Write a Little Man Computer program that determines if the number entered by a user is an odd or even number.
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Write a Little Man Computer
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- Write ARM Assembly program to do: - ISBN Checksum Validator Use Keil uvision software to write and simulate an ARM assembly program that checks the validity of a given ISBN by verifying the checksum Helping Information to solve program : 1. An ISBN consists of nine digits plus a validation digit. 2. The digits are numbered from right to left as d1, d2 .. d10, with d1 being the validation digit. 3. You can define the ISBN number in a 10-byte array in the memory. Each digit is stored in a byte. 4. To check whether the ISBN is valid or not, the following calculations are performed: Result = (10 x d10 + 9 x d9 + ... + ix di + ... + 2 x d2) % 11 Calculate d1 11 - Result If the calculated d1 equal to the d1 in the input ISBN then it is valid If d1 = 10, it is written 'X' For example, given the following ISBN: ISBN = 5123487654 ISBN= 5 1 2 3 487654 d10 d9 d8 d7 d6 d5 d4 d3 d2 d1 Result = x 10 + x 9 + x 8 + x 7 + x 6 + x 5 + x 4 + x 3 + x 2 mod 11 = 216 mod 11 =7 d1 = 11 - Result = 11 -7 =4 ,…In MPLAB PIC16F84A Design an algorithm that compares two 3-byte numbers. •Each number is large enough to require 24 bits of 3 bytes of space. (You need 3 addresses for each of them) •There are three possible outcomes (Greater, equal, and lesser). Each outcome leads to a separate subroutine. •Subroutines are not important they can be empty. •Greater Subroutine sets RB0 •Equal Subroutine sets RB1 •Lesser Subroutine sets RB2 Thank you.A19 I 320 Vec Time I A16 2 AI5 3 31 A18 300 A17 290 A14 28 A13 270 A8 A12 14 A7 A6- 6 AS 7 260 49 A4 18 250 AIl 240 A10 23 A20 2204 АЗ 19 A2 10 Al A21 12 D7 210 200 190 180 170 A0 DO 13 D6 DI 14 D5 D24 15. D4 Vss 16 D3 Top View Given the memory chip shown in this picture determine the size of the memory chip in bits? a. 32 Mbit O b. 8 Mbit 1Mbit d. 2 Mbit e. 16 Mbit f. 4 Mbit
- physcal addresses are 4s ng 4 Ame dat in a cetain compe, te addresses can be translaled without y TLB entries At most how many ditina vid the address translation peh has 12 vld The Translation Look aside Bulfer (TLB)i sine is kB and the word size iby The memory is word addresible. The pe virtual addresses are 64 bea long d th sine is miss?Java Programming Language DSA Don't use built in libraries. We want to build an address book that is capable of storing name, address & phone number ofa person. Address book provides functionality in the form of a menu. The feature list includes:• Add – to add a new person record• Delete – to delete an existing person record by name• Search – to search a person record by name· Exit – to exit from applicationBuffer overflow: break it down for me.
- What is the difference between the x84 and x64 bit versions of Windows? using citationsAssembly Language x86: (Micro Macro): templet: .386.model flat, stdcall.stack 4096ExitProcess PROTO, dwExitCode: DWORD.data day BYTE 0month BYTE 0year WORD 0 .codemain PROC main ENDPINVOKE ExitProcess, 0END main Question: What bit string repersents April 1, 2024? Examples: Date: 00111 (year) 1100 (month) 10010 (day)00111 = 18, 1100 = 12, 10010 = 30, 1980 + 30 = 2010 = 2010 Dec 18mov ax, 00111110010010b; 2010 Dec 10; 30 12 18 DAY:mov dx, ax and dx, 0000000000011111bmov day, dl (18) ; 00111 Month: mov dx, ax and dx, 00000001111000000b shr dx, 5 ; 000000000000 1100 (12) mov month, dl Year:mov dx, axshr dx, 9 ; 011110and dx, 0000000001111111bmov year, dl (30) 10010Entries in the access control list explicitly allow or deny what? Group of answer choices access to the access list copying the token list specific types of operations on the token specific types of operations on the acess list
- The memory location at address 00002001 contains the memory variable in binary form. What is the data memory variable in hexadecimal form? MEMORY 1110 1011 00002001 1110 1010 00002000 1110 1001 00001999 1110 1000 00001998 1110 0111 00001997 1110 0110 00001996 DATA ADDRESS The data memory variable in hexadecimal form is E7. a. b. The data memory variable in hexadecimal form is EA. The data memory variable in hexadecimal form is EB. C. The data memory variable in hexadecimal form is E9. Od.Code a descriptor that describes a memory segment that begins at location 0005CF00h and ends at location 00060EFFh. The memory segment is a data segment that grows upward in the memory system and can be written. The segment has a user level privilege (lowest) and has not been accessed. The descriptor is for an 80386 microprocessor.movlw movwf Cirf rw Loop: addwf btfsc incf decfsz COUNTER, 1 if temp1 started with a value 125 goto Addlw Movwf NOP 0X03 COUNTER temp2 temp1.0; status.c temp2 In the given program segment; Loop Ox02 Whatwill temp2 = ? when NOP is executed? temp1 92- а. 2 (b. 1 c. 15 DECIMAL d. 4