\documentclass{article}
\usepackage{../style}
\usepackage{../langs}
\usepackage{../grammar}
\usepackage{../graphics}


\begin{document}

\section*{Handout 7 (Compilation)}

The purpose of a compiler is to transform a program, a human
can read and write, into code the machine can run as fast as
possible. The fastest code would be machine code the CPU can
run directly, but it is often enough to improve the speed of a
program by just targeting a virtual machine. This produces not
the fastest possible code, but code that is fast enough and
has the advantage that the virtual machine takes care of
things a compiler would normally need to take care of (like
explicit memory management). Why study compilers? John Regher
gives this answer in his compiler blog:\footnote{\url{http://blog.regehr.org/archives/1419}}

\begin{quote}\it{}``We can start off with a couple of observations
  about the role of compilers. First, hardware is getting weirder
  rather than getting clocked faster: almost all processors are
  multicores and it looks like there is increasing asymmetry in
  resources across cores. Processors come with vector units, crypto
  accelerators, bit twiddling instructions, and lots of features to
  make virtualization and concurrency work. We have DSPs, GPUs,
  big.little, and Xeon Phi. This is only scratching the
  surface. Second, we’re getting tired of low-level languages and
  their associated security disasters, we want to write new code, to
  whatever extent possible, in safer, higher-level
  languages. Compilers are caught right in the middle of these
  opposing trends: one of their main jobs is to help bridge the large
  and growing gap between increasingly high-level languages and
  increasingly wacky platforms. It’s effectively a perpetual
  employment act for solid compiler hackers.''
\end{quote}  

As a first example in this module we will implement a compiler for the
very simple While-language. It will generate code for the Java Virtual
Machine (JVM). This is a stack-based virtual machine, a fact which
will make it easy to generate code for arithmetic expressions. For
example for generating code for the expression $1 + 2$ we need to
generate the following three instructions

\begin{lstlisting}[numbers=none]
ldc 1
ldc 2
iadd 
\end{lstlisting}

\noindent The first instruction loads the constant $1$ onto
the stack, the next one $2$, the third instruction adds both
numbers together replacing the top two elements of the stack
with the result $3$. For simplicity, we will throughout
consider only integer numbers and results. Therefore we can
use the JVM instructions \code{iadd}, \code{isub},
\code{imul}, \code{idiv} and so on. The \code{i} stands for
integer instructions in the JVM (alternatives are \code{d} for
doubles, \code{l} for longs and \code{f} for floats).

Recall our grammar for arithmetic expressions (\meta{E} is the
starting symbol):


\begin{plstx}[rhs style=, margin=3cm]
: \meta{E} ::= \meta{T} $+$ \meta{E}
         | \meta{T} $-$ \meta{E}
         | \meta{T}\\
: \meta{T} ::= \meta{F} $*$ \meta{T}
          | \meta{F} $\backslash$ \meta{T}
          | \meta{F}\\
: \meta{F} ::= ( \meta{E} )
          | \meta{Id}
          | \meta{Num}\\
\end{plstx}


\noindent where \meta{Id} stands for variables and \meta{Num}
for numbers. For the moment let us omit variables from
arithmetic expressions. Our parser will take this grammar and
given an input produce abstract syntax trees. For example for
the expression $1 + ((2 * 3) + (4 - 3))$ it will produce the
following tree.

\begin{center}
\begin{tikzpicture}
\Tree [.$+$ [.$1$ ] [.$+$ [.$*$ $2$ $3$ ] [.$-$ $4$ $3$ ]]]
\end{tikzpicture}
\end{center}

\noindent To generate code for this expression, we need to
traverse this tree in post-order fashion and emit code for
each node---this traversal in post-order fashion will produce
code for a stack-machine (what the JVM is). Doing so for the
tree above generates the instructions

\begin{lstlisting}[numbers=none]
ldc 1 
ldc 2 
ldc 3 
imul 
ldc 4 
ldc 3 
isub 
iadd 
iadd
\end{lstlisting}

\noindent If we ``run'' these instructions, the result $8$
will be on top of the stack (I leave this to you to verify;
the meaning of each instruction should be clear). The result
being on the top of the stack will be a convention we always
observe in our compiler, that is the results of arithmetic
expressions will always be on top of the stack. Note, that a
different bracketing of the expression, for example $(1 + (2 *
3)) + (4 - 3)$, produces a different abstract syntax tree and
thus potentially also a different list of instructions.
Generating code in this fashion is rather easy to implement:
it can be done with the following recursive
\textit{compile}-function, which takes the abstract syntax
tree as argument:

\begin{center}
\begin{tabular}{lcl}
$\textit{compile}(n)$ & $\dn$ & $\pcode{ldc}\; n$\\
$\textit{compile}(a_1 + a_2)$ & $\dn$ &
$\textit{compile}(a_1) \;@\;\textit{compile}(a_2)\;@\; \pcode{iadd}$\\
$\textit{compile}(a_1 - a_2)$ & $\dn$ & 
$\textit{compile}(a_1) \;@\; \textit{compile}(a_2)\;@\; \pcode{isub}$\\
$\textit{compile}(a_1 * a_2)$ & $\dn$ & 
$\textit{compile}(a_1) \;@\; \textit{compile}(a_2)\;@\; \pcode{imul}$\\
$\textit{compile}(a_1 \backslash a_2)$ & $\dn$ & 
$\textit{compile}(a_1) \;@\; \textit{compile}(a_2)\;@\; \pcode{idiv}$\\
\end{tabular}
\end{center}

However, our arithmetic expressions can also contain
variables. We will represent them as \emph{local variables} in
the JVM. Essentially, local variables are an array or pointers
to memory cells, containing in our case only integers. Looking
up a variable can be done with the instruction

\begin{lstlisting}[mathescape,numbers=none]
iload $index$
\end{lstlisting}

\noindent 
which places the content of the local variable $index$ onto 
the stack. Storing the top of the stack into a local variable 
can be done by the instruction

\begin{lstlisting}[mathescape,numbers=none]
istore $index$
\end{lstlisting}

\noindent Note that this also pops off the top of the stack.
One problem we have to overcome, however, is that local
variables are addressed, not by identifiers, but by numbers
(starting from $0$). Therefore our compiler needs to maintain
a kind of environment where variables are associated to
numbers. This association needs to be unique: if we muddle up
the numbers, then we essentially confuse variables and the
consequence will usually be an erroneous result. Our extended
\textit{compile}-function for arithmetic expressions will
therefore take two arguments: the abstract syntax tree and the
environment, $E$, that maps identifiers to index-numbers.

\begin{center}
\begin{tabular}{lcl}
$\textit{compile}(n, E)$ & $\dn$ & $\pcode{ldc}\;n$\\
$\textit{compile}(a_1 + a_2, E)$ & $\dn$ & 
$\textit{compile}(a_1, E) \;@\;\textit{compile}(a_2, E)\;@\; \pcode{iadd}$\\
$\textit{compile}(a_1 - a_2, E)$ & $\dn$ &
$\textit{compile}(a_1, E) \;@\; \textit{compile}(a_2, E)\;@\; \pcode{isub}$\\
$\textit{compile}(a_1 * a_2, E)$ & $\dn$ &
$\textit{compile}(a_1, E) \;@\; \textit{compile}(a_2, E)\;@\; \pcode{imul}$\\
$\textit{compile}(a_1 \backslash a_2, E)$ & $\dn$ & 
$\textit{compile}(a_1, E) \;@\; \textit{compile}(a_2, E)\;@\; \pcode{idiv}$\\
$\textit{compile}(x, E)$ & $\dn$ & $\pcode{iload}\;E(x)$\\
\end{tabular}
\end{center}

\noindent In the last line we generate the code for variables
where $E(x)$ stands for looking up the environment to which
index the variable $x$ maps to.

There is a similar \textit{compile}-function for boolean
expressions, but it includes a ``trick'' to do with
\pcode{if}- and \pcode{while}-statements. To explain the issue
let us first describe the compilation of statements of the
While-language. The clause for \pcode{skip} is trivial, since
we do not have to generate any instruction

\begin{center}
\begin{tabular}{lcl}
$\textit{compile}(\pcode{skip}, E)$ & $\dn$ & $([], E)$\\
\end{tabular}
\end{center}

\noindent $[]$ is the empty list of instructions. Note that
the \textit{compile}-function for statements returns a pair, a
list of instructions (in this case the empty list) and an
environment for variables. The reason for the environment is
that assignments in the While-language might change the
environment---clearly if a variable is used for the first
time, we need to allocate a new index and if it has been used
before, we need to be able to retrieve the associated index.
This is reflected in the clause for compiling assignments:

\begin{center}
\begin{tabular}{lcl}
$\textit{compile}(x := a, E)$ & $\dn$ & 
$(\textit{compile}(a, E) \;@\;\pcode{istore}\;index, E')$
\end{tabular}
\end{center}

\noindent We first generate code for the right-hand side of
the assignment and then add an \pcode{istore}-instruction at
the end. By convention the result of the arithmetic expression
$a$ will be on top of the stack. After the \pcode{istore}
instruction, the result will be stored in the index
corresponding to the variable $x$. If the variable $x$ has
been used before in the program, we just need to look up what
the index is and return the environment unchanged (that is in
this case $E' = E$). However, if this is the first encounter 
of the variable $x$ in the program, then we have to augment 
the environment and assign $x$ with the largest index in $E$
plus one (that is $E' = E(x \mapsto largest\_index + 1)$). 
That means for the assignment $x := x + 1$ we generate the
following code

\begin{lstlisting}[mathescape,numbers=none]
iload $n_x$
ldc 1
iadd
istore $n_x$
\end{lstlisting}

\noindent 
where $n_x$ is the index for the variable $x$.

More complicated is the code for \pcode{if}-statments, say

\begin{lstlisting}[mathescape,language={},numbers=none]
if $b$ then $cs_1$ else $cs_2$
\end{lstlisting}

\noindent where $b$ is a boolean expression and the $cs_{1/2}$
are the statements for each \pcode{if}-branch. Lets assume
we already generated code for $b$ and $cs_{1/2}$. Then in the
true-case the control-flow of the program needs to be

\begin{center}
\begin{tikzpicture}[node distance=2mm and 4mm,
 block/.style={rectangle, minimum size=1cm, draw=black, line width=1mm},
 point/.style={rectangle, inner sep=0mm, minimum size=0mm, fill=red},
 skip loop/.style={black, line width=1mm, to path={-- ++(0,-10mm) -| (\tikztotarget)}}]
\node (A1) [point] {};
\node (b) [block, right=of A1] {code of $b$};
\node (A2) [point, right=of b] {};
\node (cs1) [block, right=of A2] {code of $cs_1$};
\node (A3) [point, right=of cs1] {};
\node (cs2) [block, right=of A3] {code of $cs_2$};
\node (A4) [point, right=of cs2] {};

\draw (A1) edge [->, black, line width=1mm] (b);
\draw (b) edge [->, black, line width=1mm] (cs1);
\draw (cs1) edge [->, black, line width=1mm] (A3);
\draw (A3) edge [->, black, skip loop] (A4);
\node [below=of cs2] {\raisebox{-5mm}{\small{}jump}};
\end{tikzpicture}
\end{center}

\noindent where we start with running the code for $b$; since
we are in the true case we continue with running the code for
$cs_1$. After this however, we must not run the code for
$cs_2$, but always jump after the last instruction of $cs_2$
(the code for the \pcode{else}-branch). Note that this jump is
unconditional, meaning we always have to jump to the end of
$cs_2$. The corresponding instruction of the JVM is
\pcode{goto}. In case $b$ turns out to be false we need the
control-flow

\begin{center}
\begin{tikzpicture}[node distance=2mm and 4mm,
 block/.style={rectangle, minimum size=1cm, draw=black, line width=1mm},
 point/.style={rectangle, inner sep=0mm, minimum size=0mm, fill=red},
 skip loop/.style={black, line width=1mm, to path={-- ++(0,-10mm) -| (\tikztotarget)}}]
\node (A1) [point] {};
\node (b) [block, right=of A1] {code of $b$};
\node (A2) [point, right=of b] {};
\node (cs1) [block, right=of A2] {code of $cs_1$};
\node (A3) [point, right=of cs1] {};
\node (cs2) [block, right=of A3] {code of $cs_2$};
\node (A4) [point, right=of cs2] {};

\draw (A1) edge [->, black, line width=1mm] (b);
\draw (b) edge [->, black, line width=1mm] (A2);
\draw (A2) edge [skip loop] (A3);
\draw (A3) edge [->, black, line width=1mm] (cs2);
\draw (cs2) edge [->,black, line width=1mm] (A4);
\node [below=of cs1] {\raisebox{-5mm}{\small{}conditional jump}};
\end{tikzpicture}
\end{center}

\noindent where we now need a conditional jump (if the
if-condition is false) from the end of the code for the 
boolean to the beginning of the instructions $cs_2$. Once we 
are finished with running $cs_2$ we can continue with whatever
code comes after the if-statement.

The \pcode{goto} and the conditional jumps need addresses to
where the jump should go. Since we are generating assembly
code for the JVM, we do not actually have to give (numeric)
addresses, but can just attach (symbolic) labels to our code.
These labels specify a target for a jump. Therefore the labels
need to be unique, as otherwise it would be ambiguous where a
jump should go to. A label, say \pcode{L}, is attached to code
like

\begin{lstlisting}[mathescape,numbers=none]
L:
  $instr_1$
  $instr_2$
    $\vdots$
\end{lstlisting}
 
\noindent where a label is indicated by a colon. 
 
Recall the ``trick'' with compiling boolean expressions: the 
\textit{compile}-function for boolean expressions takes three
arguments: an abstract syntax tree, an environment for 
variable indices and also the label, $lab$, to where an conditional 
jump needs to go. The clause for the expression $a_1 = a_2$, 
for example, is as follows:

\begin{center}
\begin{tabular}{lcl}
$\textit{compile}(a_1 = a_2, E, lab)$ & $\dn$\\ 
\multicolumn{3}{l}{$\qquad\textit{compile}(a_1, E) \;@\;\textit{compile}(a_2, E)\;@\; \pcode{if_icmpne}\;lab$}
\end{tabular}
\end{center}

\noindent where we are first generating code for the
subexpressions $a_1$ and $a_2$. This will mean after running
the corresponding code there will be two integers on top of
the stack. If they are equal, we do not have to do anything
(except for popping them off from the stack) and just continue
with the next instructions (see control-flow of ifs above).
However if they are \emph{not} equal, then we need to
(conditionally) jump to the label $lab$. This can be done with
the instruction

\begin{lstlisting}[mathescape,numbers=none]
if_icmpne $lab$
\end{lstlisting}

\noindent Other jump instructions for boolean operators are

\begin{center}
\begin{tabular}{l@{\hspace{10mm}}c@{\hspace{10mm}}l}
$\not=$ & $\Rightarrow$ & \pcode{if_icmpeq}\\
$<$ & $\Rightarrow$ & \pcode{if_icmpge}\\
$\le$ & $\Rightarrow$ & \pcode{if_icmpgt}\\
\end{tabular}
\end{center}

\noindent and so on. I leave it to you to extend the
\textit{compile}-function for the other boolean expressions.
Note that we need to jump whenever the boolean is \emph{not}
true, which means we have to ``negate'' the jump
condition---equals becomes not-equal, less becomes
greater-or-equal. If you do not like this design (it can be
the source of some nasty, hard-to-detect errors), you can also
change the layout of the code and first give the code for the
else-branch and then for the if-branch. However in the case
of while-loops this way of generating code still seems
the most convenient.

We are now ready to give the compile function for 
if-statments---remember this function returns for staments a 
pair consisting of the code and an environment:

\begin{center}
\begin{tabular}{lcl}
$\textit{compile}(\pcode{if}\;b\;\pcode{then}\; cs_1\;\pcode{else}\; cs_2, E)$ & $\dn$\\ 
\multicolumn{3}{l}{$\qquad L_\textit{ifelse}\;$ (fresh label)}\\
\multicolumn{3}{l}{$\qquad L_\textit{ifend}\;$ (fresh label)}\\
\multicolumn{3}{l}{$\qquad (is_1, E') = \textit{compile}(cs_1, E)$}\\
\multicolumn{3}{l}{$\qquad (is_2, E'') = \textit{compile}(cs_2, E')$}\\
\multicolumn{3}{l}{$\qquad(\textit{compile}(b, E, L_\textit{ifelse})$}\\
\multicolumn{3}{l}{$\qquad\phantom{(}@\;is_1$}\\
\multicolumn{3}{l}{$\qquad\phantom{(}@\; \pcode{goto}\;L_\textit{ifend}$}\\
\multicolumn{3}{l}{$\qquad\phantom{(}@\;L_\textit{ifelse}:$}\\
\multicolumn{3}{l}{$\qquad\phantom{(}@\;is_2$}\\
\multicolumn{3}{l}{$\qquad\phantom{(}@\;L_\textit{ifend}:, E'')$}\\
\end{tabular}
\end{center}

\noindent In the first two lines we generate two fresh labels
for the jump addresses (just before the else-branch and just
after). In the next two lines we generate the instructions for
the two branches, $is_1$ and $is_2$. The final code will
be first the code for $b$ (including the label 
just-before-the-else-branch), then the \pcode{goto} for after
the else-branch, the label $L_\textit{ifesle}$, followed by
the instructions for the else-branch, followed by the 
after-the-else-branch label. Consider for example the 
if-statement:

\begin{lstlisting}[mathescape,numbers=none,language={}]
if 1 = 1 then x := 2 else y := 3
\end{lstlisting}

\noindent 
The generated code is as follows:

\begin{lstlisting}[mathescape,language={}]
   ldc 1
   ldc 1
   if_icmpne L_ifelse $\quad\tikz[remember picture] \node (C) {\mbox{}};$
   ldc 2
   istore 0
   goto L_ifend $\quad\tikz[remember picture] \node (A) {\mbox{}};$
L_ifelse: $\quad\tikz[remember picture] \node[] (D) {\mbox{}};$
   ldc 3
   istore 1
L_ifend: $\quad\tikz[remember picture] \node[] (B) {\mbox{}};$
\end{lstlisting}

\begin{tikzpicture}[remember picture,overlay]
  \draw[->,very thick] (A) edge [->,to path={-- ++(10mm,0mm) 
           -- ++(0mm,-17.3mm) |- (\tikztotarget)},line width=1mm] (B.east);
  \draw[->,very thick] (C) edge [->,to path={-- ++(10mm,0mm) 
           -- ++(0mm,-17.3mm) |- (\tikztotarget)},line width=1mm] (D.east);
\end{tikzpicture}

\noindent The first three lines correspond to the the boolean
expression $1 = 1$. The jump for when this boolean expression
is false is in Line~3. Lines 4-6 corresponds to the if-branch;
the else-branch is in Lines 8 and 9. Note carefully how the
environment $E$ is threaded through the recursive calls of
\textit{compile}. The function receives an environment $E$,
but it might extend it when compiling the if-branch, yielding
$E'$. This happens for example in the if-statement above
whenever the variable \code{x} has not been used before.
Similarly with the environment $E''$ for the second call to
\textit{compile}. $E''$ is also the environment that needs to
be returned as part of the answer.

The compilation of the while-loops, say 
\pcode{while} $b$ \pcode{do} $cs$, is very similar. In case
the condition is true and we need to do another iteration, 
and the control-flow needs to be as follows

\begin{center}
\begin{tikzpicture}[node distance=2mm and 4mm,
 block/.style={rectangle, minimum size=1cm, draw=black, line width=1mm},
 point/.style={rectangle, inner sep=0mm, minimum size=0mm, fill=red},
 skip loop/.style={black, line width=1mm, to path={-- ++(0,-10mm) -| (\tikztotarget)}}]
\node (A0) [point, left=of A1] {};
\node (A1) [point] {};
\node (b) [block, right=of A1] {code of $b$};
\node (A2) [point, right=of b] {};
\node (cs1) [block, right=of A2] {code of $cs$};
\node (A3) [point, right=of cs1] {};
\node (A4) [point, right=of A3] {};

\draw (A0) edge [->, black, line width=1mm] (b);
\draw (b) edge [->, black, line width=1mm] (cs1);
\draw (cs1) edge [->, black, line width=1mm] (A3);
\draw (A3) edge [->,skip loop] (A1);
\end{tikzpicture}
\end{center}

\noindent Whereas if the condition is \emph{not} true, we
need to jump out of the loop, which gives the following
control flow.

\begin{center}
\begin{tikzpicture}[node distance=2mm and 4mm,
 block/.style={rectangle, minimum size=1cm, draw=black, line width=1mm},
 point/.style={rectangle, inner sep=0mm, minimum size=0mm, fill=red},
 skip loop/.style={black, line width=1mm, to path={-- ++(0,-10mm) -| (\tikztotarget)}}]
\node (A0) [point, left=of A1] {};
\node (A1) [point] {};
\node (b) [block, right=of A1] {code of $b$};
\node (A2) [point, right=of b] {};
\node (cs1) [block, right=of A2] {code of $cs$};
\node (A3) [point, right=of cs1] {};
\node (A4) [point, right=of A3] {};

\draw (A0) edge [->, black, line width=1mm] (b);
\draw (b) edge [->, black, line width=1mm] (A2);
\draw (A2) edge [skip loop] (A3);
\draw (A3) edge [->, black, line width=1mm] (A4);
\end{tikzpicture}
\end{center}

\noindent Again we can use the \textit{compile}-function for
boolean expressions to insert the appropriate jump to the
end of the loop (label $L_{wend}$ below).

\begin{center}
\begin{tabular}{lcl}
$\textit{compile}(\pcode{while}\; b\; \pcode{do} \;cs, E)$ & $\dn$\\ 
\multicolumn{3}{l}{$\qquad L_{wbegin}\;$ (fresh label)}\\
\multicolumn{3}{l}{$\qquad L_{wend}\;$ (fresh label)}\\
\multicolumn{3}{l}{$\qquad (is, E') = \textit{compile}(cs_1, E)$}\\
\multicolumn{3}{l}{$\qquad(L_{wbegin}:$}\\
\multicolumn{3}{l}{$\qquad\phantom{(}@\;\textit{compile}(b, E, L_{wend})$}\\
\multicolumn{3}{l}{$\qquad\phantom{(}@\;is$}\\
\multicolumn{3}{l}{$\qquad\phantom{(}@\; \text{goto}\;L_{wbegin}$}\\
\multicolumn{3}{l}{$\qquad\phantom{(}@\;L_{wend}:, E')$}\\
\end{tabular}
\end{center}

\noindent I let you go through how this clause works. As an example
you can consider the while-loop

\begin{lstlisting}[mathescape,numbers=none,language={}]
while x <= 10 do x := x + 1
\end{lstlisting}

\noindent yielding the following code

\begin{lstlisting}[mathescape,language={}]
L_wbegin: $\quad\tikz[remember picture] \node[] (LB) {\mbox{}};$
   iload 0
   ldc 10
   if_icmpgt L_wend $\quad\tikz[remember picture] \node (LC) {\mbox{}};$
   iload 0
   ldc 1
   iadd
   istore 0
   goto L_wbegin $\quad\tikz[remember picture] \node (LA) {\mbox{}};$
L_wend: $\quad\tikz[remember picture] \node[] (LD) {\mbox{}};$
\end{lstlisting}
 
\begin{tikzpicture}[remember picture,overlay]
  \draw[->,very thick] (LA) edge [->,to path={-- ++(10mm,0mm) 
           -- ++(0mm,17.3mm) |- (\tikztotarget)},line width=1mm] (LB.east);
  \draw[->,very thick] (LC) edge [->,to path={-- ++(10mm,0mm) 
           -- ++(0mm,-17.3mm) |- (\tikztotarget)},line width=1mm] (LD.east);
\end{tikzpicture}


Next we need to consider the statement \pcode{write x}, which
can be used to print out the content of a variable. For this
we need to use a Java library function. In order to avoid
having to generate a lot of code for each
\pcode{write}-command, we use a separate helper-method and
just call this method with an argument (which needs to be
placed onto the stack). The code of the helper-method is as
follows.


\begin{lstlisting}[language=JVMIS]
.method public static write(I)V 
    .limit locals 1 
    .limit stack 2 
    getstatic java/lang/System/out Ljava/io/PrintStream; 
    iload 0
    invokevirtual java/io/PrintStream/println(I)V 
    return 
.end method
\end{lstlisting}

\noindent The first line marks the beginning of the method,
called \pcode{write}. It takes a single integer argument
indicated by the \pcode{(I)} and returns no result, indicated
by the \pcode{V}. Since the method has only one argument, we
only need a single local variable (Line~2) and a stack with
two cells will be sufficient (Line 3). Line 4 instructs the
JVM to get the value of the field \pcode{out} of the class
\pcode{java/lang/System}. It expects the value to be of type
\pcode{java/io/PrintStream}. A reference to this value will be
placed on the stack. Line~5 copies the integer we want to
print out onto the stack. In the next line we call the method
\pcode{println} (from the class \pcode{java/io/PrintStream}).
We want to print out an integer and do not expect anything
back (that is why the type annotation is \pcode{(I)V}). The
\pcode{return}-instruction in the next line changes the
control-flow back to the place from where \pcode{write} was
called. This method needs to be part of a header that is
included in any code we generate. The helper-method
\pcode{write} can be invoked with the two instructions

\begin{lstlisting}[mathescape,language=JVMIS]
iload $E(x)$ 
invokestatic XXX/XXX/write(I)V
\end{lstlisting}

\noindent where we first place the variable to be printed on
top of the stack and then call \pcode{write}. The \pcode{XXX}
need to be replaced by an appropriate class name (this will be
explained shortly).


\begin{figure}[t]
\begin{lstlisting}[mathescape,language=JVMIS]
.class public XXX.XXX
.super java/lang/Object

.method public <init>()V
    aload_0
    invokenonvirtual java/lang/Object/<init>()V
    return
.end method

.method public static main([Ljava/lang/String;)V
    .limit locals 200
    .limit stack 200

      $\textit{\ldots{}here comes the compiled code\ldots}$

    return
.end method
\end{lstlisting}
\caption{Boilerplate code needed for running generated code.\label{boiler}}
\end{figure}


By generating code for a While-program, we end up with a list
of (JVM assembly) instructions. Unfortunately, there is a bit
more boilerplate code needed before these instructions can be
run. The complete code is shown in Figure~\ref{boiler}. This
boilerplate code is very specific to the JVM. If we target any
other virtual machine or a machine language, then we would
need to change this code. Lines 4 to 8 in Figure~\ref{boiler}
contain a method for object creation in the JVM; this method
is called \emph{before} the \pcode{main}-method in Lines 10 to
17. Interesting are the Lines 11 and 12 where we hardwire that
the stack of our programs will never be larger than 200 and
that the maximum number of variables is also 200. This seem to
be conservative default values that allow is to run some
simple While-programs. In a real compiler, we would of course
need to work harder and find out appropriate values for the
stack and local variables.

To sum up, in Figure~\ref{test} is the complete code generated
for the slightly non-sensical program

\begin{lstlisting}[mathescape,language=While]
x := 1 + 2;
write x
\end{lstlisting}

\noindent Having this code at our disposal, we need the
assembler to translate the generated code into JVM bytecode (a
class file). This bytecode is understood by the JVM and can be
run by just invoking the \pcode{java}-program.


\begin{figure}[p]
\lstinputlisting[language=JVMIS]{../progs/test-small.j}
\caption{Generated code for a test program. This code can be 
processed by an Java assembler producing a class-file, which
can be run by the {\tt{}java}-program.\label{test}}
\end{figure}

\end{document}

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