Question

Assembly Language: Write and run a program to find the values of each destination operand:

.data

varB BYTE 65h,31h,02h,05h

varW WORD 6543h,1202h

varD DWORD 12345678h

.code

mov ax, WORD PTR [varB+2] ; a=

mov bl, BYTE PTR varD ; b=

mov bl, BYTE PTR [varW+2] ; c=

mov ax, WORD PTR [varD+2] ; d=

mov eax,DWORD PTR varW   ; e=

Answer 1

VarB :

Address	Data
0000	65h	varB
0001	31h	varB+1
0002	02h	varB+2
0003	05h	varB+3

VarW :

Offset Address	Data
0000	6543h
0002	1202h

VarD :

Offset Address	Data
0000	12345678h

a. mov ax, WORD PTR [varB+2] ;

It moves WORD stored at location varB+2 i.e. at address offset 0002.

Value is stored in reverse order in register, therefore ax will contain 0502h.

Hence, ax = 0502h

b. mov bl, BYTE PTR varD ;

Above instruction moves a byte from varD to bl register.

MSB i.e. 12 is stored at highest offset and LSB i.e. 78h is stored at lowest offset considered in byte format.

Byte format	Offset
78	0000 varD
56	0001
34	0002
12	0003

Hence, bl will store 78h.

c. mov bl, BYTE PTR [varW+2] ;

Above instruction stores a byte from varW+2 into register bl.

varW+2 points at offset address 0002 which contains 1202h.

Now, LSB of 1202h i.e. 02h is stored at least offset if considered in byte format.

Therefore, bl = 02h

d. mov ax, WORD PTR [varD+2] ;

Byte format of varD :

Byte format	Offset
78	0000	varD
56	0001	varD+1
34	0002	varD+2
12	0003	varD+3

Now, above instruction will move word into ax i.e. 1234h.

Hence, ax = 1234h

e. mov eax,DWORD PTR varW   ;

Word format	Offset
6543h	0000
1202h	0002

Above instruction moves double word to eax. Register stores data in reverse format as in memory.

Therefore, eax = 12026543h