Question

Designing a mask for bit manipulation allows us to clear, set, invert, and extract the selected bits in the bit string. Applying bitwise operations AND, OR, XOR, NOT, SHL, SHR, ROL, and ROR allow us to pack and unpack bits in the bit string. In the following instructions below illustrates how bits are being "packed" and "unpacked" from/to the bit string to the 16-bits registers AX/BX, what would be the final bit string value loaded to the memory variable "unpackedData" from the 16-bit register BX? MOV EAX, 0 MOV EBX, 0 MOV AX, 0x47E3 MOV packedData, AX MOV BX, packedData AND BX, 0xE000 ROL BX, 4 MOV unpackedData, BX

Answer 1

Answer is as follows :

MOV EAX,0 // Move data i.e. 0 to 32 bit register EAX i.e. clear the register EAX

MOV EBX,0 // Move data i.e. 0 to 32 bit register EBX i.e. clear the register EBX

MOV AX, 0x47E3 // 16 bit data 47E3 is moved to 16 bit register AX ( 16 bit LSB's of EAX)

MOV packedData , AX // Move data of AX i..e. 47E3 to variable packedData

MOV, BX, packedData //Move data in variable packedData to register BX

AND BX, 0xE000 // perform AND operation on contents of BX i.e. 47E3 and E000 and store the result back to BX.'

So

0x47E3 = 0100 1111 1110 0011

0xE000 = 1000 0000 0000 0000

After ANDing these we get , 0000 0000 0000 0000 i.e. 0 So BX contains 0

ROL BX,4 // Rotate the contents of BX by 4 bits to left. After rotating we get 0000 0000 0000 0000 i.e. again 0. So BX contains 0.

MOV unpackedData,BX  // Mov data of BX i.e. 0 to variable unpackedData.

Finally, unpackedData contains 0.