Jeremy Siek 6 gadi atpakaļ
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9cf3d6611a
1 mainītis faili ar 7 papildinājumiem un 8 dzēšanām
  1. 7 8
      book.tex

+ 7 - 8
book.tex

@@ -2850,19 +2850,18 @@ $\Rightarrow$
 
 
 The resulting program is almost an x86 program. The remaining step
 The resulting program is almost an x86 program. The remaining step
 is to apply the patch instructions pass. In this example, the trivial
 is to apply the patch instructions pass. In this example, the trivial
-move of \code{-16(\%rbp)} to itself is deleted and the addition of
-\code{-8(\%rbp)} to \key{-16(\%rbp)} is fixed by going through
-\code{rax}. The following shows the portion of the program that
-changed.
+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{rax} as follows.
 \begin{lstlisting}
 \begin{lstlisting}
-  (addq (int 4) (reg rbx))
-  (movq (deref rbp -8) (reg rax)
-  (addq (reg rax) (deref rbp -16))
+  (movq (deref rbp -16) (reg rax)
+  (addq (reg rax) (deref rbp -8))
 \end{lstlisting}
 \end{lstlisting}
+
 An overview of all of the passes involved in register allocation is
 An overview of all of the passes involved in register allocation is
 shown in Figure~\ref{fig:reg-alloc-passes}.
 shown in Figure~\ref{fig:reg-alloc-passes}.
 
 
-\begin{figure}[p]
+\begin{figure}[tbp]
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
 \begin{tikzpicture}[baseline=(current  bounding  box.center)]
 \node (R1) at (0,2)  {\large $R_1$};
 \node (R1) at (0,2)  {\large $R_1$};
 \node (R1-2) at (3,2)  {\large $R_1$};
 \node (R1-2) at (3,2)  {\large $R_1$};