The following code will jump to the label named Target mov eax, -50 cmp eax, -30
ID: 3816501 • Letter: T
Question
The following code will jump to the label named Target mov eax, -50 cmp eax, -30 jg Target a. True b. False After executing the following Instructions: movax, 7654h movdx, 234Bh shrd ax, dx, 4 a. ax = 0xB765h b. dx = 0x4234h c ax = 0x7652h d. none of the above Which statement is true after executing the following instructions: mov ecx, 12340000h push cx, mov ecx, 4455h pop cx a. exc = 12340000h b. ecx = 0 c. ecx = 4455h d. none of the above For question 4 and 5, assume that your clock has: 40 seconds instead of 60 seconds 25 minutes instead of 60 minutes 10 hrs instead of 12 hours How many bits do you need for the seconds? a. 5 bits b. 4 bites c 6 bits d. None of the above How many bits do you need for the seconds, minutes, and the hours? a. 15 bits b. 18 bits c. 17 bitsExplanation / Answer
1.
It is False.
-As JG is Jump if greater and -50<-30.
-So it will not jump to the Target and it is false.
2.
It is
a. ax=0XB765 h
-as shrd ax, dx ,4 =>places the second operand to the left of the first, shifts the whole(given or number) bit string right, and updates only the first operand.
-so it will give by shifting 4 bits right it will give
ax=0XB765 h.
3.
It will be c. ecx=4455h
- as it is last value will be referred here in the ecx.
4.
c. 6 bits
as 2^5=32 and 2^6=64
so we will require 6 bits as
=>40>2^5 and
=>40<2^64
- SO we will require 6 bits.
5.
a. 15 bits
-as seconds--6 bits
- minutes--25<2^5(32) ==>5 bits
- hours-- 10<2^4(16)==>4 bits
total bits=6+5+4=15 bits.
So 15 bits will be required for representation.
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