CHAPTE 12 . For each part of this problem, assume the \"before the requested aft
ID: 3701237 • Letter: C
Question
CHAPTE 12 . For each part of this problem, assume the "before the requested after" values Exercises 4.3 After Instruction EAX, EDX, CF, OF mul ebx al EAX: FF FF FFE4 EBX: 00 00 00 02 mul value EAX, EDX, CF, OF o) AX: FF FF FF E4 doubleword at value mul ax AX, DX, CF, OF AX: FF FF (d) AL: OF H: 4C mul bh mul bh imul ec:x imul ebx AX, CF, OF el AL: F BH: C4 AX, CF, OF EAX: 00 00 00 17 EAX, EDX, CF, OF EAX, EDX, CF, OF EAX, EDX, CF, OF EAX, EDX, CF, OF ECX: 00 00 00 B2 EAX: FF FF FFE4 EBX: 00 0004 C2 h) EAX: FF FF FF E4 doubleword at value:imul F FF F 34 ) EAX: FF FF FF F value imul eax 0 AL: OF BH: 4C (k) AL: FO BH: C4 imul bh AX, CF, OF () EAX: 0000 00 0A EBX: FF FF FF FC imul bh AX, CF, OF EDX: FF03 FF 01 (m) EAX: 00 00 00 OA imul ebx EAX, EBX, EDX, CF, OF EBX: FF FF FF FC EDX: FF 03 FF 01 mul ebx EAX, EBX, EDX CF0FExplanation / Answer
a. EAX = FF FF FF E4
EBX = 00 00 00 02
mul ebx ; performs eax * ebx and stores result in edx:eax = 00000001 FF FF FF C8 , The upper have contains significant digits, therefore CF = 1.
After instruction:
EDX = 00 00 00 01 , EAX = FF FF FFC8, CF = 1 and OF = 0
d. AL= 0F BH= 4C
instruction :mul bh ; values in AL , BH are multiplied and result is stored in AX
0F * 4C = 0474
After instruction:
AX = 0474, CF = 0, OF = 0
g. EAX = FF FF FF E4
EBX = 00 00 00 C2
imul ebx ; it is a signed integer multiply. Therefore OF flag will be 1 if the EDX is not a signed extension of EAX, otherwise it will be 0
EAX * EBX = C1 FFFFEAC8, stored in EDX : EAX
After the instruction:
CF = 1,OF = 1,EDX = 00 00 00 C1, EAX=FFFFEAC8
h. EAX = FF FF FF E4
value = FF FF FF 3A
instruction :
imul value; eax * value is stored in edx : eax = FF FF FF 1E 00 00 18 00
After instruction:
EDX = FF FF FF 1E , EAX = 00 00 18 00, CF = 1, OF = 0