatakan dengan bilangan heksa desimal
• contoh
• lokasi memori 90 pada memori dengan ruang memori
  sebesar 64K (65536 216) dinyatakan dengan
  alamat
• 0x005A
• jika ruang memori sebesar 232 (4G)
• 0x0000005A

29
Penyimpanan data multi-byte (Little Endian)
int i 90 90 0x5A 0000 0000 0000 0000 0000
0000 0101 1010
i
j
int j 987700 987700 0x000F1234 0000 0000
0000 1111 0001 0010 0011 0100
30

• Addition of Positive Numbers

31
One-Bit Full Adder (1/2)

• Example Binary Addition

Carries

• Thus for any bit of addition
• The inputs are ai, bi, CarryIni
• The outputs are Sumi, CarryOuti
• Note CarryIni1 CarryOuti

32
One-Bit Full Adder (2/2)

• To create one-bit full adder
• implement gates for Sum
• implement gates for CarryOut
• connect all inputs with same name

33
Ripple-Carry Adders adding n-bits numbers
CarryIn0
A0
1-bit FA
Sum0
B0
CarryOut0
CarryIn1
A1
1-bit FA
Sum1
B1
CarryOut1
CarryIn2
A2
1-bit FA
Sum2
B2
CarryOut2
CarryIn3
A3
1-bit FA
Sum3
B3
CarryOut3

• Kinerja operasi penjumlahan (dan juga
  operasi-operasi aritmatika lainnya) akan
  bergantung pada besar unit data dan konfigurasi
  Adder (Arithmetic Logical Unit) yang digunakan

34

• Signed Numbers

35
How to Represent Negative Numbers?

• So far, unsigned numbers
• Obvious solution define leftmost bit to be sign!
  
• 0 gt , 1 gt -
• Rest of bits can be numerical value of number
• Representation called sign and magnitude

36
Shortcomings of sign and magnitude?

• Arithmetic circuit more complicated
• Special steps depending whether signs are the
  same or not
• Also, Two zeros
• 0x00000000 0ten
• 0x80000000 -0ten
• What would it mean for programming?
• Sign and magnitude abandoned

37
Another try complement the bits

• Example 710 001112 -710 110002
• Called ones Complement
• Note postive numbers have leading 0s, negative
  numbers have leadings 1s.

• What is -00000 ?
• How many positive numbers in N bits?
• How many negative ones?

38
Shortcomings of ones complement?

• Arithmetic not too hard
• Still two zeros
• 0x00000000 0ten
• 0xFFFFFFFF -0ten
• What would it mean for programming?
• Ones complement eventually abandoned because
  another solution was better

39
Search for Negative Number Representation

• Obvious solution didnt work, find another
• What is result for unsigned numbers if tried to
  subtract large number from a small one?
• Would try to borrow from string of leading 0s,
  so result would have a string of leading 1s
• With no obvious better alternative, pick
  representation that made the hardware simple
  leading 0s ? positive, leading 1s ? negative
• 000000...xxx is gt0, 111111...xxx is lt 0
• This representation called twos complement

40
Twos Complement Number line
00000
11111
00001

• 2N-1 non-negatives
• 2N-1 negatives
• one zero
• how many positives?
• comparison?
• overflow?

11110
00010
0
-1
1
2
-2
. . .
. . .
15
-15
-16
01111
10001
10000
41
Twos Complement Numbers

• 0000 ... 0000 0000 0000 0000two
  0ten0000 ... 0000 0000 0000 0001two
  1ten0000 ... 0000 0000 0000 0010two
  2ten. . .0111 ... 1111 1111 1111 1101two
  2,147,483,645ten0111 ... 1111 1111 1111
  1110two 2,147,483,646ten0111 ... 1111 1111
  1111 1111two 2,147,483,647ten1000 ... 0000
  0000 0000 0000two 2,147,483,648ten1000 ...
  0000 0000 0000 0001two 2,147,483,647ten100
  0 ... 0000 0000 0000 0010two
  2,147,483,646ten. . . 1111 ... 1111 1111
  1111 1101two 3ten1111 ... 1111 1111 1111
  1110two 2ten1111 ... 1111 1111 1111
  1111two 1ten
• One zero, 1st bit is called sign bit
• but one negative with no positive
  2,147,483,648ten

42
Twos Complement Formula

• Can represent positive and negative numbers in
  terms of the bit value times a power of 2
• d31 x -231 d30 x 230 ... d2 x 22 d1 x 21
  d0 x 20
• Example1111 1111 1111 1111 1111 1111 1111
  1100two
• 1x-231 1x230 1x229... 1x220x210x20
• -231 230 229 ... 22 0 0
• -2,147,483,648ten 2,147,483,644ten
• -4ten
• Note need to specify width we use 32 bits

43
Twos complement shortcut Negation

• Invert every 0 to 1 and every 1 to 0, then add 1
  to the result
• Sum of number and its ones complement must be
  111...111two
• 111...111two -1ten
• Let x mean the inverted representation of x
• Then x x -1 ? x x 1 0 ? x 1 -x
• Example -4 to 4 to -4
• x 1111 1111 1111 1111 1111 1111 1111 1100two
• x 0000 0000 0000 0000 0000 0000 0000 0011two
• 1 0000 0000 0000 0000 0000 0000 0000 0100two
• () 1111 1111 1111 1111 1111 1111 1111 1011two
• 1 1111 1111 1111 1111 1111 1111 1111 1100two

44
Twos comp. shortcut Sign extension

• Convert 2s complement number using n bits to
  more than n bits
• Simply replicate the most significant bit (sign
  bit) of smaller to fill new bits
• 2s comp. positive number has infinite 0s
• 2s comp. negative number has infinite 1s
• Bit representation hides leading bits sign
  extension restores some of them
• 16-bit -4ten to 32-bit
• 1111 1111 1111 1100two
• 1111 1111 1111 1111 1111 1111 1111 1100two

45

• Addition Subtraction of Signed Numbers

46
Addition Subtraction Operations

• Subtraction
• Form 2s complement of the subtrahend
• Add the two numbers as in Addition

• Addition
• Just add the two numbers
• Ignore the Carry-out from MSB
• Result will be correct, provided theres no
  overflow

0 1 0 1 (5)0 0 1 0 (2) 0 1 1 1 (7)
0 1 0 1 (5)1 0 1 0 (-6) 1 1 1 1 (-1)
0 0 1 0 (2) 0 0 1 0?0 1 0 0 (4) 1 1 0
0 (-4) 1 1 1 0 (-2)
1 0 1 1 (-5)1 1 1 0 (-2)11 0 0 1 (-7)
0 1 1 1 (7)1 1 0 1 (-3)10 1 0 0 (4)
1 1 1 0 (-2) 1 1 1 0?1 0 1 1 (-5) 0 1 0
1 (5) 10 0 1 1 (3)
47
Overflow

• Examples 7 3 10 but ...
  
• - 4 5 - 9 but ...

1
1
1
0
1
0
1
1
1
1
1
0
0
7
4
3
5
0
0
1
1

1
0
1
1

1
0
1
0
0
1
1
1
6
7
48
Overflow Detection

• Overflow the result is too large (or too small)
  to represent properly
• Example - 8 lt 4-bit binary number lt 7
• When adding operands with different signs,
  overflow cannot occur!
• Overflow occurs when adding
• 2 positive numbers and the sum is negative
• 2 negative numbers and the sum is positive
• Overflow can be detected by evaluating
• Carry into MSB ? Carry out of MSB

49

• Arithmetic Branching Conditions

50
Condition Codes

• CC Flags will be set/cleared by arithmetic
  operations
• N (negative) 1 if result is negative (MSB 1),
  otherwise 0
• C (carry) 1 if carry-out(borrow) is generated,
  otherwise 0
• V (overflow) 1 if overflow occurs, otherwise 0
• Z (zero) 1 if result is zero, otherwise 0

0 1 0 1 (5)0 1 0 0 (4) 1 0 0 1 (-7)
0 1 0 1 (5)1 0 1 0 (-6) 1 1 1 1 (-1)
0 0 1 1 (3)1 1 0 1 (-3)10 0 0 0 (0)
0 1 1 1 (7)1 1 0 1 (-3)10 1 0 0 (4)