guide for assembly

8/18/2019 guide for assembly

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CS 40A FINAL HOWARD DACHSLAGER

ASSUME NO SYNTAX ERRORS AND THE LOGIC IS CORRECT. NONE OF THE

PROGRAMS ARE COMPLETE.

Unless otherwise indicated, all numbers are assume unsigned (natural ).

NAME __ STUDENT #

1. For the finite ring R 2 = {00000000, 00000001, ...,

} find:

11010111!10101010 = N2 = _00101101__

2. Compute (2354327* 25517 )mod(3117) = N7 = _15_____

For the finite BYTE of octal numbers R 8 = {0, 1, 2,3,4,5,6,7, 10, 11, ...,, N } ,

answer question 3, and 4:

3. N = _377 _____

4. The number in the BYTE R 8 : N8 = (78)5 M0D(4008) = __67_____

5. For the dword :

The additive inverse in hexadecimal is

6. Assume in an AL program:

x DWORD

The additive inverse of x is

2B 07 FC FF

15 39 04 01

0011 0101 1110 1010 1000 1101 0110 1101


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7. Complete the following tables using hexidecimal numbers only :

8.

Let R = {0,1,2,3,..., 255}be decimal byte ring. For this ring find the following value of

N = 100 + (- 235*59 + 222)*133 = _24________

9.

Complete the following table (Make sure all celled are filled):

mov eax , 278901d

EAX

10. Complete the following table using binary numbers only. (Make sure all celled are filled):

1101 1011 0010 0110 1000 0011 1010 0011

32 25 24 17 16 9 8 1

INSTRUCTIONS

mov eax, 3F293567h 3 F 2 9 3 5 6 7

mov ax, 9BCh 3 F 2 9 0 9 B C

mov ax, 3D32h 3 F 2 9 3 D 3 2

mov ax, 5h 3 F 2 9 0 0 0 5

mov ax, 3h 3 F 2 9 0 0 0 3

mov eax, 1267h 0 0 0 0 1 2 6 7

BASE

16:

0 0 0 4 4 1 7 5

BASE 2: 0 0 0 0 0 0 0 0 0 0 0 0 0100 0100 0001 0111 0101

BASE 8: 0 1 0 4 0 5 6 5


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11.

Complete the following table with HEXADECIMAL NUMBERS ONLY:

For problems 12, 13 complete the following tables using hexadecimal numbers only :

32 25 24 17 16 9 8 1

INSTRUCTIONS

mov eax, 112937234d 0000 0110 1011 1011 0100 1001 0001 0010

mov ax, 9BCh 0000 0110 1011 1011 0100 1001 1011 1100

mov al, 5 0000 0110 1011 1011 0100 1001 0000 0101

mov ah, 0Eh 0000 0110 1011 1011 0000 1110 0000 0101

mov al, 2d 0000 0110 1011 1011 0000 1110 0000 0010

AL PSEUDO CODE AL CODE X Y EAX EBX

X := 23o mov x, 23o 13 - - -

Y := 57o mov y, 57o 13 2F - -

EAX := X mov eax, x 13 2F 13 -

EBX := Y mov ebx, y 13 2F 13 2F

X := EBX mov x, ebx 2F 2F 13 2F

Y := EAX mov y, eax 2F 13 13 2F

Make sure all cells are filled.

12. BASE 16


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14. The largest signed integer number, base 8, that can be store in a variable of type BYTE is

____127__________

The smallest signed integer number, base 8, that can be store in a variable of type BYTE is

___-127____

15. Assume the decimal byte ring {0, 1, 2, ..., 255} is considered signed numbers.

Then the relation 210 < 4 is (a). true (b) false

32

25

24

17

16

9

8 1

INSTRUCTIONS

mov eax, 293567h 0 0 2 9 3 5 6 7

mov ax, 9BCh 0 0 2 9 0 9 B C

mov ax, 3D32h 0 0 2 9 3 D 3 2

mov ax, 5h 0 0 2 9 0 0 0 5

mov ax, 3h 0 0 2 9 0 0 0 3

mov eax, 1267 0 0 0 0 1 2 6 7

13. 32 25 24 17 16 9 8 1

INSTRUCTIONS

mov eax, 112937234d 0 6 B B 4 9 1 2

mov ax, 9BCh 0 6 B B 0 9 B C

mov al, 5 0 6 B B 0 9 0 5mov ah, 0Eh 0 6 B B 0 E 0 5

mov al, 2 0 6 B B 0 E 0 2


5/7

____false_______

16. K := 0

R := 2259

WHILE R > 0

BEGIN

K := K +1

R :=2258 - K*100

END

R := R + 55

For the above program, the final value stored in R is __-7________

17.

For the instruction: mov ecx , 725715041d complete the following table below:

ECX .

18. Complete the following table USING HEXADECIMAL NUMBERS ONLY :

BASE 2: 1010 1101 0000 0110 0010 0001 1000 0001

BASE 16: 2 B 4 1 8 8 6 1

BASE 9: 1 7 7 6 5 0 3 7 8 5

AL PSEUDO CODE AL CODE X Y EAX EBX

X := 29d mov x, 29 1D - - -


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19. Complete the following table using BINARY NUMBERS ONLY (Make sure all cells

that have numbers are filled)

20. Complete the following table in DECIMAL NUMBERS ONLY. (Make sure all cells that

have numbers are filled)

Y := 59d mov y, 59 1D 3B - -

EAX := X mov eax, x 1D 3B 1D -

EBX := Y mov ebx, y 1D 3B 1D 3B

X := EBX mov x, ebx 3B 3B 1D 3B

Y := EAX mov y, eax 3B 1D 1D 3B

32 25 24 17 16 9 8 1

INSTRUCTIONS

mov eax, 293567h 0000 0000 0010 1001 0011 0101 0110 0111

mov ax, 9BCh 0000 0000 0010 1001 0000 1001 1011 1100

mov ax, 3D32h 0000 0000 0010 1001 0011 1101 0011 0010

mov ax, 5h 0000 0000 0010 1001 0000 0000 0000 0101

mov ax, 3h 0000 0000 0010 1001 0000 0000 0000 0011

mov eax, 1267 0000 0000 0000 0000 0001 0010 0110 0111


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INSTRUCTIONS CYCLE OF INSTRUC-

TIONS

N10 A N7 K

N10 := 489 N10 := 489 1E9 - - -

K : = 1 K : = 1 1E9 - - 1

N7 := 0 N7 := 0 1E9 - 0 1

WHILE N10 0 WHILE N10 0

BEGIN BEGIN

A := N10 MOD 7 A := N10 MOD 7 1E9 6 0 1

N7 := N7 + A*K N7 := N7 + A*K 1E9 6 6 1

N10 := N10÷7 N10 := N10÷7 45 6 6 1

K := 10*K K := 10*K 45 6 6 10

A := N10 MOD 7 45 6 6 10

N7 := N7 + A*K 45 6 42 10

N10 := N10÷7 9 6 42 10

K := 10*K 9 6 42 64

A := N10 MOD 7 9 2 42 64

N7 := N7 + A*K 9 2 10A 64

N10 := N10÷7 1 2 10A 64

K := 10*K 1 2 10A 3E8

A := N10 MOD 7 1 1 10A 3E8

N7 := N7 + A*K 1 1 4F2 3E8

N10 := N10÷7 0 1 4F2 3E8

K := 10*K 0 1 4F2 2710

END END

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