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- Use the following data declarations: .data byte Val word Val sbyte 1, 2, 3, FCh word 1000h, 2000h, 3000h, 4000h dwordVal dword 34567890h, 90785634h, 0Ah, 33445566h Show the value of the final destination operand after each of the following code fragments has executed: (If any instruction/s is invalid, indicate "INV" as the answer and briefly explain why) a. mov bh,byteVal+2 b. mov edx,1 C. add dx,[wordVal+4] mov ecx,5 xchg ecx,[dwordVal+12] d. mov ah, byte Val+3 sub ah,[ byte Val+0] sub ah,[ byte Val +2] e. mov eax, dword ptr dwordVal+7 f. movsx cx,byteVal+3 answer bh= answer edx= answer ecx= answer ah= answer eax= answer (show your answer in binary) CX=Q 4. State whether the following instructions are valid or not. If valid then define properly, if not, then state the reason of their invalidity: MOV [BX + SI + 2], AL XCHG AL, CX ADD 23H, AL5. Load the register (CL) from the memory location [050OH] then subtract the content of this register from the accumulator (AL). Correct the result as a (BCD) numbers. Let [0500H] 12H & AL 3FH %3D
- 4-Write a program to add the following 32-bit numbers. use the DAD, LHLD and SHLD instructions. 1F20D43A h + B7248D6C hLIST P = 16F877AUsing INCLUDE "P16F877A.INC" and PWM (Pulse Width Modulation), write the microprocessor code that enables the fan to operate at 80% if the button on the 5th bit is pressed, and 30% if the button on the third bit is pressed.This flag is used by the instructions that perform BCD (binary .coded decimal) arithmetic Carry Flag O Parity Flag O Auxiliary Carry Flag O the Direction Flag O
- Topic: Assembly Language Write a program called bit_check.asm that jumps to a label if either bit 4, 5, or 6 is set in the BL register. If these bits are set the program should print "Bits 4, 5, or 6 are set." followed by a new line character. If they are not set, jump to a label to print "Bits 4, 5, and 6 are not set." followed by a new line character. There should be 4 tests, test with each bit position set and with none of them set. Use the print_string procedure to print your output. Remember, you can jump to the exit label from anywhere in your program.MOV AL, 5 ADD AL, 4 The flag register is affected by the above addition and the carry flag will equal * to 1 MOV AL,OF5H ADD AL,OBH After the addition AL will contain FF BB 00 11 IKQuesti on 1: This problem covers 4-bit binary multiplication. Fill in the table for the Product, Multiplier and Multiplicand for each step. You need to provide the DESCRIPTION of the step being performed (shift left, shift right, add, no add). The value of M (Multiplicand) is 1100, Q (Multiplier) is initially 0111. Product Multiplicand Multiplier Description Step 0000 0000 0000 1100 0111 Initial Values Step 0 Step 1 Step 2 Step 3 Step 4 Step 5 Step 6 Step 7 Step 8. Step 9 Step 10 Step 11 Step 12 Step 13 Step 14 Step 15
- 2.- If the binary code for LOAD is (0011), for ADD is (0010), for STORE is (0110), for JMP is (0001) and for END is (0111). Assuming a 16 bits format, write the object code (binary representation) of the following program: Note: use 4 bits for the op-code and 12 bits for the ADDR.Assignment for Computer Architecture! this is about hamming codes write the code IN MIPS ASSEMBLY LANGUAGE calculating hamming codes; The key to the Hamming Code is the use of extra parity bits to allow the identification of a single error. Create the code word as follows: Mark all bit positions that are powers of two as parity bits. (positions 1, 2, 4, 8, 16, 32, 64, etc.) All other bit positions are for the data to be encoded. (positions 3, 5, 6, 7, 9, 10, 11, 12, 13, 14, 15, 17, etc.) Each parity bit calculates the parity for some of the bits in the code word. The position of the parity bit determines the sequence of bits that it alternately checks and skips.Position 1: check 1 bit, skip 1 bit, check 1 bit, skip 1 bit, etc. (1,3,5,7,9,11,13,15,...)Position 2: check 2 bits, skip 2 bits, check 2 bits, skip 2 bits, etc. (2,3,6,7,10,11,14,15,...)Position 4: check 4 bits, skip 4 bits, check 4 bits, skip 4 bits, etc. (4,5,6,7,12,13,14,15,20,21,22,23,...)Position 8: check 8 bits,…Complete the table below to show the numerical results from applying the indicated operation code to the data shown. Opcodes are six-bit numbers defined as shown below. R = 1 for Rotate 0 for Shift; D = 1 for Right 0 for Left; F is fill, and A2-A0 define the number of bits. R D F A2 A1 A0 Input(Base10) Input(Base2/8-bit) OpCode Output(Base10) Output(Base2/8-bit)) 47 00101111 110011 229 11100101 80 110111 17 011001 123 100011
