Question

Given an 8-bit binary number/counter, why is it that 11111111 converts to -1 in decimal format?

Answer 1

Good question =)
Well notice that there are different methods for that.
It will be -1 in two's complement system.
Basically you'd ask yourself, what number added to 1 will result in 0?
Let's look in 4 bits, it will be easier.
Remember we're limited only to 4 bits so any carry is dropped.
$$
\begin{array}{lr}
&0001\\
+&1111 \\
\hline
&1 \boxed{0000}
\end{array}
$$
Likewise -2 will be the number that added to 2 will make zero so:
$$
\begin{array}{lr}
&0010\\
+&1110 \\
\hline
&1 \boxed{0000}
\end{array}
$$so 1110 is -2, and so on.

More generally you can say that to convert a number to its negative you have to first invert all the digits, because then we have:
$$
\begin{array}{lr}
&0101\\
+&1010 \\
\hline
&1111
\end{array}
$$And then by adding one to the inverted we'd get 0 instead of 1111 so:
$$
\begin{array}{lr}
&0101\\
+&1011 \\
\hline
&1 \boxed{0000}
\end{array}
$$So the negative of 0101 (5 in decimal) is 1011
Notice that this is depended on the size of your register. so the negative of 000101 is 111011

Answer 2

Thank you!

Answer 3

You're welcome