It is simple hexa-decimal addition.
8+F = 17
Explanation:
8+15 = 23 (which is 17 in hexadecimal)
Further explanation:
8+15 = 23
23-16 = 7 (here 16 is hexadecimal base) 1 is carried
so it becomes 17
I hope now you can understand why 8+F = 7 written in the answer above.
Sorry for the little mistake, I mistakenly wrote 01FF as 011F. Please check now.
This calculation can be expressed as: Physical address = SS (hex) x 16 + SP (hex).
SS (7698)h x 16 results as 76980
same in the second question.
Answer 1.
The value of the stack segment register (SS) = 7698h
The value of the stack pointer (SP) which is Offset = 01FFh
Thus, Physical address of the top of the stack is:
7
6
9
8
0
+
0
1
F
F
7
6
B
7
F
This calculation can be expressed as: Physical address = SS (hex) x 16 + SP (hex). Answer 2.
The value of the code segment register (CS) = 5526h
The value of the Instruction Pointer, holding address of the instruction = 8874h
Physical address of the instruction
Physical Address = CS (Hex) x 16 + IP
aksit
i am also finding the ans of this qsn how you wrote 8+F=7 plz tell
admin
It is simple hexa-decimal addition.
8+F = 17
Explanation:
8+15 = 23 (which is 17 in hexadecimal)
Further explanation:
8+15 = 23
23-16 = 7 (here 16 is hexadecimal base)
1 is carried
so it becomes 17
I hope now you can understand why 8+F = 7 written in the answer above.
Vandana
SS=76980
SP=01FF
01FF kaise 011F ho gaya
9+1=10=A hota hai but apne B kyu likha
Similarly 8+1=9 =9 he hona chahey
Plz admin explain ?
admin
Sorry for the little mistake, I mistakenly wrote 01FF as 011F. Please check now.
This calculation can be expressed as: Physical address = SS (hex) x 16 + SP (hex).
SS (7698)h x 16 results as 76980
same in the second question.
Vandana
Thanks
admin
Answer 1.
The value of the stack segment register (SS) = 7698h
The value of the stack pointer (SP) which is Offset = 01FFh
Thus, Physical address of the top of the stack is:
This calculation can be expressed as: Physical address = SS (hex) x 16 + SP (hex).
Answer 2.
The value of the code segment register (CS) = 5526h
The value of the Instruction Pointer, holding address of the instruction = 8874h
Physical address of the instruction
Physical Address = CS (Hex) x 16 + IP