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- Make a quadruples address code out of the statement below: A = - A * (A + B ) - (B – C) / DQuestion Write an assembly code to implement the y = (x1+x2) * (x3+x4) expression on 2-address machine, and then display the value of y on the screen. Assume that the values of the variables are known. Hence, do not worry about their values in your code. The assembly instructions that are available in this machine are the following: Load b, a Load the value of a to b Add b, a Add the value of a to the value of b and place the result in b Subt b, a Subtract the value of a from the value of b and place the result in b Mult b, a Multiply the values found in a and b and place the result in b Store b, a Store the value of a in b. Output a Display the value of a on the screen Halt Stop the program Note that a or b could be either a register or a variable. Moreover, you can use the temporary registers R1 & R2 in your instructions to prevent changing the values of the variables (x1,x2,x3,x4) in the expression. In…Q1- Write a program in assembly language for the 8085 microprocessor to send 10 bytes of data located at the memory address (3000H to 3009H) using SOD at a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz. When sending each of the required bytes, you must adhere to the following: The two high bits of the start bits must be sent, after that the data bits are sent, after that the low bit of the stop bit is sent. The following flowchart will help you, but you should notice that this flowchart deals with one byte, and you are required to deal with 10 bytes. The solution must be integrated and include the calculation of the baudrate delay time Transmit No Set up Character Bit Counter Send Start Bit Wait Bit Time Get Character in Accumulator Output Bit Using Do Wait Bit Time Rotate Next Bit in Do Decrement Bit Counter Is It Last Bit? Yes Add Parity if Necessary • Send Two Stop Bits Return (a)
- 38. Implement the following expression in assembly language, using 32-bit integers (you may modify any registers you wish): eax = -dword1 + (edx - ecx) + 1 You can use this data definition for testing your code: dword1 DWORD 10h 39. se the following data declarations to write an assembly language loop that copies the string from source to target. Use indexed addressing with EDI, and use the LOOP instruction source BYTE "String to be copied",0 target BYTE SIZEOF source DUP(0),0Translate the following C code into MIPS code. Here X and Y are stored in $s1 and $s2 registers respectively and the base address of A and B are in $s4 and $s5 registers respectively. X = 7* A[6] -4*B[7] Y = 17*(X+A[9]) A[X+1] = Y Please solve this with #proper explanationWrite the following expression using 3 address format, 2 address format, 1 address format and zero address format. W = ((M + C) / (C+ 5))*2 %3D
- Create three address code for given expression z := a + a * (b - c) + (b - c) * dWrite three address code and quadruple for the expression a[i] = -b * (a[k--] – y[ k--] /2)For the following code segment write the machine language representation of each instruction in binary. The instruction codes are add->32,beq->4,addi->8, lw->35,j->2. Asume that Loop has the address of Ox4CB23 Loop: beq $t1, $t2, done lw $s1, 0(($t0) add $s0, $s1, $s0 addi $t1, $t1, 1 j Loop done:
- Q5.Write a multiplication an Intel 8085 assembly program to multiply 2 numbers. The numbers are stored in memory locations 3000H & 3001H. Store the result in memory locations 3002H & 3003H. Show your flow chart and the assembly code. Q6. Answer True or False for the followings: a) Machine code is the assembly code b) Data field is 16 bit while address field is 8 bit c) Trainerkitcanbeused for implementing assembly code d) ADo bus can be used for addressing and datatransfer e) WR and RD pins are on the same pinQ1- Write a program in assembly language for the 8085 microprocessor to receive 10 bytes of data via the SID and store it at the memory address (3000H to 3009H) using a baud rate of 1200. Information: The 8085 processor operates at a frequency of 3.072 MHz. When you receive each byte of the required bytes, you must adhere to the following: The bits of two high bits will be received at the beginning of the reception (start bits), after that the data bits will be received, after that the low bit of the stop bit will be received (stop bit). The following flowchart will help you, but you should notice that this flowchart deals with one byte, and you are required to deal with 10 bytes The solution must be integrated and include the calculation of the baudrate delay time Of+CD!HID+[00 Yes SIDATA Read SID Start Bit? Wait for Half-Bit Time Set up Bit Counter Wait Bit Time Read SID Save Bit Decrement Bit Counter All Bits Received? Add Bit to Previous Bits Go Back to Get Next Bit Return IMUNI1. Given the following C code: short ARR [50]; // short is 2 bytes for (short i = 1; i<= 50; i++) ARR [i] = i * i; Convert this code into assembly. Write two versions of the code by using two different addressing modes.