Sign Up to our social questions and Answers Engine to ask questions, answer people’s questions, and connect with other people.
Login to our social questions & Answers Engine to ask questions answer people’s questions & connect with other people.
Lost your password? Please enter your email address. You will receive a link and will create a new password via email.
Please briefly explain why you feel this question should be reported.
Please briefly explain why you feel this answer should be reported.
Please briefly explain why you feel this user should be reported.
My EAX register contains the xxxxxx9D value and I have the following assembly code:
C0C8 14 --> ROR AL,14
To me, it means that the last 8 bits of the EAX‘s 32 bits value are rotated bitwise by 14 mod 8 = 6 positions
0x9D = b1001 1101
will be transformed into
b0111 0110 = 0x76
However, OllyDbg tells me that EAX = xxxxxxD9, which means EAX has been rotated bitwise by 4 bits!
Where am I wrong?
You are trying to rotate an 8-bit register by 20 positions. That’s a bit much, rotating by 8 produces the same value. Rotating by 9 is the same as rotating by 1. Etcetera. The processor will thus rotate by 20 mod 8 = 4 positions.