Essential-of-Compilation/book.tex

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\newcommand{\itm}[1]{\ensuremath{\mathit{#1}}}
\newcommand{\Stmt}{\itm{stmt}}
\newcommand{\Exp}{\itm{exp}}
\newcommand{\Ins}{\itm{instr}}
\newcommand{\Prog}{\itm{prog}}
\newcommand{\Arg}{\itm{arg}}
\newcommand{\Int}{\itm{int}}
\newcommand{\Var}{\itm{var}}
\newcommand{\Op}{\itm{op}}
\newcommand{\key}[1]{\texttt{#1}}
\newcommand{\READ}{(\key{read})}
\newcommand{\UNIOP}[2]{(\key{#1}\,#2)}
\newcommand{\BINOP}[3]{(\key{#1}\,#2\,#3)}
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\title{\Huge \textbf{Essentials of Compilation} \\ 
  \huge An Incremental Approach}

\author{\textsc{Jeremy G. Siek}
   \thanks{\url{http://homes.soic.indiana.edu/jsiek/}}
   }

\begin{document}

\frontmatter
\maketitle

\begin{dedication}
This book is dedicated to the programming languages group at Indiana University.
\end{dedication}

\tableofcontents
%\listoffigures
%\listoftables

\mainmatter

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter*{Preface}

\cite{Sarkar:2004fk}
\cite{Keep:2012aa}
\cite{Ghuloum:2006bh}

%\section*{Structure of book}
% You might want to add short description about each chapter in this book.

%\section*{About the companion website}
%The website\footnote{\url{https://github.com/amberj/latex-book-template}} for %this file contains:
%\begin{itemize}
%  \item A link to (freely downlodable) latest version of this document.
%  \item Link to download LaTeX source for this document.
%  \item Miscellaneous material (e.g. suggested readings etc).
%\end{itemize}

\section*{Acknowledgements}

Need to give thanks to 
\begin{itemize}
\item Kent Dybvig
\item Daniel P. Freidman
\item Oscar Waddell
\item Abdulaziz Ghuloum
\item Dipanwita Sarkar
\end{itemize}

%\mbox{}\\
%\noindent Amber Jain \\
%\noindent \url{http://amberj.devio.us/}

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
\chapter{Integers and Variables}

%\begin{chapquote}{Author's name, \textit{Source of this quote}}
%``This is a quote and I don't know who said this.''
%\end{chapquote}

The $S_0$ language includes integers, operations on integers,
(arithmetic and input), and variable definitions.  The syntax of the
$S_0$ language is defined by the grammar in
Figure~\ref{fig:s0-syntax}. This language is rich enough to exhibit
several compilation techniques but simple enough so that we can
implement a compiler for it in two weeks of hard work.  To give the
reader a feeling for the scale of this first compiler, the instructor
solution for the $S_0$ compiler consists of 6 recursive functions and
a few small helper functions that together span 256 lines of code.

\begin{figure}[htbp]
\centering
\fbox{
\begin{minipage}{0.85\textwidth}
\[
\begin{array}{lcl}
  \Op  &::=& \key{+} \mid \key{-} \mid \key{*} \mid \key{read} \\
  \Exp &::=& \Int \mid (\Op \; \Exp^{+}) \mid \Var \mid \LET{\Var}{\Exp}{\Exp}
\end{array}
\]
\end{minipage}
}
\caption{The syntax of the $S_0$ language. The abbreviation \Op{} is
  short for operator, \Exp{} is short for expression, \Int{} for integer,
  and \Var{} for variable.}
\label{fig:s0-syntax}
\end{figure}

The result of evaluating an expression is a value.  For $S_0$, values
are integers. To make it straightforward to map these integers onto
x86-64 assembly~\citep{Matz:2013aa}, we restrict the integers to just
those representable with 64-bits, the range $-2^{63}$ to $2^{63}$.

We will walk through some examples of $S_0$ programs, commenting on
aspects of the language that will be relevant to compiling it.  We
start with one of the simplest $S_0$ programs; it adds two integers.
\[
\BINOP{+}{10}{32}
\]
The result is $42$, as you might expected. 
%
The next example demonstrates that expressions may be nested within
eachother, in this case nesting several additions and negations.
\[
\BINOP{+}{10}{ \UNIOP{-}{ \BINOP{+}{12}{20} } }
\]
What is the result of the above program?

The \key{let} construct stores a value in a variable which can then be
used within the body of the \key{let}. So the following program stores
$32$ in $x$ and then computes $\BINOP{+}{10}{x}$, producing $42$.
\[
\LET{x}{ \BINOP{+}{12}{20} }{ \BINOP{+}{10}{x} } 
\]
When there are multiple \key{let}'s for the same variable, the closest
enclosing \key{let} is used. Consider the following program with two
\key{let}'s that define variables named $x$.
\[
\LET{x}{32}{ \BINOP{+}{ \LET{x}{10}{x} }{ x } }
\]
For the purposes of showing which variable uses correspond to which
definitions, the following shows the $x$'s annotated with subscripts
to distinguish them.
\[
\LET{x_1}{32}{ \BINOP{+}{ \LET{x_2}{10}{x_2} }{ x_1 } }
\]

The \key{read} operation prompts the user of the program for an
integer. Given an input of $10$, the following program produces $42$.
\[
\BINOP{+}{(\key{read})}{32}
\]
We include the \key{read} operation in $S_0$ to demonstrate that order
of evaluation can make a different. Given the input $52$ then $10$,
the following produces $42$ (and not $-42$).
\[
\LET{x}{\READ}{ \LET{y}{\READ}{ \BINOP{-}{x}{y} } }
\]
The initializing expression is always evaluated before the body of the
\key{let}, so in the above, the \key{read} for $x$ is performed before
the \key{read} for $y$.
%
The behavior of the following program is somewhat subtle because
Scheme does not specify an evaluation order for arguments of an
operator such as $-$.
\[
\BINOP{-}{\READ}{\READ}
\]
Given the input $42$ then $10$, the above program can result in either
$42$ or $-42$, depending on the whims of the Scheme implementation.

The goal for this chapter is to implement a compiler that translates
any program $p \in S_0$ into a x86-64 assembly program $p'$ such that
the assembly program exhibits the same behavior on Intel hardward as
the $S_0$ program running in a Scheme implementation.
\[
\xymatrix{
p \in S_0  \ar[rr]^{\text{compile}} \ar[drr]_{\text{run in Scheme}\quad}   &&  p' \in \text{x86-64} \ar[d]^{\quad\text{run on an x86 machine}}\\
& & n \in \mathbb{Z}   
}
\]
In the next section we introduce enough of the x86-64 assembly
language to compile $S_0$.

\section{x86-64 Assembly}

An x86-64 program is a sequence of instructions. The instructions
manipulate 16 variables called \emph{registers} and can also load and
store values into \emph{memory}. Memory is a mapping of 64-bit
addresses to 64-bit values. The syntax $n(r)$ is used to read the
address $a$ stored in register $r$ and then offset it by $n$ bytes (8
bits), producing the address $a + n$. The arithmetic instructions,
such as $\key{addq}\,s\,d$, read from the source $s$ and destination
argument $d$, apply the arithmetic operation, then stores the result
in the destination $d$. In this case, computing $d \gets d + s$.  The
move instruction, $\key{movq}\,s\,d$ reads from $s$ and stores the
result in $d$. The $\key{callq}\,\mathit{label}$ instruction executes
the procedure specified by the label, which we shall use to implement
\key{read}. Figure~\ref{fig:x86-a} defines the syntax for this subset
of the x86-64 assembly language.

\begin{figure}[tbp]
\fbox{
\begin{minipage}{0.96\textwidth}
\[
\begin{array}{lcl}
\itm{register} &::=& \key{rsp} \mid \key{rbp} \mid \key{rax} \mid \key{rbx} \mid \key{rcx}
              \mid \key{rdx} \mid \key{rsi} \mid \key{rdi} \mid \\
              && \key{r8} \mid \key{r9} \mid \key{r10}
              \mid \key{r11} \mid \key{r12} \mid \key{r13}
              \mid \key{r14} \mid \key{r15} \\
\Arg &::=&  \key{\$}\Int \mid \key{\%}\itm{register} \mid \Int(\key{\%}\itm{register}) \\ 
\Ins &::=& \key{addq} \; \Arg \; \Arg \mid 
      \key{subq} \; \Arg \; \Arg \mid 
      \key{imulq} \; \Arg \; \Arg \mid 
      \key{negq} \; \Arg \mid \\
  && \key{movq} \; \Arg \; \Arg \mid 
      \key{callq} \; \mathit{label} \mid
      \key{pushq}\;\Arg \mid \key{popq};\Arg \mid \key{retq} \\
\Prog &::= & \key{.globl \_main}\\
      &    & \key{\_main:} \; \Ins^{+} 
\end{array}
\]
\end{minipage}
}
\caption{A subset of the x86-64 assembly language.}
\label{fig:x86-a}
\end{figure}

Figure~\ref{fig:p0-x86} depicts an x86-64 program that is equivalent
to $\BINOP{+}{10}{32}$. The \key{globl} directive says that the
\key{\_main} procedure is externally visible, which is necessary so
that the operating system can call it. The label \key{\_main:}
indicates the beginning of the \key{\_main} procedure.  The
instruction $\key{movq}\,\$10, \%\key{rax}$ puts $10$ into the
register \key{rax}. The following instruction $\key{addq}\,\key{\$}32,
\key{\%rax}$ adds $32$ to the $10$ in \key{rax} and puts the result,
$42$, back into \key{rax}. The instruction \key{retq} finishes the
\key{\_main} function by returning the integer in the \key{rax}
register to the operating system.

\begin{figure}[htbp]
\centering
\begin{minipage}{0.6\textwidth}
\begin{lstlisting}
	.globl _main
_main:
	movq	$10, %rax
	addq	$32, %rax
	retq
\end{lstlisting}
\end{minipage}
\caption{A simple x86-64 program equivalent to $\BINOP{+}{10}{32}$.}
\label{fig:p0-x86}
\end{figure}

The next example exhibits the use of memory.  Figure~\ref{fig:p1-x86}
lists an x86-64 program that is equivalent to $\BINOP{+}{52}{
  \UNIOP{-}{10} }$. To understand how this x86-64 program uses memory,
we need to explain a region of memory called called the
\emph{procedure call stack} (\emph{stack} for short). The stack
consists of a separate \emph{frame} for each procedure call. The
memory layout for an individual frame is shown in
Figure~\ref{fig:frame}.  The register \key{rsp} is called the
\emph{stack pointer} and points to the item at the top of the
stack. The stack grows downward in memory, so we increase the size of
the stack by subtracting from the stack pointer. The frame size is
required to be a multiple of 16 bytes. The register \key{rbp} is the
\emph{base pointer} which serves two purposes: 1) it saves the
location of the stack pointer for the procedure that called the
current one and 2) it is used to access variables associated with the
current procedure. We number the variables from $1$ to $n$. Variable
$1$ is stored at address $-8\key{(\%rbp)}$, variable $2$ at
$-16\key{(\%rbp)}$, etc.

\begin{figure}
\centering
\begin{minipage}{0.6\textwidth}
\begin{lstlisting}
	.globl _main
_main:
	pushq	%rbp
	movq	%rsp, %rbp
	subq	$16, %rsp

	movq	$10, -8(%rbp)
	negq	-8(%rbp)
	movq	$52, %rax
	addq	-8(%rbp), %rax

	addq	$16, %rsp
	popq	%rbp
	retq
\end{lstlisting}
\end{minipage}
\caption{An x86-64 program equivalent to $\BINOP{+}{52}{\UNIOP{-}{10} }$.}
\label{fig:p1-x86}
\end{figure}


\begin{figure}
\centering
\begin{tabular}{|r|l|} \hline
Position & Contents \\ \hline
8(\key{\%rbp}) & return address \\
0(\key{\%rbp}) & old \key{rbp} \\
-8(\key{\%rbp}) & variable $1$ \\
-16(\key{\%rbp}) & variable $2$ \\
 \ldots  & \ldots \\
0(\key{\%rsp}) & variable $n$\\ \hline
\end{tabular}

\caption{Memory layout of a frame.}
\label{fig:frame}
\end{figure}

Getting back to the program in Figure~\ref{fig:p1-x86}, the first
three instructions are the typical prelude for a procedure.  The
instruction \key{pushq \%rbp} saves the base pointer for the procedure
that called the current one onto the stack and subtracts $8$ from the
stack pointer. The second instruction \key{movq \%rsp, \%rbp} changes
the base pointer to the top of the stack. The instruction \key{subq
  \$16, \%rsp} moves the stack pointer down to make enough room for
storing variables.  This program just needs one variable ($8$ bytes)
but because the frame size is required to be a multiple of 16 bytes,
it rounds to 16 bytes.

The next four instructions carry out the work of computing
$\BINOP{+}{52}{\UNIOP{-}{10} }$. The first instruction \key{movq \$10,
  -8(\%rbp)} stores $10$ in variable $1$. The instruction \key{negq
  -8(\%rbp)} changes variable $1$ to $-10$. The \key{movq \$52, \%rax}
places $52$ in the register \key{rax} and \key{addq -8(\%rbp), \%rax}
adds the contents of variable $1$ to \key{rax}, at which point
\key{rax} contains $42$.

The last three instructions are the typical conclusion of a procedure.
The \key{addq \$16, \%rsp} instruction moves the stack pointer back to
point at the old base pointer. The amount added here needs to match
the amount that was subtracted in the prelude of the procedure.  Then
\key{popq \%rbp} returns the old base pointer to \key{rbp} and adds
$8$ to the stack pointer.  The \key{retq} instruction jumps back to
the procedure that called this one and subtracts 8 from the stack
pointer.

\section{Planning the route from $S_0$ to x86-64}

To compile one language to another it helps to focus on the
differences between the two languages. It is these differences that
the compiler will need to bridge. What are the differences between
$S_0$ and x86-64 assembly? Here we list some of the most important the
differences.

\begin{enumerate}
\item Variables in $S_0$ can overshadow other variables with the same
  name. The registers and memory locations of x86-64 all have unique
  names.

\item An argument to an $S_0$ operator can be any expression, whereas
  x86-64 instructions restrict their arguments to integers, registers,
  and memory locations.

\item x86-64 arithmetic instructions typically take two arguments and
  update the second argument in place. In contrast, $S_0$ arithmetic
  operations only read their arguments and produce a new value.

\item An $S_0$ program can have any number of variables whereas x86-64
  has only 16 registers.
\end{enumerate}



\section{An intermediate C-like language}

\[
\begin{array}{lcl}
\Arg &::=& \Int \mid \Var \\
\Exp &::=& \Arg \mid (\Op \; \Arg^{*})\\
\Stmt &::=& (\key{assign} \; \Var \; \Exp) \mid (\key{return}\; \Exp)
\end{array}
\]


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