Two's Complement Wrap-Around

In this section, we give an example showing how temporary overflow in two's complement fixed-point causes no ill effects.

In 3-bit signed fixed-point arithmetic, the available numbers are as shown in Table 9.1.

Table 9.1: Three-bit two's-complement binary fixed-point numbers.

Decimal	Binary
-4	100
-3	101
-2	110
-1	111
0	000
1	001
2	010
3	011

Let's perform the sum $3+3-4 = 2$, which gives a temporary overflow ($3+3=6$, which wraps around to $-2$), but a final result ($2$) which is in the allowed range $[-4,3]$:

$$\begin{eqnarray*} 011 + 011 &=& 110 \qquad \mbox{$(3+3=-2\;\left(\mbox{mod}\;8\r... ...100 &=& 010 \qquad \mbox{$(-2-4=2\;\left(\mbox{mod}\;8\right))$} \end{eqnarray*}$$

Now let's do $1+3-2 = 2$ in three-bit two's complement:

$$\begin{eqnarray*} 001 + 011 &=& 100 \qquad \mbox{$(1+3=-4\;\left(\mbox{mod}\;8\r... ...110 &=& 010 \qquad \mbox{$(-4-2=2\;\left(\mbox{mod}\;8\right))$} \end{eqnarray*}$$

Thus, in both cases, the intermediate result overflows, but the final result is correct. Another way to state what happened is that a positive wrap-around in the first addition is canceled by a negative wrap-around in the second addition.