al status of the registers, after execution of the instruction JZ 0102h (a) CS-0
ID: 2292786 • Letter: A
Question
al status of the registers, after execution of the instruction JZ 0102h (a) CS-072A, IP 0100 (b) CS-072A, IP-0102 (c) CS 072C, IP-0100 (d) CS-072C, IP-0102 AX-1234 BX-5678 CX-A1B2 DX40000 SP-FFFE BP-C3D4 1sNGHZ DS-072A ES-072A 072A :0100 50 DI-2018 Ss-072A CS-072A IP-0100 PUSH AX 34. After running the instruction PUSH AX , (a) IP 0100, SP-0001 (b) IP-0100, SP-FFFC (c) IP-0101, SP-0001 (d) IP-0101, SP-FFFC 35. After running the instruction, 1234 can be located in memory using the debug comman (d)-P (a) -D (b) -DSS: SP (c)-T 36. If the next instruction POP BX is run (a) SP-FFFE, BX:1234 b) SP-FFFC, BX 5678 (C) SP-FFFE, BX-5678 (d) SP-FFFC, BX-1234 MoV Cx, 0 MOV AX, 0 here: INC AX LOOP here 37. Number of the iterations to be run by the codes is (d) 2416 (c) 2A8 (b) Infinite 38. After the program, (c) AX-2416 (d) AX 2A16-1 (b) AX-1 (a) AX-0Explanation / Answer
33. JZ means jump if zero. that means the control jumps to the specified address. that address is loaded into IP. at the same time CS(code segment) address is also incremented. hence option D is right.
34. option d; while pushing the data to stack, stack pointer(SP) is decremented for every byte transfer. Here AX means 16bytes. so SP should become FFFC hence options a,c are wrong. After pushing instruction pointer (IP) is incremented for next instruction. hence option D is correct.
35. option B; since the data gets stored in stack segment(ss)
36.while POP operation SP gets incremented for every byte. hence SP should FFFE, and BX is loaded with contents of accumulator i.e. 1234 hence option A is right.
37.LOOP means decrement and jump if no zero. initially AX is 0 then incremented 1 and again decremented to 0 and it never jumps to here since there is a zero. so no iterations happened hence option A is correct.
38. due to decrement operation performed by LOOP instruction AX is 0 hence option A is correct.
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