me you are given an array of four 16-Bit humbers stored in with a sta Using PL- elative addressing, write an LC-3 machine language program that will copy the same four 16-bit numbers in reverse rder, starting at memory address x3120. For example, if the following are the four 16-bit values stored in x3110, x3110 12 x3111 23

i have this code .ORIG x3000SetupLEA R0, InfoPUTS ADD R6, R6, #5; Init Main Loop counter (loop 5 times) ;Print first 2 Fib values before starting;----- Your Code Here -----;LD R1, #0   JSR printNum   LD R2, #1   JSR printNumJSR printNum    ; Print the value in R2   JSR printCommaSpace ; Print comma and space   ;----- End Your Code Here -----;;Loop and call subroutines to calculate each subsequent value in the sequence;Exit the loop once all single-digit values in the sequence have been printedMainLoop ;----- Your Code Here -----;   JSR calcNextFib   JSR printNum   JSR printCommaSpaceADD R6, R6, #-1BRz Done    BR MainLoop   ;----- End Your Code Here -----; Done HALT ;-----------------------------------;Subroutines;----------------------------------- ;Uses R1 and R2 to calc next value in Fibonacci sequence;When complete, Fib2 (R2) will contain the new Fib number;and Fib1 (R1) will contain the previous value of Fib2 (R2)calcNextFib;----- Your Code Here -----;ST R1, SaveR3 ; Save R1ST R2,…

Write a C99 program compiled with GCC on the RPi 3b/3b+ that uses two 8-bit unsigned integers (uint8_t) variables named “a” and “b”. The initial value of a=0 and b = 6. The program should contain a single for loop that increments the “a” variable a total of 512 iterations (the counter will repeat in the middle). The program should output 512 values in the format below:a a == b a < b a > b0 0 1 01 0 1 0....While the C program uses a 8-bit integer and the hardware circuit uses a 4-bit integer, the modulo nature of the counter is the same. The comparator operation isalso the same.

Using MIPS assembly language, write a MIPS programs that the determines what the ECC code should be for a given number (an 8-bit byte).The codes you create are to work for 8-bit positive numbers as these are simpler to work with than larger numbers. The program is to request the user to enter a byte of data (a positive integer in the range of 0 to 255 in decimal) and then create the 12-bit Hamming code as described in your text (see above). The program is to then output this (with an appropriate label) in hex. the following code is not working and has errors: # Prompt user for inputli $v0, 4       # system call for printing stringla $a0, prompt  # prompt stringsyscall # Read input byteli $v0, 5       # system call for reading integersyscallmove $t0, $v0   # store the byte in register $t0 # Calculate ECC Hamming codesrl $t1, $t0, 4   # extract the first 4 bitsxor $t1, $t1, $t0 # calculate parityand $t1, $t1, 0b1011 # mask out irrelevant bitssll $t1, $t1, 8   # shift left by 8 bitssrl…

Assume you now have 1kB of memory, i.e. the memory address space runs from 0 to 1023. The starting address of the first word is 0, the second word is 4, the third word is 8, and so on. The last word comprising 4 bytes resides in addresses 1020, 1021, 1022, 1023. Thus, the last word starts at 1020, which is a multiple of 4.  Now assume the same 1kB of memory but now, word size is 64 bits. The starting address of the first word is 0, the second, third, and last word starts at? my answer: (correct) the second is equal to  = 8 the third is equal to  = 16       help me  find the last word

Question/ program give below run on 8086 used to add two 32-bit number. Fill the missing code to store the result immediately in location start with 3004? 3000 12 4000 | 12 3004 | 24 3001 11 4001 | 10 3005| 21 3002 02 4002 | 21 3006 | 23 3003 | 0L 4003 | 04 3007 05 Mov AX.3000h Fig (1) memory data segment Mov DS, AX Mov AX,4000 Mov ES, AX Mov S0000k, Mov DL0000k Mov CX.04h NEXT: Mov AX, [ST] Mov BX, (DI} ADD AX, BX

Write an ALP to generate a bill for 6 items purchased in a LULU hypermarketwhose prices are represented as BCD numbers 12h ,14h,18h,22h and 35h and 42hOMR. Save the bill amount in the memory address 8500h

You are to write an Intel x86 assembly language program which makes useof a loop to access the individual elements of the array containing the ASCIIcharacters. You are to initialize the array with the following db directive: message db 'juMping JAck flaSh #1',10, 0 The ASCII character represented by decimal value 10 (0x0A) is the line feedcharacter (similar to endl in C++). The byte with all zeros (the so called NULL character in ASCII) is used to represent the end of the string.   Before entering the loop, which performs the case conversion, the program is to display on the screen the (original) contents of the string. Within the loop, the program is to determine whether the current character represents a lower case character. If the character is lower case, it is to be converted into its uppercase version. Non-alphabetic characters are to remain as is. Upon convertingall the lower case characters, the program is to display the string on the screen. Here is an example of a correct…

Consider two integer arrays B and D of 100 elements. Let X2 and X3 contain the base address of array "B" and "D", respectively, and the index į is in X0, h is in X10. Write LEGV8 code to perform the following operations: (a) Get B[15] from memory and put it in register X5. (b) Put the value 132 in D[6]. (c) Translate the following C code into LEGv8 instructions int j = 50, h; do { D[2*j] = (B[j] + B[j+2])*9 - h; j^j -2; while {j >0}; DO NOT use multiply instruction in your code, use shift instruction for multiplication instead! (d)How many instructions are executed during the running of the code in (c), and how many memory data references are made during executio'n?

Suppose we store an array of two 32-bit numbers {0x12345678, 0x9ABCDEF0} starting at memory address 0x100. If we stored this big-endian, which byte of our data would be stored at memory address 0x104 ? A. 9A B. 12 C. FO D. 78 O E. None of the above (or N/A)