Made easy

Question

In it’s solution how are they obtaining x with the given operation? Should we not have 1 more register to keep x separate?

Given program below uses 6 temporary variables
\[
\begin{array}{l}
\quad u, v, w, x, y, z \\
u=1 \\
v=10 \\
w=20 \\
x=u+v \\
y=w+x \\
z=w+y \\
v=w+y \\
y=v+z \\
z=5+y \\
\text { return } x+5
\end{array}
\]
Activate Windows

Assume that all operation take their operands from registers, what is the minimum number of register needed to execute this program withoutspilling? ings to activate Winds

Solution :
(c)
\[
\begin{array}{lll}
R_{1} \leftarrow u & \cdots \cdots \cdots & (u=1) \\
R_{2} \leftarrow v & \cdots \cdots \cdots & (v=10) \\
R_{3} \leftarrow w & \cdots \cdots \cdots & (w=20) \\
R_{1} \leftarrow R_{1}+R_{2} & \cdots \cdots \cdots & (x=u+v) \\
R_{1} \leftarrow R_{3}+R_{1} & \cdots \cdots \cdots & (y=w+x) \\
R_{2} \leftarrow R_{3}+R_{1} & \cdots \cdots \cdots & (z=w+y) \\
R_{3} \leftarrow R_{3}+R_{1} & \cdots \cdots \cdots & (v=w+y) \\
R_{1} \leftarrow R_{2}+R_{3} & \cdots \cdots \cdots & (y=v+z) \\
R_{3} \leftarrow 5+R_{1} & \cdots \cdots \cdots & (x=5+y) \\
\text { return }\left(R_{2}+R_{3}\right) & \cdots \cdots \cdots & \text { return }(x+5)
\end{array}
\]

Hence 3 register needed only.