Reputation: 93
We have three 16-bit words:
0110011001100000
0101010101010101
1000111100001100
sum of the first two
0110011001100000
0101010101010101
-----------------
1011101110110101
adding the sum to the third
1000111100001100
1011101110110101
-------------------
10100101011000001
but the book says for that part that it's:
0100101011000010
It says that the last addition had overflow which was wrapped around but i don't understand.
After that it obtains the 1st complement:
1011010100111101
which becomes the checksum.
I don't understand the adding the sum to the third part. Can anyone explain?
Upvotes: 3
Views: 7369
Reputation: 12401
Here's adding the sum to the third value.
Note the indentation. The overflow bit is the leftmost bit.
1000111100001100
1011101110110101
-----------------
10100101011000001
^
Add the overflow to the truncated result:
0100101011000001
0000000000000001
-----------------
0100101011000010
Which is the desired result for that step.
Upvotes: 4