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some stuff

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NikolajDanger
2022-04-30 09:46:03 +02:00
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(* ABSTRACT SYNTAX *)
type VALUE = INT of int
type BINOP = BPLUS | BMINUS | BTIMES
type RANGEOP = RSUM | RPROD | RMAX | RARGMAX
type EXP =
| CONSTANT of VALUE
| VARIABLE of string
| OPERATE of BINOP * EXP * EXP
| LET_IN of string * EXP * EXP
| OVER of RANGEOP * string * EXP * EXP * EXP
(* A list mapping variable names to their values. *)
type SymTab = (string * VALUE) list
(* Inserts a variable into the variable table. *)
let bind v k vtab = (v, k) :: vtab : SymTab
(* Lookup a the value of a variable. *)
let rec lookup v = function
| (v', k) :: vtab -> if v = v' then k else lookup v vtab
| (_ : SymTab) -> failwith ("unbound variable : " + v)
(* HELP FUNCTIONS *)
let VALUEtoint = function INT x -> x
(* EVALUATION *)
(*****************************************************
* TODO: Task 2: complete the definition of `eval`. You may also
* define any auxiliary functions as you see fit, to help express `eval`
* as cleanly and concisely as possible.
****************************************************)
(* The evaluation function. *)
let rec eval (vtab : SymTab) (e : EXP) : VALUE =
match e with
| CONSTANT n -> n
| VARIABLE v ->
lookup v vtab
| OPERATE (op, e1, e2) ->
let exp1 = VALUEtoint (eval vtab e1)
let exp2 = VALUEtoint (eval vtab e2)
match op with
| BPLUS -> INT (exp1 + exp2)
| BMINUS -> INT (exp1 - exp2)
| BTIMES -> INT (exp1 * exp2)
| LET_IN (var, e1, e2) ->
let exp1 = eval vtab e1
let vtab = bind var exp1 vtab
eval vtab e2
| OVER (rop, var, e1, e2, e3) ->
let exp1 = VALUEtoint (eval vtab e1)
let exp2 = VALUEtoint (eval vtab e2)
let results = [
for i in exp1..exp2 ->
let vtab = bind var (INT i) vtab
eval vtab e3
]
match rop with
| RSUM -> List.fold (fun (INT acc) (INT elem) -> INT (acc + elem)) (INT 0) results
| RPROD -> List.fold (fun (INT acc) (INT elem) -> INT (acc * elem)) (INT 1) results
| RMAX -> List.fold (fun (INT acc) (INT elem) -> INT (max acc elem)) results[0] results
| RARGMAX ->
let max_elem = List.fold (fun (INT acc) (INT elem) -> INT (max acc elem)) results[0] results
INT ((List.findIndex ((=) max_elem) results) + exp1)
(* YOU SHOULDN'T NEED TO MODIFY ANYTHING IN THE REMAINDER OF THIS FILE,
BUT YOU ARE WELCOME TO LOOK AT IT. *)
(* LEXER
**********
* A lexer, is a program which reads a stream/list of chars,
* and transforms them into a list of tokens.
**********
* This example code is very simple and naive, but it should not
* introduce anything which you have not seen on the PoP course.
* Furhter more, we hope that you will read it, and get an intuition
* about what is going on
**********
*)
type TOKEN =
| NUM of int | ID of string
| PLUS | MULT | MINUS | EQ
| LPAR | RPAR | LET | IN
| SUM | PROD | MAX | ARGMAX | OF | TO
let explode (s : string) = [for c in s -> c]
let isA t c = List.exists (fun c' -> c = c') t
let digit = ['0'..'9']
let letter = ['a'..'z']
let symbol = explode "+-*/=()"
let whitespace = explode " \t\n"
let keyword = ["let"; "in"; "sum"; "prod"; "max"; "argmax"; "of"; "to"]
let rec lex = function
| (c :: _) as cs when c |> isA digit -> lexNum "" cs
| (c :: _) as cs when c |> isA letter -> lexWord "" cs
| (c :: cs) when c |> isA symbol -> lexSymbol c :: lex cs
| (c :: cs) when c |> isA whitespace -> lex cs
| (c :: cs) -> failwith ("invalid character : " + string c)
| (_ : char list) -> ([] : TOKEN list)
and lexNum s = function
| (c :: cs) when c |> isA digit -> lexNum (s + string c) cs
| cs -> (NUM (System.Int32.Parse s)) :: lex cs
and lexWord s = function
| (c :: cs) when c |> isA (letter @ digit) -> lexWord (s + string c) cs
| cs ->
(if s |> isA keyword then getKeyWord s else ID s) :: lex cs
and lexSymbol = function
| '+' -> PLUS
| '-' -> MINUS
| '*' -> MULT
| '=' -> EQ
| '(' -> LPAR
| ')' -> RPAR
| sym -> failwith ("Unknown Symbol : " + string sym)
and getKeyWord = function
| "let" -> LET
| "in" -> IN
| "sum" -> SUM
| "prod" -> PROD
| "max" -> MAX
| "argmax" -> ARGMAX
| "to" -> TO
| "of" -> OF
| word -> failwith ("Invalid Keyword : " + word)
(* PARSER
**********
* A parser transforms the list of tokens to an abstract syntax tree,
* which reprecents the syntax of an expression.
**********
* The example code uses a somewhat ad hoc approach.
* Later on in the course, we will see more systematic ways of
* constructing parsers.
*)
(* Returns tail of ts, assuming head of ts equals t *)
let expect t ts =
match ts with
| t' :: ts1 when t' = t -> ts1
| _ -> failwith "parse error"
(* Each parser function returns the expression parsed from the beginning
of the token list, together with any remaining tokens. *)
let rec parse_exp ts =
match ts with
| LET :: ID v :: EQ :: ts1 ->
let (e1, ts2) = parse_exp ts1
let ts3 = expect IN ts2
let (e2, ts4) = parse_exp ts3
(LET_IN (v,e1,e2), ts4)
| r :: ID v :: EQ :: ts1 when r |> isA [SUM; PROD; MAX; ARGMAX] ->
let (e1, ts2) = parse_exp ts1
let ts3 = expect TO ts2
let (e2, ts4) = parse_exp ts3
let ts5 = expect OF ts4
let (e3, ts6) = parse_exp ts5
let rop = match r with
| SUM -> RSUM
| PROD -> RPROD
| MAX -> RMAX
| ARGMAX -> RARGMAX
| _ -> failwith ("Unknown range operation : " + string r)
(OVER (rop, v, e1, e2, e3), ts6)
| _ -> parse_arith ts
and parse_arith ts =
let (e1, ts1) = parse_term ts
parse_terms e1 ts1
and parse_terms e1 ts =
match ts with
| PLUS :: ts1 -> let (e2, ts2) = parse_term ts1
parse_terms (OPERATE (BPLUS, e1,e2)) ts2
| MINUS :: ts1 -> let (e2, ts2) = parse_term ts1
parse_terms (OPERATE (BMINUS, e1,e2)) ts2
| _ -> (e1, ts)
and parse_term ts =
let (e1, ts1) = parse_factor ts
parse_factors e1 ts1
and parse_factors e1 ts =
match ts with
| MULT :: ts1 -> let (e2, ts2) = parse_factor ts1
parse_factors (OPERATE (BTIMES, e1,e2)) ts2
| _ -> (e1, ts)
and parse_factor ts =
match ts with
| ID v :: ts1 -> (VARIABLE v, ts1)
| NUM n :: ts1 -> (CONSTANT (INT n), ts1)
| LPAR :: ts1 ->
let (e, ts2) = parse_exp ts1
let ts3 = expect RPAR ts2
(e, ts3)
| _ -> failwith ("parse error")
let parse ts =
let (e, ts1) = parse_exp ts
if ts1 = [] then e else failwith "parse error"
(*************
* When interpreting code, we lex it into tokens, parse the tokens into an
* abstract syntax tree, and evaluate the tree into some value.
* The definition of what it means to be a value, and an expression (tree),
* can be found in the parser.
*************
* In this code, the evaluation of such a tree has been reprecented, should
* be implemented in the below function [eval] {~_^}
*)
(* FOR RUNNING A PROGRAM *)
let run program = (lex >> parse >> eval []) (explode program)
(* Tests *)
let eval_test_constant = (eval [] (CONSTANT (INT 4)) = INT 4)
//let eval_test_lookup = (eval [("x", INT 4)] (VARIABLE "x") = INT 4)
let program0 = "1 - 2 - 3"
let program1 = "let x = 4 in x + 3"
let program2 = ("let x0 = 2 in \
let x1 = x0 * x0 in \
let x2 = x1 * x1 in x2 * x2")
let program3 = "sum x = 1 to 4 of x*x"
let program4 = "max x = 0 to 10 of 5 * x - x * x"
let program5 = "argmax x = 0 to 10 of 5 * x - x * x"
(*
let eval_test_arithmetic = (run program0 = INT -4)
let eval_test_let = (run program1 = INT 7)
let eval_test_nested_let = (run program2 = INT 30)
let eval_test_max = (run program4 = INT 6)
let eval_test_argmax = (run program4 = INT 2)
*)
(* THIS FUNCTION CALLS RUN ON USER INPUT IN AN INTERACTIVE LOOP *)
let interpreter () =
let mutable running = true
printfn "Welcome to the calculator! Type \"exit\" to stop."
while running do
printf "Input an expression : "
let program = System.Console.ReadLine()
try
if program = "" then ()
else if program = "exit" then running <- false
else printfn "Evaluation result : %A" (run program)
with
| Failure (msg) -> printfn "%s" msg
interpreter ()

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\relax
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\documentclass[a4paper]{article}
\renewcommand{\thesection}{Task \arabic{section}}
\begin{document}
\section{}
\begin{enumerate}
\item I am a DIKU Computer Science BSc, general profile
\item I have little to no proficiency with F\#
\item I have mininal knowledge of assembly programming
\item
\end{enumerate}
\section{}
\section{}
\end{document}