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- Modify the following program in lc-3 to get the input(numbers) from the keyboard and print the result to console. .ORIG x3000Loop LD R0, number1 ; load number1 into R0LDR R1, number2 ; load number2 into R1ST R2, SaveR2 ; save register R2LD R5, goSUB ; load address of SUB into R4JSRR R5 ; go to subroutine whose address in R5STR R3, result ; store resultLD R2, SaveR2 ; restore old value R2HALTnumber1 .FILL #10number2 .FILL # -8goSUB .FILL SUB ; initialize goSUB to address of SUBSaveR2 .BLKW #1; reserve space SaveR2 and SaveR3result .BLKW #1SUB NOT R1, R1ADD R2, R1, #1ADD R3, R0, R2RET.ENDConvert given code to LEGv8 code:int f, g, y //global 64-bit variablesint sum (int a, int b) { // at memory address X0+1000.return (a +b)} int main (void) // at memory address X0 + 800{f=2;g=3;y= sum (f, g);return y;}Convert this code, making valid assumptions about registers and register use. Notethat brackets and global variable declarations are not affecting the addresses of the instructionsin memory.Given the following data definitions, the address of the first variable var1 is given at 0x1001 1000 (hexadecimal). Using little endian, show the memory dump below. Fill each box below with a byte of the allocated memory in hexadecimal. You may use the ASCII table provided. .data var1: .byte 3, -2, ‘A’ var2: .half 1, 256, 0xFFFF var3: .word 0x3DE1C74, 0xFF .align 3 str1: .asciiz "CPS2390" DATA SEGMENT: ADDRESS
- In C, both long x and int *y are 8-byte values. How can we can tell which one is which in the assembly compiled from the C? (I.e., what is true of one but not the other? If several answers seem correct, pick the most precise option.) A. x is stored in a register, y in memory B. x is modified using addq, y using addl C. x may be an argument to add but y won't be D. When an argument of mov, a register storing x will never be inside parentheses E. When an argument of mov, a register storing y will always be inside parenthesesWrite a program in HACK assembly, without using symbols, that computes thebitwise exclusive or (XOR) of the values stored in RAM[1] and the value of thememory location with address stored in RAM[2]. The result of the computationshould be stored in RAM[0].You can think of RAM[2] as being a pointer to where the second operand of the XORis stored.Let's say that p is a pointer to memory and the next four bytes in memory (in hex) beginning at p's address are: 89 AB CD EF. After the following code is run on a little-endian computer what are the values of x, y, and the four bytes in memory beginning at p's address. Answer for x and y in four digits of upper-case hexadecimal (eg, 0AC6). Answer for memory as eight digits of upper-case hexadecimal with a single space between each byte (eg, 89 AB CD EF). x: y: mem at p:
- Q1- Write a program in assembly language for the 8085 microprocessor to send one byte of data located at the memory address (3000H ) using SOD at a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz . When sending the required byte, you must adhere to the following: The two high bits of the start bits(1 1) must be sent, after that the data bits are sent, after that the low bit of the stop bit (0) is sent. The following flowchart will help you. The solution must be integrated and include the calculation of the baudrate delay time(Practice) a. Using Figure 2.14 and assuming the variable name rate is assigned to the byte at memory address 159, determine the addresses corresponding to each variable declared in the following statements. Also, fill in the correct number of bytes with the initialization data included in the declaration statements. (Use letters for the characters, not the computer codes that would actually be stored.) floatrate; charch1=M,ch2=E,ch3=L,ch4=T; doubletaxes; intnum,count=0; b. Repeat Exercise 9a, but substitute the actual byte patterns that a computer using the ASCII code would use to store characters in the variables ch1, ch2, ch3, and ch4. (Hint: Use Appendix B.)Write a service routine which resets all elements of an array that resides in memory location from A000 H to A0FF H with DS equal to 0000 H. The service routine address is CS:IP where CS is 2000 H and IP is 0100H. Assume the interrupt type that is called is 50 (x8086- nano)
- Solve the 8085 Write a program to load twenty memory locations starting from 8005H, where each location's content should increases by 2 over the previous one, however, the first location should contain 04H, assuming that the programs start at memory location 9009H?49. Which of the following is/are restriction(s) in classless addressing ? a. The number of addresses needs to be a power of 2 b. The mask needs to be included in the address to define the block c. The starting address must be divisible by the number of addresses in the block d. All of the aboveIf the value 8 is stored in the memory location designated by address 5, what is the functional difference between writing the value 5 into cell number 6 and copying the contents of cell 5 into cell 6?