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- Translate the following machine code into English. There is no need to trace the execution of the program. Write one English sentence per instruction. For example, for instruction 0xB103, you would write, "Compare register 1 to register 0; if they're the same, jump to address 0x03." 0x10AA 0x2FB0 0xBF00 0x7521 0xA504 0x40E1 0x50FFComputer Science Write an 8051 assembly language program to determine if a sequence of bytes is a palendrome. The number of bytes in the sequence will be stored in memory location 0x20 and the sequence starting at location 0x21. For example MOV 0x20 #0x05 MOV 0x21 #0x00 MOV 0x21 #0x10 MOV 0x21 #0x20 MOV 0x21 #0x10 MOV 0x21 #0x00 could be used to set up the data. Upon completion register R0 should contain a value of 0x01 if the sequence is a palendrome and 0x00 if it is not. Make sure you test your code on various input sequences (length 0, length 1, even length palendrome, even length non-palendrome, odd length palendrome, odd length non-palendrome. Deliverables Assembly code source file screen shot(s) of running program1. T/F - if (B)=006000 (PC)=003600 (X)=000090, for the machine instruction 0x032026, the target address is 003000.2. T/F – PC register stores the return address for subroutine jump.3. T/F – S register contains a variety of information such as condition code.4. T/F – INPUT WORD 1034 – This means Operating system should reserve 1034 bytes in memory5. T/F - In a two pass assembler, adding literals to literal table and address resolution of local symbol are done using first pass and second pass respectively.
- 1.The Instruction that sets all odd bits (D0, D2, D4, ... ) of register r1 to one and keeps its even bits unchanged is: Group of answer choices A.ORR r1, r1, #0x55555555 B.ORR r1, r1, #0xAAAAAAAA C.AND r1, r1, #0x55555555 D.ORR r1, r1, #0x01010101 2. Suppose Mask = 0x00000F0F and P = 0xDCBADCBA. What is the result of the following bitwise operation? Q = ~Mask Group of answer choices a. Q = 0xFFFFF0F0 b. Q = 0x00000F0F c. Q = 0x00000C0A d. Q = 0xDCBAD0B0Single instruction computer (SIC) has only one instruction that for all operations our MIPS does. The instruction has the following format. sbn a, b, c # Mems[a]=Mem[a]- Mem[b]; if (Mem[a]<0) go to PC+c For example, here is the program to copy a number from location a to location b: Start: sbn temp, temp, 1 sbn temp, a, 1 sbn b, b, 1 sbn b, temp, 1 So build SIC program to add a and b, leaving the result in a and leaving b unmodified.Write an assembly language program for the 8085 Microprocessor to subtract three 8-bit numbers 12H, FFH, 03H. Store the result in the memory location 2050H and carry (borrow) in 2051H.
- For example, instruction like this. 0011 0000 0000 0000 ; start at x30000001 001 001 1 00000 ;x3000 set N CC if R1 < 00000 100 0 0000 0001 ;x3001 if N goto x3003 R2+=50000 111 0 0000 0001 ;x3002 Not N goto x30040001 010 010 1 00101 ;x3003 R2 += 5;x3004 doneQuestion 36 Find the Binary Representation for each of the following Decimal numbers 85 Group of answer choices 1100101 1.010101 1010101 none of them Question 37 Assume that EBX and ECX have the following values EBX: FF FF FF 75 ECX: 00 00 01 A2 After the execution of the instruction XCHG EBX, ECX The Value in ECX is _________ Group of answer choices FFFFFF75 000001A2 00000117 none of themWrite an assembly language instruction that has five WORD size variables in its data section as follows: num1 WORD 5 num2 WORD 7 num3 WORD 8 num4 WORD 9 result WORD ? Write an assembly language program that adds num1 + num2 + num3 + num4 and places the result in result. Note that do not add two memory locations in one instruction.
- Write a MARIE program simulation the following Python code. The MARIE program must generate the output similar to Python's. The program must have a loop. Python code: num = [1,2,3,4,5,6]for i in num: print (i) This is what I have so far but it only prints from 1 to infinity: ORG 100START, LOAD NUM STORE COUNT OUTPUT LOAD NUM ADD ONE STORE NUM LOAD NUM SUBT COUNT SKIPCOND 400 // Salta si COUNT < NUM JUMP START HALT ONE, DEC 1NUM, DEC 1COUNT, DEC 1 END STARTWrite an assembly language program to define an array of word size and initialize it with 5 numbers. Use shift instruction to check the last bit of each of the numbers(to check if the number is positive or negative). If the last tit is a 1, it means the number is negative. Count the positive numbers that this array contains.Write a new program in MIPS Assembly language using your LFSR function from the first problem. Start from state 0x00000001, then run a loop counting all the possible states in the LFSR before it cycles back to 0x00000001. This is shown by the following pseudocode: count = 0; initial = state = 0x00000001; do { state = LFSR(state); count++; while (state != initial); print count; Verify that the number of states is (2^15) - 1 = 32767 . Your program should complete within 30 seconds. If it takes longer then you likely have entered an infinite loop.