change example

book.tex

@@ -2478,33 +2478,32 @@ stack locations or registers.
 \begin{figure}
 \begin{minipage}{0.45\textwidth}
 Example $R_1$ program:
-% s0_22.rkt
+% s0_28.rkt
 \begin{lstlisting}
 (let ([v 1])
-(let ([w 46])
-(let ([x (+ v 7)])
-(let ([y (+ 4 x)])
-(let ([z (+ x w)])
-     (+ z (- y)))))))
+  (let ([w 42])
+    (let ([x (+ v 7)])
+      (let ([y x])
+	(let ([z (+ x w)])
+	  (+ z (- y)))))))
 \end{lstlisting}
 \end{minipage}
 \begin{minipage}{0.45\textwidth}
 After instruction selection:
 \begin{lstlisting}
-locals: v w x y z t.1
+locals: (v w x y z t)
 start:
     movq $1, v
-    movq $46, w
+    movq $42, w
     movq v, x
     addq $7, x
-    movq $4, y
-    addq x, y
+    movq x, y
     movq x, z
     addq w, z
-    movq y, t.1
-    negq t.1
+    movq y, t
+    negq t
     movq z, %rax
-    addq t.1, %rax
+    addq t, %rax
     jmp conclusion
 \end{lstlisting}
 \end{minipage}
@@ -2681,29 +2680,27 @@ sets shown between each instruction to make the figure easy to read.
 \begin{lstlisting}
                        |$\{\}$|    
     movq $1, v
-                       |$\{v\}$|  
-    movq $46, w
+                       |$\{v\}$|    
+    movq $42, w
                        |$\{v,w\}$|    
     movq v, x
                        |$\{w,x\}$|    
     addq $7, x
                        |$\{w,x\}$|    
-    movq $4, y
-                       |$\{w,x,y\}$|    
-    addq x, y
+    movq x, y
                        |$\{w,x,y\}$|    
     movq x, z
                        |$\{w,y,z\}$|    
     addq w, z
                        |$\{y,z\}$|    
-    movq y, t.1
-                       |$\{t.1,z\}$|    
-    negq t.1
-                       |$\{t.1,z\}$|    
+    movq y, t
+                       |$\{t,z\}$|    
+    negq t
+                       |$\{t,z\}$|    
     movq z, %rax
-                       |$\{t.1\}$|    
-    addq t.1, %rax
-                       |$\{\}$|    
+                       |$\{t\}$|    
+    addq t, %rax
+                       |$\{\}$|
     jmp conclusion
                        |$\{\}$|
 \end{lstlisting}
@@ -2788,17 +2785,16 @@ the following interference for each instruction.
 \begin{quote}
 \begin{tabular}{ll}
 \lstinline{movq $1, v}& no interference by rule 3,\\
-\lstinline{movq $46, w}& $w$ interferes with $v$ by rule 3,\\
+\lstinline{movq $42, w}& $w$ interferes with $v$ by rule 3,\\
 \lstinline{movq v, x}& $x$ interferes with $w$ by rule 3,\\
 \lstinline{addq $7, x}& $x$ interferes with $w$ by rule 1,\\
-\lstinline{movq $4, y}& $y$ interferes with $w$ and $x$ by rule 3,\\
-\lstinline{addq x, y}& $y$ interferes with $w$ and $x$ by rule 1,\\
+\lstinline{movq x, y}& $y$ interferes with $w$ but not $x$ by rule 3,\\
 \lstinline{movq x, z}& $z$ interferes with $w$ and $y$ by rule 3,\\
 \lstinline{addq w, z}& $z$ interferes with $y$ by rule 1, \\
-\lstinline{movq y, t.1}& $t.1$ interferes with $z$ by rule 3, \\
-\lstinline{negq t.1}& $t.1$ interferes with $z$ by rule 1, \\
+\lstinline{movq y, t}& $t$ interferes with $z$ by rule 3, \\
+\lstinline{negq t}& $t$ interferes with $z$ by rule 1, \\
 \lstinline{movq z, %rax}   & no interference (ignore rax), \\
-\lstinline{addq t.1, %rax} & no interference (ignore rax). \\
+\lstinline{addq t, %rax} & no interference (ignore rax). \\
   \lstinline{jmp conclusion}& no interference.
 \end{tabular}
 \end{quote}
@@ -2809,7 +2805,7 @@ Figure~\ref{fig:interfere}.
 \large
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (t1) at (0,2) {$t.1$};
+\node (t1) at (0,2) {$t$};
 \node (z) at (3,2)  {$z$};
 \node (x) at (6,2)  {$x$};
 \node (y) at (3,0)  {$y$};
@@ -2821,7 +2817,6 @@ Figure~\ref{fig:interfere}.
 \draw (z) to (y);
 \draw (z) to (w);
 \draw (x) to (w);
-\draw (y) to (x);
 \draw (y) to (w);
 \draw (v) to (w);
 \end{tikzpicture}
@@ -2983,7 +2978,7 @@ colored and they are unsaturated, so we annotate each of them with a
 dash for their color and an empty set for the saturation.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (t1) at (0,2) {$t.1:-,\{\}$};
+\node (t1) at (0,2) {$t:-,\{\}$};
 \node (z) at (3,2)  {$z:-,\{\}$};
 \node (x) at (6,2)  {$x:-,\{\}$};
 \node (y) at (3,0)  {$y:-,\{\}$};
@@ -2994,18 +2989,17 @@ dash for their color and an empty set for the saturation.
 \draw (z) to (y);
 \draw (z) to (w);
 \draw (x) to (w);
-\draw (y) to (x);
 \draw (y) to (w);
 \draw (v) to (w);
 \end{tikzpicture}
 \]
 The algorithm says to select a maximally saturated vertex and color it
 $0$. In this case we have a 6-way tie, so we arbitrarily pick
-$t.1$. We then mark color $0$ as no longer available for $z$ because
-it interferes with $t.1$.
+$t$. We then mark color $0$ as no longer available for $z$ because
+it interferes with $t$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (t1) at (0,2) {$t.1:0,\{\}$};
+\node (t1) at (0,2) {$t:0,\{\}$};
 \node (z) at (3,2)  {$z:-,\{0\}$};
 \node (x) at (6,2)  {$x:-,\{\}$};
 \node (y) at (3,0)  {$y:-,\{\}$};
@@ -3016,7 +3010,6 @@ it interferes with $t.1$.
 \draw (z) to (y);
 \draw (z) to (w);
 \draw (x) to (w);
-\draw (y) to (x);
 \draw (y) to (w);
 \draw (v) to (w);
 \end{tikzpicture}
@@ -3026,7 +3019,7 @@ vertex, which is $z$, and color it with the first available number,
 which is $1$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (t1) at (0,2) {$t.1:0,\{\}$};
+\node (t1) at (0,2) {$t:0,\{1\}$};
 \node (z) at (3,2)  {$z:1,\{0\}$};
 \node (x) at (6,2)  {$x:-,\{\}$};
 \node (y) at (3,0)  {$y:-,\{1\}$};
@@ -3037,87 +3030,82 @@ which is $1$.
 \draw (z) to (y);
 \draw (z) to (w);
 \draw (x) to (w);
-\draw (y) to (x);
 \draw (y) to (w);
 \draw (v) to (w);
 \end{tikzpicture}
 \]
-The most saturated vertices are now $w$ and $y$. We color $y$ with the
+The most saturated vertices are now $w$ and $y$. We color $w$ with the
 first available color, which is $0$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (t1) at (0,2) {$t.1:0,\{\}$};
+\node (t1) at (0,2) {$t:0,\{1\}$};
 \node (z) at (3,2)  {$z:1,\{0\}$};
 \node (x) at (6,2)  {$x:-,\{0\}$};
-\node (y) at (3,0)  {$y:0,\{1\}$};
-\node (w) at (6,0)  {$w:-,\{0,1\}$};
-\node (v) at (9,0)  {$v:-,\{\}$};
+\node (y) at (3,0)  {$y:-,\{0,1\}$};
+\node (w) at (6,0)  {$w:0,\{1\}$};
+\node (v) at (9,0)  {$v:-,\{0\}$};
 
 \draw (t1) to (z);
 \draw (z) to (y);
 \draw (z) to (w);
 \draw (x) to (w);
-\draw (y) to (x);
 \draw (y) to (w);
 \draw (v) to (w);
 \end{tikzpicture}
 \]
-Vertex $w$ is now the most highly saturated, so we color $w$ with $2$.
-We cannot choose $0$ or $1$ because those numbers are in $w$'s
-saturation set. Indeed, $w$ interferes with $y$ and $z$, whose colors
+Vertex $y$ is now the most highly saturated, so we color $y$ with $2$.
+We cannot choose $0$ or $1$ because those numbers are in $y$'s
+saturation set. Indeed, $y$ interferes with $w$ and $z$, whose colors
 are $0$ and $1$ respectively.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (t1) at (0,2) {$t.1:0,\{\}$};
+\node (t1) at (0,2) {$t:0,\{1\}$};
 \node (z) at (3,2)  {$z:1,\{0,2\}$};
-\node (x) at (6,2)  {$x:-,\{0,2\}$};
-\node (y) at (3,0)  {$y:0,\{1,2\}$};
-\node (w) at (6,0)  {$w:2,\{0,1\}$};
-\node (v) at (9,0)  {$v:-,\{2\}$};
+\node (x) at (6,2)  {$x:-,\{0\}$};
+\node (y) at (3,0)  {$y:2,\{0,1\}$};
+\node (w) at (6,0)  {$w:0,\{1,2\}$};
+\node (v) at (9,0)  {$v:-,\{0\}$};
 
 \draw (t1) to (z);
 \draw (z) to (y);
 \draw (z) to (w);
 \draw (x) to (w);
-\draw (y) to (x);
 \draw (y) to (w);
 \draw (v) to (w);
 \end{tikzpicture}
 \]
-Now $x$ has the highest saturation, so we color it $1$.
+Now $x$ and $v$ are the most saturated, so we color $v$ it $1$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (t1) at (0,2) {$t.1:0,\{\}$};
+\node (t1) at (0,2) {$t:0,\{1\}$};
 \node (z) at (3,2)  {$z:1,\{0,2\}$};
-\node (x) at (6,2)  {$x:1,\{0,2\}$};
-\node (y) at (3,0)  {$y:0,\{1,2\}$};
-\node (w) at (6,0)  {$w:2,\{0,1\}$};
-\node (v) at (9,0)  {$v:-,\{2\}$};
+\node (x) at (6,2)  {$x:-,\{0\}$};
+\node (y) at (3,0)  {$y:2,\{0,1\}$};
+\node (w) at (6,0)  {$w:0,\{1,2\}$};
+\node (v) at (9,0)  {$v:1,\{0\}$};
 
 \draw (t1) to (z);
 \draw (z) to (y);
 \draw (z) to (w);
 \draw (x) to (w);
-\draw (y) to (x);
 \draw (y) to (w);
 \draw (v) to (w);
 \end{tikzpicture}
 \]
-In the last step of the algorithm, we color $v$ with $0$.
+In the last step of the algorithm, we color $x$ with $1$.
 \[
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
-\node (t1) at (0,2) {$t.1:0,\{\}$};
+\node (t1) at (0,2) {$t:0,\{1,\}$};
 \node (z) at (3,2)  {$z:1,\{0,2\}$};
-\node (x) at (6,2)  {$x:1,\{0,2\}$};
-\node (y) at (3,0)  {$y:0,\{1,2\}$};
-\node (w) at (6,0)  {$w:2,\{0,1\}$};
-\node (v) at (9,0)  {$v:0,\{2\}$};
+\node (x) at (6,2)  {$x:1,\{0\}$};
+\node (y) at (3,0)  {$y:2,\{0,1\}$};
+\node (w) at (6,0)  {$w:0,\{1,2\}$};
+\node (v) at (9,0)  {$v:1,\{0\}$};
 
 \draw (t1) to (z);
 \draw (z) to (y);
 \draw (z) to (w);
 \draw (x) to (w);
-\draw (y) to (x);
 \draw (y) to (w);
 \draw (v) to (w);
 \end{tikzpicture}
@@ -3137,11 +3125,11 @@ variables, we arrive at the following assignment of variables to
 registers and stack locations.
 \begin{gather*}
   \{ v \mapsto \key{\%rcx}, \,
-  w \mapsto \key{-16(\%rbp)},  \,
-  x \mapsto \key{-8(\%rbp)}, \\
-  y \mapsto \key{\%rcx},  \,
-  z\mapsto \key{-8(\%rbp)}, 
-  t.1\mapsto \key{\%rcx} \}
+     w \mapsto \key{\%rcx},  \,
+     x \mapsto \key{-8(\%rbp)}, \\
+     y \mapsto \key{-16(\%rbp)},  \,
+     z\mapsto \key{-8(\%rbp)}, 
+     t\mapsto \key{\%rcx} \}
 \end{gather*}
 Applying this assignment to our running example, on the left, yields
 the program on the right.
@@ -3150,17 +3138,16 @@ the program on the right.
   \begin{minipage}{0.3\textwidth}
 \begin{lstlisting}
 movq $1, v
-movq $46, w
+movq $42, w
 movq v, x
 addq $7, x
-movq $4, y
-addq x, y
+movq x, y
 movq x, z
 addq w, z
-movq y, t.1
-negq t.1
+movq y, t
+negq t
 movq z, %rax
-addq t.1, %rax
+addq t, %rax
 jmp conclusion
 \end{lstlisting}
 \end{minipage}
@@ -3168,14 +3155,13 @@ $\Rightarrow\qquad$
 \begin{minipage}{0.45\textwidth}
 \begin{lstlisting}
 movq $1, %rcx
-movq $46, -16(%rbp)
+movq $42, %rcx
 movq %rcx, -8(%rbp)
 addq $7, -8(%rbp)
-movq $4, %rcx
-addq -8(%rbp), %rcx
+movq -8(%rbp), -16(%rbp)
 movq -8(%rbp), -8(%rbp)
-addq -16(%rbp), -8(%rbp)
-movq %rcx, %rcx
+addq %rcx, -8(%rbp)
+movq -16(%rbp), %rcx
 negq %rcx
 movq -8(%rbp), %rax
 addq %rcx, %rax
@@ -3187,11 +3173,11 @@ jmp conclusion
 The resulting program is almost an x86 program. The remaining step is
 the patch instructions pass. In this example, the trivial move of
 \code{-8(\%rbp)} to itself is deleted and the addition of
-\code{-16(\%rbp)} to \key{-8(\%rbp)} is fixed by going through
+\code{-8(\%rbp)} to \key{-16(\%rbp)} is fixed by going through
 \code{rax} as follows.
 \begin{lstlisting}
-movq -16(%rbp), %rax
-addq %rax, -8(%rbp)
+movq -8(%rbp), %rax
+addq %rax, -16(%rbp)
 \end{lstlisting}
 
 An overview of all of the passes involved in register allocation is
@@ -3303,10 +3289,10 @@ movq x, y
 addq $4, y
 movq x, z
 addq w, z
-movq y, t.1
-negq t.1
+movq y, t
+negq t
 movq z, %rax
-addq t.1, %rax
+addq t, %rax
 jmp conclusion
 \end{lstlisting}
 \end{minipage}