# Write and execute assembly programs that give register AX = 5 , BX = 8, CX = 9.Then swap the value to be AX = 8 ,BX = 9, CX =5. AX = 5
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![# Write and execute assembly programs that give register AX = 5 , BX
= 8, CX = 9.Then swap the value to be AX = 8 ,BX = 9, CX =5.
AX = 5
BX = 8
Cx-9
# Write and execute assembly programs that
- Define num1 as byte with a value 6
- Define num1 as byte with a value 5
- num1 = numl + 2[ Using inc ]
- num2 = num2 - 3 [ Using dec ]
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Solved in 2 steps
- 1. 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.Using the information below, in assembly language: x BYTE -2, -3, 2, 1 z WORD 3000h, 4000h, 5000h, 6000h y WORD -14, 32 bx starts with a value of 2222h, what is the hex value of bx when the following instructions execute in sequence? 1. mov bl, x 2. mov bh, [x+4]5. When two words are multiplied (one in BX), the most significant word of result will be in _______, and the least significant word will be in __________ (1 Mark) a. BX, CX b. CX, BX c. AX, DX d. DX, AX 6. Compare and contrast SUB and CMP instructions? (1 Mark) 7. Write the contents of AH and BL after execution of the program. (1 Mark) MOV AH,40h SAL AH, 01 MOV BL,80h SAR BL,01 HLT AH= ____________ BL= ____________ 8. Write the contents of AL and BL register after execution of the program (1 Mark) MOV AL, FFh MOV BL, AL CMP AL, F0H SUB BL,10H HLT AL= ____________ BL= ____________ 9. Identify and correct the mistakes. (1 Mark) MOV AX, 32H ____________________ MOV BL, 2424H ____________________ DAS AX, BL ____________________ HLT ____________________ 10. Write an ALP to rotate right the contents of AL register 4 times including carry and save the result in DH register (1 Mark)
- 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),03) The 8-bit register AR, BR, CR, and DR initially have the following values: [5]AR = 11010110; BR = 11100111; CR = 10110001; DR = 10111010Determine the 8-bit values in each register after the execution of the following sequenceof microoperations.AR AR + BR Add BR + ARCR CR AND DR, BR BR + 1 AND DR to CR, Increment BRAR AR - CR Subtract CR from ARQuestion 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…
- 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 accordance…What will be the value of BL (in Hexadecimal) after the execution of the following instructions: MOV CL,52 MOV BL, 19 CMP BL, CL MOV CL,2 RCR BL, CLWhat will be the value of BL (in Hexadecimal) after execution of the following instructions: MOV CH,26 MOV BL, 17 XOR BL,CH SHR BL, 1 SHR BL,1
- Read the Assembly program carefully, understand its working/functionality and answer the below given questions. MOV DX, 0090 MOV DS, DX MOV DX, 0019 MOV AX, 0 MOV CX, FFFFUP: ADD CX, 0002 INC AX SUB DX, CX JNZ UP ;JNZ is Jump if ZF=0 in the result of SUB instruction INT 10 INT 7 a. What will be the values of below given registers after execution of the program. DX: CX: b. What will be displayed on the LCD after the execution of the program?For the MIPS assembly instructions below, what is the corresponding C statement?Assume that the variables f, g, h, i, and j are assigned to registers $s0, $s1, $s2, $s3, and$s4, respectively. Assume that the base address of the arrays A and B are in registers $s6 and $s7, respectively. sll $t1, $s1, 2add $t1, $t1, $s6lw $t1, 0($t1)sub $t0, $s3, $s4sll $t0, $t0, 2add $t0, $t0, $s7lw $t0, 0($t0)add $t1, $t1, $t0sll $t0, $s0, 2add $t0, $t0, $s7sw $t1, 0($t0)2 Write assembly program lists for XORing two Binary numbers. Y=A+A' B' ORG 200 ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... ....................................................................... .......................................................................…