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\begin{document}

\lstset{morekeywords={def,if,then,else,write,read},keywordstyle=\color{codepurple}\bfseries}

\section*{Handout 8 (A Functional Language)}

The language we looked at in the previous lecture was rather

primitive and the compiler rather crude---everything was

essentially compiled into a big monolithic chunk of code

inside the main function. In this handout we like to have a

look at a slightly more comfortable language, which I call

Fun-language, and a tiny-teeny bit more realistic compiler.

collection of programs we want to be able to write and compile

is as follows:

\begin{lstlisting}[numbers=none]

def fib(n) = if n == 0 then 0 

             else if n == 1 then 1 

             else fib(n - 1) + fib(n - 2);

def fact(n) = if n == 0 then 1 else n * fact(n - 1);

def ack(m, n) = if m == 0 then n + 1

                else if n == 0 then ack(m - 1, 1)

                else ack(m - 1, ack(m, n - 1));

def gcd(a, b) = if b == 0 then a else gcd(b, a % b);                

\end{lstlisting}

The grammar of the Fun-language is slightly simpler than the

While-language, because the main syntactic category are

expressions (we do not have statements). The grammar rules are

as follows:

\begin{plstx}[rhs style=,margin=1.5cm]

: \meta{Exp} ::= \meta{Id} | \meta{Num} {\hspace{3.7cm}}

             |   \meta{Exp} + \meta{Exp} | ... | (\meta{Exp}) {\hspace{3.7cm}}

             |   \code{write} \meta{Exp} {\hspace{5cm}}

             |   \meta{Exp} ; \meta{Exp}  {\hspace{5cm}}

             |   \textit{FunName} (\meta{Exp}, ..., \meta{Exp})\\

: \meta{BExp} ::= ...\\

: \meta{Decl} ::= \meta{Def} ; \meta{Decl}

             | \meta{Exp}\\

: \meta{Def} ::= \code{def} \textit{FunName} ($x_1$, ..., $x_n$) = \meta{Exp}\\               

\end{plstx}

\noindent where, as usual, \meta{Id} stands for variables and

\meta{Num} for numbers. We can call a function by applying the

arguments to a function name (as shown in the last clause of

\meta{Exp}). The arguments in such a function call can be

again expressions, including other function calls. In

contrast, when defining a function (see \meta{Def}-clause) the

arguments need to be variables, say $x_1$ to $x_n$. We call

the expression on the right of = in a function definition as

the \emph{body of the function}. We have the restriction that

the variables inside a function body can only be those that

are mentioned as arguments of the function. A Fun-program is

then a sequence of function definitions separated by

semicolons, and a final ``main'' call of a function that

starts the computation in the program. For example

\begin{lstlisting}[numbers=none]

def fact(n) = if n == 0 then 1 else n * fact(n - 1);

write(fact(5))

\end{lstlisting}

\noindent would be a valid Fun-program. The parser of the

Fun-language produces abstract syntax trees which in Scala

can be represented as follows:

{\small

\begin{lstlisting}[numbers=none]

abstract class Exp

abstract class BExp 

abstract class Decl

case class Var(s: String) extends Exp

case class Num(i: Int) extends Exp

case class Aop(o: String, a1: Exp, a2: Exp) extends Exp

case class If(a: BExp, e1: Exp, e2: Exp) extends Exp

case class Write(e: Exp) extends Exp

case class Sequ(e1: Exp, e2: Exp) extends Exp

case class Call(name: String, args: List[Exp]) extends Exp

case class Bop(o: String, a1: Exp, a2: Exp) extends BExp

case class Def(name: String, 

               args: List[String], 

               body: Exp) extends Decl

case class Main(e: Exp) extends Decl

\end{lstlisting}}

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

$\textrm{\textit{compile}}($exp1$)$

pop

$\textrm{\textit{compile}}($exp2$)$

\end{lstlisting}

\noindent In effect we ``forget'' about the result the first

expression calculates. I leave you to think about why this

sequence operator is still useful in the Fun-language, even 

if the first result is just ``discarded''. 

There is also one small modification we have to perform when

calling the write method. Remember in the Fun-language we have

the convention that every expression needs to return an

integer as a result (located on the top of the stack). Our

helper function implementing write, however, ``consumes'' the

top element of the stack and violates this convention.

Therefore before we call, say, \pcode{write(1+2)}, we need to

duplicate the top element of the stack like so

\begin{figure}[t]

\begin{lstlisting}[language=JVMIS, 

                   xleftmargin=2mm, 

                   numbers=none]

.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}

\caption{The helper function for printing out integers.\label{write}}

\end{figure}

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

$\textrm{\textit{compile}}($1+2$)$

dup

invokestatic XXX/XXX/write(I)V

\end{lstlisting}

\noindent We also need to first generate code for the

argument-expression of write, which in the While-language was

only allowed to be a single variable.

Most of the new code in the compiler for the Fun-language

comes from function definitions and function calls. For this

have a look again at the helper function in

Figure~\ref{write}. Assuming we have a function definition

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

def fname (x1, ... , xn) = ...

\end{lstlisting}

\noindent then we have to generate

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

.method public static fname (I...I)I

   .limit locals ??

   .limit stack ?? 

   ...

  ireturn

.method end  

\end{lstlisting}

\noindent where the number of \pcode{I}s corresponds to the

number of arguments the function has, say \pcode{x1} to

\pcode{xn}. The final \pcode{I} is needed in order to indicate

that the function returns an integer. Therefore we also have

to use \pcode{ireturn} instead of \pcode{return}. However,

more interesting are the two \pcode{.limit} lines.

\pcode{Locals} refers to the local variables of the method,

which can be queried and overwritten using the JVM

instructions \pcode{iload} and \pcode{istore}, respectively.

Before we call a function with, say, three arguments, we need

to ensure that these three arguments are pushed onto the stack

(we will come to the corresponding code shortly). Once we are

inside the method, the arguments on the stack turn into local

variables. So in case we have three arguments on the stack, we 

will have inside the function three local variables that can 

be referenced by the indices $0..2$. Determining the limit for

local variables is the easy bit. Harder is the stack limit.

Calculating how much stack a program needs is equivalent to 

the Halting problem, and thus undecidable in general. 

Fortunately, we are only asked how much stack a \emph{single} 

call of the function requires. This can be relatively easily

compiled by recursively analysing which instructions we 

generate and how much stack they might require.

\begin{center}

\begin{tabular}{lcl}

$\textit{estimate}(n)$ & $\dn$ & $1$\\

$\textit{estimate}(x)$ & $\dn$ & $1$\\

$\textit{estimate}(a_1\;aop\;a_2)$ & $\dn$ &

$\textit{estimate}(a_1) + \textit{estimate}(a_2)$\\

$\textit{estimate}(\pcode{if}\;b\;\pcode{then}\;e_1\;\pcode{else}\;e_2)$ & $\dn$ & 

$\textit{estimate}(b) +$\\ 

& & $\quad max(\textit{estimate}(e_1), \textit{estimate}(e_2))$\\

$\textit{estimate}(\pcode{write}(e))$ & $\dn$ & 

$\textit{estimate}(e) + 1$\\

$\textit{estimate}(e_1 ; e_2)$ & $\dn$ & 

$max(\textit{estimate}(e_1), \textit{estimate}(e_2))$\\

$\textit{estimate}(f(e_1, ..., e_n))$ & $\dn$ & 

$\sum_{i = 1..n}\;estimate(e_i)$\medskip\\

$\textit{estimate}(a_1\;bop\;a_2)$ & $\dn$ &

$\textit{estimate}(a_1) + \textit{estimate}(a_2)$\\

\end{tabular}

\end{center} 

\noindent This function overestimates the stack size, for

example, in the case of \pcode{if}s. Since we cannot predict

which branch will be run, we have to allocate the maximum 

of stack each branch might take. I leave you also to think 

about whether the estimate in case of function calls is the 

best possible estimate. Note also that in case of \pcode{write}

we need to add one, because we duplicate the top-most element

in the stack.

With this all in place, we can start generating code, for

example, for the two functions:

\begin{lstlisting}[numbers=none]

def suc(x) = x + 1;

def add(x, y) = if x == 0 then y

                else suc(add(x - 1, y));

\end{lstlisting}

\noindent The successor function is a simple loading of the

argument $x$ (index $0$) onto the stack, as well as the number

$1$. Then we add both elements leaving the result of the 

addition on top of the stack. This value will be returned by 

the \pcode{suc}-function. See below:

\begin{lstlisting}[language=JVMIS]

.method public static suc(I)I 

.limit locals 1

.limit stack 2

  iload 0

  ldc 1

  iadd

  ireturn

.end method 

\end{lstlisting}

\noindent The addition function is a bit more interesting

since in the last line we have to call the function

recursively and ``wrap around'' a call to the successor

function. The code is as follows:

\begin{lstlisting}[language=JVMIS]

.method public static add(II)I 

.limit locals 2

.limit stack 5

  iload 0

  ldc 0

  if_icmpne If_else

  iload 1

  goto If_end

If_else:

  iload 0

  ldc 1

  isub

  iload 1

  invokestatic XXX/XXX/add(II)I

  invokestatic XXX/XXX/suc(I)I

If_end:

  ireturn

.end method

\end{lstlisting}

\noindent The locals limit is 2 because add takes two arguments.

The stack limit is a simple calculation using the estimate

function. We first generate code for the boolean expression

\pcode{x == 0}, that is loading the local variable 0 and the

number 0 onto the stack (Lines 4 and 5). If the not-equality

test fails, we continue with returning $y$, which is the local

variable 1 (followed by a jump to the return instruction). If

the not-equality test succeeds, then we jump to the label 

\pcode{If_else} (Line 9). After that label is the code for

\pcode{suc(add(x - 1, y))}. We first have to evaluate the

argument of the suc-function. But this means we first have to

evaluate the two arguments of the add-function. This means

loading $x$ and $1$ onto the stack and subtracting them.

Then loading $y$ onto the stack. We can then make a recursive 

call to add (its two arguments are on the stack). When this

call returns we have the result of the addition on the top of 

the stack and just need to call suc. Finally, we can return 

the result on top of the stack as the result of the 

add-function.

\end{document}

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