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\section*{Handout 7 (Compilation)}

The purpose of a compiler is to transform a program, a human

can 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 care of things a

compiler would normally need to take care of (like explicit

memory management).

As an example we will implement a compiler for the very simple

While-language. We will be generating code for the Java

Virtual Machine. 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 elements of the stack with

the result $3$. For simplicity, we will throughout consider

only integer numbers and results. Therefore we can use the

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 ($E$ 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

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 \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 explain first 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 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}

$\text{compile}(x := a, E)$ & $\dn$ & 

$(\textit{compile}(a, E) \;@\;\pcode{istore}\;index, E')$

\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 $cs_i$

are the instructions 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 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 addresses, but need

to attach 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. 

A labels, say \pcode{L}, is attached to code like

\begin{lstlisting}[mathescape,numbers=none]

L:

  $instr_1$

  $instr_2$

    $\vdots$

\end{lstlisting}

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 

We are 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 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}

$=$ & $\Rightarrow$ & \pcode{if_icmpne}\\

$\not=$ & $\Rightarrow$ & \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---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.

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