Question

) Identify the hexadecimal value moved to the destination by each of the following instruction below based on the given data declaration. Assume that the offset of byteVal is 00000000H data byteVal wordVal dwordVal aString BYTE 1,2,3,4 WORD 1000H,2000H,3000H,4000H DWORD 12345678H,34567890H BYTE "ABCDEFG",0 ax,offset byteVal i. mov dx,wordVal i. mov al,aString+2 i11. mov dx,wordVal+4 iv. mov mov eax,offset byteVal+2 V. CO2 (10 marks) d) The incomplete program below used to find the multiplication product of the given numbers by 4 from memory location 6100H and store the result from memory location 7000H as shown in Figure Q2(d). Identify the missing programming code and fill in the blanks Initialize source memory pointer BX, 6100H : Initialize destination memory pointer LEA CX MOV AX, BX: Get the number Multiply AX with 4 BACK: SHL AX,2 MOV CX,AX : store the result at destination memory location : Increment source memory pointer : Increment destination memory pointer MOV AX,BX CMP AX,FFH: Check for last number : If not repeat JNZ HLT CO2 (10 marks

Answer 1

Answer :- c)i) ax = 1 i.e. the first byte value of the data array byteVal.

Answer :- c)ii) dx = 1000H i.e. the first word value of the data array wordVal.

Answer :- c)iii) al = 'C' i.e. the 2d byte value of the data array stringVal. Count starts from zero.

Answer :- c)iv) dx = 3000H i.e. the 2nd word value of the data array. The 0th word has (0,1), first has (2, 3). Hence 2nd word has (4, 5). So second word is copied to dx register.

Answer :- c)v) eax = 0x00000003 i.e. the 2d byte value of the data array byteVal.