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Cleanup signed, unsigned, adding, & overflow

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John Winans 2020-08-18 16:04:52 -05:00
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book/binary/chapter.tex
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@ -564,46 +564,42 @@ Binary: 1 1 1 1 1 1 1 1
\ldots because: $-128+64+32+16+8+4+2+1=-1$.
This format has the virtue of allowing the same addition logic discussed above to be
used to calculate the sums of signed numbers as unsigned numbers.
Calculating $4+5 = 9$
Calculating the signed addition: $4+5 = 9$
\begin{verbatim}
1 <== carries
000100 <== 4
+000101 <== 5
------
001001 <== 9
1 <== carries
000100 <== 4 = 0 + 0 + 0 + 4 + 0 + 0
+000101 <== 5 = 0 + 0 + 0 + 4 + 0 + 1
-------
001001 <== 9 = 0 + 0 + 8 + 0 + 0 + 1
\end{verbatim}
Calculating $-4+ -5 = -9$
Calculating the signed addition: $-4+ -5 = -9$
\begin{verbatim}
1 11 <== carries
111100 <== -4
+111011 <== -5
1 11 <== carries
111100 <== -4 = -32 + 16 + 8 + 4 + 0 + 0
+111011 <== -5 = -32 + 16 + 8 + 0 + 2 + 1
---------
1 110111 <== -9 (with a truncation)
-32 16 8 4 2 1
1 1 0 1 1 1
-32 + 16 + 4 + 2 + 1 = -9
1 110111 <== -9 (with a truncation) = -32 + 16 + 4 + 2 + 1 = -9
\end{verbatim}
This format has the virtue of allowing the same addition logic
discussed above to be used to calculate $-1+1=0$.
Calculating the signed addition: $-1+1=0$
\begin{verbatim}
-128 64 32 16 8 4 2 1 <== place value
1 1 1 1 1 1 1 1 0 <== carries
1 1 1 1 1 1 1 1 <== carries
1 1 1 1 1 1 1 1 <== addend (-1)
+ 0 0 0 0 0 0 0 1 <== addend (1)
----------------------
1 0 0 0 0 0 0 0 0 <== sum (0 with a truncation)
\end{verbatim}
In order for this to work, the carry out of the sum of the MSBs is
ignored.
{\em In order for this to work, the carry out of the sum of the MSBs {\bfseries must} be discarded.}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Converting between Positive and Negative}
@ -612,8 +608,9 @@ Changing the sign on two's complement numbers can be described as
inverting all of the bits (which is also known as the {\em one's complement})
and then add one.
For example, inverting the number four:
For example, negating the number four:
\begin{minipage}{\textwidth}
\begin{verbatim}
-128 64 32 16 8 4 2 1
0 0 0 0 0 1 0 0 <== 4
@ -624,23 +621,24 @@ For example, inverting the number four:
----------------------
1 1 1 1 1 1 0 0 <== -4
\end{verbatim}
\end{minipage}
This can be verified by adding 5 to the result and observe that
the sum is 1:
\begin{verbatim}
-128 64 32 16 8 4 2 1
1 1 1 1 1 <== carries
1 1 1 1 1 1 <== carries
1 1 1 1 1 1 0 0 <== -4
+ 0 0 0 0 0 1 0 1 <== 5
----------------------
1 0 0 0 0 0 0 0 1
1 0 0 0 0 0 0 0 1 <== 1 (with a truncation)
\end{verbatim}
Note that the changing of the sign using this method is symmetric
in that it is identical when converting from negative to positive
and when converting from positive to negative: flip the bits and
add 1.
and when converting from positive to negative: {\em flip the bits and
add 1.}
For example, changing the value -4 to 4 to illustrate the
reverse of the conversion above:
@ -661,45 +659,56 @@ reverse of the conversion above:
\subsection{Subtraction of Binary Numbers}
Subtraction of binary numbers is performed by first negating
the subtrahend and then adding the two numbers. Due to the
nature of two's complement numbers this will work for both
signed and unsigned numbers.
Subtraction%
\enote{This section needs more examples of subtracting
signed an unsigned numbers and a discussion on how
signedness is not relevant until the results are interpreted.
For example adding $-4+ -8=-12$ using two 8-bit numbers
is the same as adding $252+248=500$ and truncating the result
to 244.}
of binary numbers is performed by first negating
the subtrahend and then adding the two numbers. Due to the
nature of two's complement numbers this method will work for both
signed and unsigned numbers!
To calculate $-4-8 = -12$
Observation: Since we always have a carry-in of zero into the LSB when
adding, we can take advantage of that fact by (ab)using that carry input
to perform that adding the extra 1 to the subtrahend as part of
changing its sign in the examples below.
\enote{This example is unclear. That the adding of one to the subtrahend
has to be done as part of the same operation as the sum of the two values.
otherwise adding 1000 to 0001 will {\em not} result in a proper overflow
staus as discussed below.}
An example showing the subtraction of two {\em signed} binary numbers: $-4-8 = -12$
\begin{verbatim}
-128 64 32 16 8 4 2 1
1 1 1 1 1 1 0 0 <== -4 (minuend)
- 0 0 0 0 1 0 0 0 <== 8 (subtrahend)
------------------------
1 1 1 <== carries
1 1 1 1 0 1 1 1 <== one's complement of -8
+ 0 0 0 0 0 0 0 1 <== plus 1
----------------------
1 1 1 1 1 0 0 0 <== -8
1 1 1 1 <== carries
1 1 1 1 1 1 1 1 1 <== carries
1 1 1 1 1 1 0 0 <== -4
+ 1 1 1 1 1 0 0 0 <== -8
----------------------
1 1 1 1 1 0 1 0 0 < == -12
+ 1 1 1 1 0 1 1 1 <== one's complement of -8
------------------------
1 1 1 1 1 0 1 0 0 <== -12
\end{verbatim}
%An example showing the subtraction of two {\em unsigned} binary numbers: $252+248=500$
%
%\begin{verbatim}
% 128 64 32 16 8 4 2 1
%
% 1 1 1 1 1 <== carries
% 1 1 1 1 1 1 0 0 <== 252
% + 1 1 1 1 1 0 0 0 <== 248
% ----------------------
% 1 1 1 1 1 0 1 0 0 < == 500 (if we do NOT truncate the MSB)
%\end{verbatim}
%
%An example showing the subtraction of two {\em unsigned} binary numbers: $252+248=500$
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsection{Truncation}
@ -707,10 +716,64 @@ staus as discussed below.}
\index{overflow}
\index{carry}
So far we have been ignoring (truncating) the carries that can come from
the MSBs when adding and subtracting. We have also been ignoring the
potential impact of a carry causing a signed number to change its sign in
a destructive way.
Discarding the carry bit that can be generated from the MSB is called {\em truncation}.
So far we have been ignoring the carries that can come from the MSBs when adding and subtracting.
We have also been ignoring the potential impact of a carry causing a signed number to change
its sign in an unexpected way.
In the examples above, truncating the results either had 1) no impact on the calculated sums
or 2) was absolutely necessary to correct the sum in cases such as: $-4 + 5$.
For example, note what happens when we try to subtract 1 from the most
negative value that we can represent in a 4 bit two's complement number:
\begin{verbatim}
-8 4 2 1
1 0 0 0 <== -8 (minuend)
- 0 0 0 1 <== 1 (subtrahend)
------------
1 1 <== carries
1 0 0 0 <== -8
+ 1 1 1 0 <== one's complement of 1
----------
1 0 1 1 1 <== this SHOULD be -9 but with truncation it is 7
\end{verbatim}
The problem with this example is that we can not represent $-9_{10}$ using a 4-bit
two's complement number.
Granted, if we would have used 5 bit numbers, then the ``answer'' would have fit OK.
But the same problem would return when trying to calculate $-16 - 1$.
So simply ``making more room'' does not solve this problem.
%However, as calculating $-1+1=0$ has demonmstrated above, it was necessary for that
%case to discard the carry out of the MSB to get the correct result.
%In the case of calculating $-1+1=0$ the addends and result all fit into same-sized
%(8-bit) values. When calculating $-8-1=-9$ the addends each can fit into 4-bit
%two's complement numbers but the result would require a 5-bit number.
This is not just a problem when subtracting, nor is it just a problem with
signed numbers.
The same situation can happen {\em unsigned} numbers.
For example:
\begin{verbatim}
8 4 2 1
1 1 1 0 0 <== carries
1 1 1 0 <== 14 (addend)
+ 0 0 1 1 <== 3 (addend)
------------
1 0 0 0 1 <== this SHOULD be 17 but with truncation it is 1
\end{verbatim}
How to handle such a truncation depends on whether the {\em original} values
being added are signed or unsigned.
The RV ISA refers to the discarding the carry out of the MSB after an
add (or subtract) of two {\em unsigned} numbers as an {\em unsigned overflow}%
@ -728,38 +791,66 @@ When adding {\em unsigned} numbers, an overflow only occurs when there
is a carry out of the MSB resulting in a sum that is truncated to fit
into the number of bits allocated to contain the result.
When subtracting {\em unsigned} numbers, an overflow only occurs when the
difference is negative (because there are no negative unsigned numbers.)
\autoref{sum:240+17} illustrates an unsigned overflow.
\autoref{sum:240+17} illustrates an unsigned overflow during addition:
\begin{figure}[H]
\centering
\begin{BVerbatim}
1 1 1 1 0 0 0 0 <== carries
1 1 1 1 0 0 0 0 <== 240
+ 0 0 0 1 0 0 0 1 <== 17
1 1 1 1 0 0 0 0 0 <== carries
1 1 1 1 0 0 0 0 <== 240
+ 0 0 0 1 0 0 0 1 <== 17
---------------------
0 0 0 0 0 0 0 1 <== sum = 1
1 0 0 0 0 0 0 0 1 <== sum = 1
\end{BVerbatim}
%{\captionof{figure}{$240+16=0$ (overflow)}\label{sum:240+17}}
\caption{$240+17=1$ (overflow)}
\label{sum:240+17}
\end{figure}
\enote{Need to add an example of an unsigned overflow after a subtraction.
When subtracting by adding the two's complement of the subtrahend, the unsigned
overflow status is represented by a 0 carry out of the most significant bit!}
Some times an overflow like this is referred to as a {\em wrap around}
because of the way that successive additions will result in a value that
increases until it {\em wraps} back {\em around} to zero and then
returns to increasing in value until it, again, wraps around again.
\begin{tcolorbox}
An {\em unsigned overflow} occurs when ever there is a carry
When adding, {\em unsigned overflow} occurs when ever there is a carry
{\em out of} the most significant bit.
\end{tcolorbox}
When subtracting {\em unsigned} numbers, an overflow only occurs when the
subtrahend is greater than the minuend (because in those cases the
different would have to be negative and there are no negative values
that can be represented with an unsigned binary number.)
\autoref{sum:3-4} illustrates an unsigned overflow during subtraction:
\begin{figure}[H]
\centering
\begin{BVerbatim}
0 0 0 0 0 1 1 <== 3 (minuend)
- 0 0 0 0 1 0 0 <== 4 (subtrahend)
---------------
0 0 0 0 0 1 1 1 <== carries
0 0 0 0 0 1 1 <== 3
+ 1 1 1 1 0 1 1 <== one's complement of 4
---------------
1 1 1 1 1 1 1 <== 255 (overflow)
\end{BVerbatim}
\caption{$3-4=255$ (overflow)}
\label{sum:3-4}
\end{figure}
\begin{tcolorbox}
When subtracting, {\em unsigned overflow} occurs when ever there is {\em not} a carry
{\em out of} the most significant bit (IFF the carry-in on the LSB is used to add the
extra 1 to the subtrahend when changing its sign.)
\end{tcolorbox}
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\subsubsection{Signed Overflow}
\index{overflow!signed}
@ -781,7 +872,7 @@ while looking more closely at the carry values.
\begin{figure}[H]
\centering
\begin{BVerbatim}
0 1 0 0 0 0 0 0 <== carries
0 1 0 0 0 0 0 0 0 <== carries
0 1 0 0 0 0 0 0 <== 64
+ 0 1 0 0 0 0 0 0 <== 64
---------------------
@ -811,7 +902,7 @@ We say that this result has been {\em truncated}.
\begin{figure}[H]
\centering
\begin{BVerbatim}
1 0 0 0 0 0 0 0 <== carries
1 0 0 0 0 0 0 0 0 <== carries
1 0 0 0 0 0 0 0 <== -128
+ 1 0 0 0 0 0 0 0 <== -128
---------------------
@ -830,7 +921,7 @@ do not have the same sign.
\begin{figure}[H]
\centering
\begin{BVerbatim}
1 1 1 1 1 1 1 1 <== carries
1 1 1 1 1 1 1 1 0 <== carries
1 1 1 1 1 1 0 1 <== -3
+ 1 1 1 1 1 0 1 1 <== -5
---------------------
@ -843,7 +934,7 @@ do not have the same sign.
\begin{figure}[H]
\centering
\begin{BVerbatim}
1 1 1 1 1 1 1 0 <== carries
1 1 1 1 1 1 1 0 0 <== carries
1 1 1 1 1 1 1 0 <== -2
+ 0 0 0 0 1 0 1 0 <== 10
---------------------
@ -862,7 +953,7 @@ the most negative value as shown in \autoref{sum:127+1}.
\begin{figure}[H]
\centering
\begin{BVerbatim}
0 1 1 1 1 1 1 1 <== carries
0 1 1 1 1 1 1 1 0 <== carries
0 1 1 1 1 1 1 1 <== 127
+ 0 0 0 0 1 0 0 1 <== 1
---------------------