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\fnote{\copyright{} Christian Urban, King's College London, 2017, 2018, 2019}

\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

good enough for improving the speed of a program to target a

virtual machine. This produces not the fastest possible code, but code

that is often pretty fast. This way of producing code has the advantage that

the virtual machine takes care of things a compiler would normally need

to take care of (like explicit memory management). 

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). Unfortunately the Java ecosystem does not come with an

assembler which would be handy for our  compiler-endeavour  (unlike

Microsoft's  Common Language Infrastructure for the .Net platform which

has an assembler out-of-the-box). As a substitute we use in this module

the 3rd-party programs Jasmin and Krakatau

\begin{itemize}

  \item \url{http://jasmin.sourceforge.net}

  \item \url{https://github.com/Storyyeller/Krakatau}

\end{itemize}

\noindent

The first is a Java program and the second a program written in Python.

Each of them allow us to generate \emph{assembly} files that are still

readable by humans, as opposed to class-files which are pretty much just

(horrible) zeros and ones. Jasmin (respectively Krakatau) will then take

an assembly file as input and generate the corresponding class file for

us. 

Good about the JVM is that it is a stack-based virtual machine, a fact

which will make it easy to generate code for arithmetic expressions. For

example when compiling the expression $1 + 2$ we need to generate the

following three instructions

\begin{lstlisting}[language=JVMIS,numbers=none]

ldc 1

ldc 2

iadd 

\end{lstlisting}

\noindent The first instruction loads the constant $1$ onto

the stack, the next one loads $2$, 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 we will obtain for the

expression $1 + ((2 * 3) + (4 - 3))$ the following tree.

\begin{center}

\begin{tikzpicture}

\Tree [.$+$ [.$1$ ] [.$+$ [.$*$ $2$ $3$ ] [.$-$ $4$ $3$ ]]]

\end{tikzpicture}

\end{center}

\noindent To generate JVM 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}[language=JVMIS,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 an important convention we always observe in our compiler. Note,

that a different bracketing of the expression, for example $(1 + (2 *

3)) + (4 - 3)$, produces a different abstract syntax tree and thus also

a different list of instructions. Generating code in this

post-order-traversal 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}[language=JVMIS,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}[language=JVMIS,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. This is similar to an interpreter,

which also needs an environment: the difference is that the 

interpreter maintains a mapping from variables to current values (what is the

currently the value of a variable), while compilers need a mapping

from variables to memory locations (where can I find the current 

value for the variable in memory).

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 whereby $[]$ 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, then we need to be able to retrieve the associated index.

This is reflected in the clause for compiling assignments, say

$\textit{x := a}$:

\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 $E' = E(x \mapsto largest\_index + 1)$). 

This means for the assignment $x := x + 1$ we generate the

following code

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

iload $n_x$

ldc 1

iadd

istore $n_x$

\end{lstlisting}

\noindent 

where $n_x$ is the index (or pointer to the memory) for the variable $x$ . The code for 

looking-up the index for the variable is as follow:

\begin{center}

\begin{tabular}{lcl}

$index \;=\; E\textit{.getOrElse}(x, |E|)$

\end{tabular}

\end{center}

\noindent

In case the environment $E$ contains an index for $x$, we return it.

Otherwise we ``create'' a new index by returning the size $|E|$ of the

environment (that will be an index that is guaranteed to be not used

yet).

A bit more complicated is the generation of code for \pcode{if}-statements, 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 of the \pcode{if}-branches. 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