30. Repeat Question 29 for BX equal to 9AOOH What are the opcode, data type, and

Question

30. Repeat Question 29 for BX equal to 9AOOH

What are the opcode, data type, and operand(s) in this instruction: MOV AH,7 5. What is meant by byte-swapping? 6. Which of these assembler directives produce data: ORG. DB. DW. DD. DF. SEG. END? 7. What does a linker do? 8. List the seven basic instruction groups. 9. Identify the source and destination addressing mode in each erf these instructions: MOV AX.BX MOV AH,7 MOV [DI]AL MOV AX.(BP) MOV AL,[SI+6] JNZ XYZ CBW 10. Why arc the flags so important in a control-type program? 11. What does this two-instruction sequence do? XCHG AX. BX XCHC BX.CX 12. What does this instruction accomplish? BSWAP ESX 13. Memory locations 00490H through 00493H contain, respectively. 0A. 9C. B2. and 78. What does AX contain after each instruction? (Assume that SI contains 00490H and that BP contains 0002.) (a) MOV AX.[SI] (b) MOV AX.[SI+l] (C) MOV AX.[SI][BP] 14. Registers AX. BX. CX. and DX contain, respectively. 1111H. 2222H. 3333H. and 4444H. What are the contents of each register after this sequence of instructions? PUSH AX PUSH CX PUSH BX POP DX POP AX POP BX 15. What it the difference between LDS and LES? When should each be used? 16. Show the instructions needed to scan a 200-byte string for the byte 25H. 17. Repeat Question 16 for the word value 9A25H. 18. Redo Example 3.32 for auto-decrement copying. 19. Redo Example 3.36 by using byte operation instead. !0. Explain how the processor switches from 16-bit operands to 32-bit operands, when operating in the real mode. 21. What are the simplified segment directives discussed in this chapter? 22. If EAX contains 00000200H. EBX contains 00000003H. and the data segment con¬tains 1000H, what is the effective address generated by these instructions? (a) MOV ECX, [EAX] (b) MOV ECX, [EBX][EAX] (c) MOV ECX.[EAX][EBX*8) (d) MOV ECX.[ESI][EDI]

Explanation / Answer

#4

Opcode is MOV, Datatype is byte, operands are AH (register) and 7 (immediate value)

#12

It is a byteswap intruction.. so if EBX =1234H then after the instruction it would become 3412H

#14

DX=BX

AX=CX

BX=AX

DX=2222H AX=3333H BX=1111H CX=3333h

#22

a) Move the contents of memory address stored in EAX to ECX

b) Move the contents of memory address stored in (EAX +EBX) to ECX

c) Move the contents of memory address stored in (EAX +EBX*8) to ECX

d) Move the contents of memory address stored in (ESI +EDI) to ECX

#30

MOVS is move with sign extended. since the last bit is 1 hence 1 will be appended

hence EBX=FFFF 9A00 H

MOVZ is move with Zero extended. hence 0 will be appended

hence EBX=0000 9A00 H

Related Questions

Navigate

• Browse (All)
• Browse 3
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.