Given the following code fragment: int * SW_Ptr; *SW_Ptr = 0x00000987; int Nib2 = (*SW_Ptr >>...
Given the following code fragment:
int * SW_Ptr;
*SW_Ptr = 0x00000987;
int Nib2 = (*SW_Ptr >> 8) & 0xF;
int Nib1 = (*SW_Ptr >> 4) & 0xF;
int Nib0 = (*SW_Ptr) & 0xF;
Which is the correct value of Nib1, Nib2, Nib3.
Microcontrollers using c prog
Homework Answers
Solution:
Given *SW_Ptr = 0
00000987
(Value is in hexadecimal format)
(a)
Finding the value of Nib2 :
=> As in the code given Nib2 = (*SW_Ptr >>8) &
0F;
Converting the hexadecimal values to binary :
=> *SW_Ptr = 000000987
= 0000 0000 0000 0000 0000 1001 1000 0111
=> 0F
= 1111
We can see >> is a right shift operator so performing right shift with *SW_Ptr by 8 bits.
=> 0000 0000 0000 0000 0000 1001 1000 0111 >> 8 => 0000 0000 0000 0000 0000 0000 0000 1001
Now we can see & is bitwise AND operator so performing bitwise AND
=> 0000 0000 0000 0000 0000 0000 0000 1001 & 1111 => 0000 0000 0000 0000 0000 0000 0000 1001
So finally the result is 0000 0000 0000 0000 0000 0000 0000 1001 in binary so converting the result into hexadecimal.
=> 0000 0000 0000 0000 0000 0000 0000 1001 = 000000009
So the value of Nib2 = 000000009
(b)
Finding the value of Nib1:
=> As in the code given Nib1 = (*SW_Ptr >>4) &
0F;
Converting the hexadecimal values to binary :
=> *SW_Ptr = 000000987
= 0000 0000 0000 0000 0000 1001 1000 0111
=> 0F
= 1111
We can see >> is a right shift operator so performing right shift with *SW_Ptr by 4 bits
=> 0000 0000 0000 0000 0000 1001 1000 0111 >> 4 => 0000 0000 0000 0000 0000 0000 1001 1000
Now we can see & is bitwise AND operator so performing bitwise AND
=> 0000 0000 0000 0000 0000 0000 1001 1000 & 1111 = 0000 0000 0000 0000 0000 0000 0000 1000
So finally the result is 0000 0000 0000 0000 0000 0000 0000 1000 in binary so converting the result into hexadecimal.
=> 0000 0000 0000 0000 0000 0000 0000 1000 = 000000008
So the value of Nib1 = 000000008
(c)
Finding the value of Nib3:
=> Given in the code its Nib0 and I think it's written by mistake so I am assuming it Nib3
Nib3 = (*SW_Ptr ) & 0F;
Converting the hexadecimal values to binary :
=> *SW_Ptr = 000000987
= 0000 0000 0000 0000 0000 1001 1000 0111
=> 0F
= 1111
We can see & is bitwise AND operator so performing bitwise AND
=> 0000 0000 0000 0000 0000 1001 1000 0111 & 1111 = 0000 0000 0000 0000 0000 0000 0000 0111
So finally the result is 0000 0000 0000 0000 0000 0000 0000 0111 in binary so converting the result into hexadecimal.
=> 0000 0000 0000 0000 0000 0000 0000 0111 = 000000007
So the value of Nib3 = 000000007
I have explained each and every part of the question with the help of the operator understanding and performing the operations as well.
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