MP2 - Interpreter

MP2 - Interpreter
• revision: 2.0
Objectives
The objective for this MP is to write an interpreter for a language with both
expressions and statements. In particular, your job is to implement the E
(evaluate) part of the read-eval-print loop (REPL.)
Goals
• Become familiar with environments and closures
• More practice with ADTs, pattern matching, recursion, and other functional
programming concepts
• Learn how to use Hash Maps in Haskell
Useful Reading
We are using Data.HashMap.Strict for this MP and moving forward. It is a
good idea to familiarize yourself with its interface. Of particular interest for this
assignment are
empty :: HashMap k v,
fromList :: (Eq k, Hashable k) => [(k, v)] -> HashMap k v,
insert :: (Eq k, Hashable k) => k -> v -> HashMap k v -> HashMap
k v, and
lookup :: (Eq k, Hashable k) => k -> HashMap k v -> Maybe v.
Getting Started
Relevant Files
In the directory src you’ll find Lib.hs with all the relevant code. In this file
you will find all of the data definitions, the primitive function maps, the parser,
stubbed out lifting functions, and stubbed out evaluation functions. In the
directory app you will find Main.hs which contains the code for a REPL of
our language. You are only responsible for making changes to the lifting and
evaluation functions in Lib.hs.
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Running Code
To run your code, start GHCi with stack ghci. From here, you can test
individual functions, or you can run the REPL by calling main:
$ stack ghci
... More Output ...
Ok, modules loaded: Main.
*Main> main
Welcome to your interpreter!
> quit;
Bye!
("",fromList [],fromList [])
To run the REPL directly, build the executable with stack build and run it
with stack exec main:
$ stack build
mp2-interpreter-0.1.0.0: build
Preprocessing executable 'main' for mp2-interpreter-0.1.0.0...
... More Output ...
$ stack exec main
Welcome to your interpreter!
> quit;
Bye!
You can exit the REPL with the command quit;.
Testing Your Code
As in MP1, you will be able to run the test-suite with stack test:
$ stack test
HOWEVER since this MP has a substantially more complicated test spec, (look
upon it with horror if you dare) We have included two ways of running the tests.
The first is
$ stack test interpreter\:test\:friendly-test
This will run the tests (first unit tests, then property based tests on randomly
generated input) and output failures in a semi-readable manner. You should use
this test-suite while solving the MP.
The second way is
$ stack test interpreter\:test\:grader-test
This is the test ultimately run by the grader on PrairieLearn. It will fail for a
given exercise if any of the associated unit tests or property tests fail. This will
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not give very readable output in case of failure. Use this test-suite only once you
have finished the MP and you are ready to submit it. We have included these
tests so you can see exactly what gets run by the grader.
The friendly test spec can be found in test/friendly/FriendlySpec.hs The
Unit tests are enumerated in test/UnitTests.hs. The property tests will most
likely not be helpful, but they are located in test/PropertyTests.hs.
Given Code
You do not need to (and should not have to) change any of the following portions
of the given code; however, it will be helpful to understand why each part of the
code has been provided.
Data Types
The given code defines several data types for use by our interpreter. We have
a datatype that represents values calculated during evaluation, a datatype to
represent program expressions that can be evaluated, and a datatype to represent
program statements that can be executed. We also define types (actually type
synonyms) to represent environments, as well the results of statement executions.
Environments and Results
Environments are a series of mappings from identifiers (variable names, function
names, etc) to “values” - “values” here possibly being actual values resulting from
evaluating expressions, but which can be any Haskell datatype. For instance we
may have a mapping of procedure names to procedure bodies.
Here we have declared Env, the type of a value environment, which maps the
variables in scope to their current values. We also have PEnv, the type of a
procedure environment, which maps procedure names to procedure bodies, for
use when we want to call a procedure.
The Result type contains the result of executing a statement - a triple containing
the output that we wish to display from evaluating the statement, the procedure
environment at that point, and the value environment at that point.
type Env = H.HashMap String Val
type PEnv = H.HashMap String Stmt
type Result = (String, PEnv, Env)
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Values
We have a few kinds of values: IntVal and BoolVal for integers and booleans,
and CloVal to represent closures. We also have a value for exceptions which
we’ll call ExnVal.
Closures, as you may recall, have two parts: the function, and the environment
from when we created the closure. They allow us to maintain the state of the
program from when it was created. For example, if there were global variables
that existed at the time, we want to have access to the original copies of them
in case they’re referenced in the function and are possibly modified after the
creation of the closure. Here, we split up the function part of the closure into
two parts: the parameters, and the function body.
data Val = IntVal Int
| BoolVal Bool
| CloVal [String] Exp Env
| ExnVal String
deriving (Eq)
instance Show Val where
show (IntVal i) = show i
show (BoolVal i) = show i
show (CloVal xs body env) = "<" ++ show xs ++ ", "
++ show body ++ ", "
++ show env ++ ">"
show (ExnVal s) = "exn: " ++ s
We’ve also defined a Show instance for Val, so that they can be pretty-printed
by GHC and GHCi.
Expressions
Expressions are evaluated to become values. We have IntExp for integers,
BoolExp for booleans, FunExp for functions, LetExp for let expressions, AppExp
for function applications, IfExp for if expressions, IntOpExp for binary integer
operations (such as addition), BoolOpExp for binary boolean operations (such
as && and ||), CompOpExp for comparisons between integers, and VarExp for
variables.
data Exp = IntExp Int
| BoolExp Bool
| FunExp [String] Exp
| LetExp [(String,Exp)] Exp
| AppExp Exp [Exp]
| IfExp Exp Exp Exp
| IntOpExp String Exp Exp
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| BoolOpExp String Exp Exp
| CompOpExp String Exp Exp
| VarExp String
deriving (Show, Eq)
Statements
A statement is an operation intended to yield a side effect. We have SetStmt for
variable assignment, PrintStmt for printing, QuitStmt to exit the interpreter,
IfStmt for conditional statements, ProcedureStmt for procedure definitions,
CallStmt for procedure calls, and SeqStmt to sequence statements, executing
one after the other (just like the semicolon in some languages).
data Stmt = SetStmt String Exp
| PrintStmt Exp
| QuitStmt
| IfStmt Exp Stmt Stmt
| ProcedureStmt String [String] Stmt
| CallStmt String [Exp]
| SeqStmt [Stmt]
deriving (Show, Eq)
Primitive Functions
The language has a number of primitive functions, such as addition and various
comparison operators. The following map the names of those functions to Haskell
functions which we can use to do the actual computations.
intOps :: H.HashMap String (Int -> Int -> Int)
intOps = H.fromList [ ("+", (+))
, ("-", (-))
, ("*", (*))
, ("/", (div))
]
boolOps :: H.HashMap String (Bool -> Bool -> Bool)
boolOps = H.fromList [ ("and", (&&))
, ("or", (||))
]
compOps :: H.HashMap String (Int -> Int -> Bool)
compOps = H.fromList [ ("<", (<))
, (">", (>))
, ("<=", (<=))
, (">=", (>=))
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, ("/=", (/=))
, ("==", (==))
]
Parser
We have given you the entire parser this time around. The parser takes the
command you type into the REPL (a String) and converts this string into a
Stmt so that you can much more easily execute it. While it isn’t important that
you understand how the parser works right now, it may be interesting to take
a look at it to see how much you can figure out. At the very least, this parser
will be a good example for you to look at for future assignments, so keep that in
mind.
It is worth noting that this language does not have the same syntax as Haskell.
To prevent overwhelming you with the language’s full grammar, the syntax will
be shown by examples in the problems. You can also look at the parser to see
how statements and expressions are formed.
REPL
Next is the REPL (and a main function which calls the REPL with empty
environments). repl waits for a line of input, then calls the parser code on
this input to convert it into a Stmt. It then proceeds to call exec on the Stmt,
printing any result and updating the environments. This loops until you input
quit;.
Problems
Your task is to implement the rest of the interpreter; in particular, the lifting
functions, the expression evaluator, and the statement executor.
Lifting Functions
Since we are not directly interacting with Haskell’s primitive values (but with
values of type Val which represent our custom language’s set of primitive values),
we cannot directly apply Haskell’s builtin functions either. For example, we
cannot directly evaluate IntVal 5 + IntVal 8. However we do want a function
can add two IntVals and get back an IntVal. To do this, we could write a
function like:
intValAdd :: Val -> Val -> Val
intValAdd (IntVal x) (IntVal y) = IntVal (x + y)
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This unpacks the two parameter IntVals using pattern matching, adds them
using the builtin +, and then puts the sum back into an IntVal. However, we
have a lot of builtin primitive functions for our language, and writing a function
for each of them would be tedious, and it would be more difficult to add primitive
functions to the language.
Thus, we have the lifting functions which “lift” a builtin Haskell function to act
on Vals instead. There are three lifting function for the relevant function types:
liftIntOp lifts binary integer functions like +, liftBoolOp lifts binary boolean
functions like &&, and liftCompOp lifts comparison functions like <=. Note that
even though Haskell can compare booleans, we can only compare integers in our
language. If the types of the values passed into a lifting function are incorrect
you should return the exception ExnVal "Cannot lift".
liftIntOp is written for you. You must write liftBoolOp and liftCompOp.
Note that the output appears as if it were not a Val type because we provided a
Show instance for Val.
*Main> let intValAdd = liftIntOp (+)
*Main> intValAdd (IntVal 5) (IntVal 4)
9
*Main> liftIntOp mod (IntVal 13) (IntVal 5)
3
*Main> liftBoolOp (&&) (BoolVal True) (BoolVal False)
False
*Main> liftBoolOp (||) (IntVal 1) (IntVal 0)
exn: Cannot lift
*Main> liftCompOp (<=) (IntVal 5) (IntVal 4)
False
*Main> liftCompOp (==) (BoolVal True) (BoolVal True)
exn: Cannot lift
Eval
The eval :: Exp -> Env -> Val function takes an expression and the current
environment (the values stored in the variables in scope), and evaluates the
expression given that environment to get a value. You will need to implement
the eval function for each possible type of expression. The rules describing how
to do so, the semantics, are in semantics.pdf. You can test the function by
calling eval directly, or by calling print on an expression in the REPL.
Constants
Before we can do any evaluation, we’ll need to define some basic expressions in
Exp for eval. In particular, modify eval to handle both IntExps and BoolExps.
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*Main> eval (IntExp 5) H.empty
5
*Main> eval (BoolExp True) H.empty
True
Welcome to your interpreter!
> print 5;
5
> print true;
True
> quit;
Bye!
Variables
Modify eval to handle VarExps. (Notice that we have no way to add variables
to the environment in the REPL yet, but we can call repl directly with a
non-empty environment.)
*Main> let env = H.fromList [("x", IntVal 3), ("y", IntVal 5)]
*Main> eval (VarExp "x") H.empty
exn: No match in env
*Main> eval (VarExp "x") env
3
*Main> eval (VarExp "y") env
5
*Main> repl H.empty env [] ""
> print x;
3
> print x + y;
8
> print z;
exn: No match in env
Arithmetic
Modify eval to handle IntOpExp so that we can evaluate arithmetic expressions.
Note that division must be handled specially to throw an exception in the case
of a division by zero. liftIntOp will come in handy.
Note that for this (and the following problem) if we are using the REPL, the
String representing the operator we want to apply will always be a valid operator
that can be found in one of the primitive function maps - otherwise the expression
would not have made it past the parser. Thus it is okay (for this assignment) to
forgo handling a failed lookup for the operator. You’ll notice there are semantic
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rules for this though, so you can handle failed lookup if you want to (it won’t be
tested).
*Main> eval (IntOpExp "+" (IntExp 5) (IntExp 4)) H.empty
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*Main> eval (IntOpExp "-" (IntOpExp "*" (IntExp 3) (IntExp 10)) (IntExp 7)) H.empty
23
*Main> eval (IntOpExp "/" (IntExp 6) (IntExp 2)) H.empty
3
*Main> eval (IntOpExp "/" (IntExp 6) (IntExp 0)) H.empty
exn: Division by 0
*Main> eval (IntOpExp "+" (IntExp 6) (IntOpExp "/" (IntExp 4) (IntExp 0))) H.empty
exn: Cannot lift
Welcome to your interpreter!
> print 5 + 4;
9
> print (3 * 10) - 7;
23
> print 6 / 0;
exn: Division by 0
Boolean and Comparison Operators
Modify eval to handle BoolOpExp and CompOpExp.
*Main> eval (BoolOpExp "and" (BoolExp True) (BoolExp False)) H.empty
False
*Main> eval (CompOpExp "/=" (IntExp 4) (IntExp 6)) H.empty
True
Welcome to your interpreter!
> print true and true;
True
> print 3 < 4;
True
> print ((3 * 5) >= (20 - 6)) or false;
True
If Expressions
Modify eval to handle IfExp.
*Main> eval (IfExp (BoolExp True) (IntExp 5) (IntExp 10)) H.empty
5
*Main> eval (IfExp (BoolExp False) (IntExp 5) (IntExp 10)) H.empty
10
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*Main> eval (IfExp (IntExp 1) (IntExp 5) (IntExp 10)) H.empty
exn: Condition is not a Bool
Welcome to your interpreter!
> print if true then 5 else 10 fi;
5
> print if false then 5 else 10 fi;
10
> print if false then 5 / 0 else 5 / 1 fi;
5
> print if 1 then 5 else 10 fi;
exn: Condition is not a Bool
Functions and Function Application
Modify eval to handle FunExp, allowing us to create functions. Note that
this creates a closure, which encapsulates both the function definition and the
environment at the time of creation. Then, modify eval to handle AppExp. This,
in conjunction with FunExp, will allow us to do function application.
Note: You can assume for the purposes of this assignment that the number of
arguments passed when a function is applied the same as the number that it
needs.
*Main> let env = H.fromList [("x", IntVal 3)]
*Main> let fun1 = FunExp ["a", "b"] (IntOpExp "+" (VarExp "a") (VarExp "b"))
*Main> eval fun1 H.empty
<["a","b"], IntOpExp "+" (VarExp "a") (VarExp "b"), fromList []>
*Main> let fun2 = FunExp ["k"] (IntOpExp "*" (VarExp "k") (VarExp "x"))
*Main> eval fun2 env
<["k"], IntOpExp "*" (VarExp "k") (VarExp "x"), fromList [("x",3)]>
*Main> eval (AppExp fun1 [IntExp 5, IntExp 4]) H.empty
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*Main> eval (AppExp fun2 [IntExp 5]) env
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*Main> let envf = H.fromList [("f", eval fun1 H.empty), ("g", eval fun2 env)]
*Main> eval (AppExp (VarExp "g") [IntExp 4]) envf
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*Main> eval (AppExp (IntExp 5) []) H.empty
exn: Apply to non-closure
Welcome to your interpreter!
> print fn [x] 2 * x end;
<["x"], IntOpExp "*" (IntExp 2) (VarExp "x"), fromList []>
> print apply fn [x] 2 * x end (4);
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> print fn [a, b] if a <= b then a else b fi end;
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<["a","b"], IfExp (CompOpExp "<=" (VarExp "a") (VarExp "b")) (VarExp "a") (VarExp "b"), from> print apply fn [a, b] if a <= b then a else b fi end (5, 7);
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> print apply fn [a, b] if a <= b then a else b fi end (7, 5);
5
> print apply 5 ();
exn: Apply to non-closure
Let Expressions
Modify eval to handle LetExp.
*Main> eval (LetExp [] (IntExp 5)) H.empty
5
*Main> eval (LetExp [("x", IntExp 5)] (VarExp "x")) H.empty
5
*Main> eval (LetExp [("x", IntOpExp "+" (IntExp 5) (IntExp 4))] (IntOpExp "*" (VarExp "x") (18
Welcome to your interpreter!
> print let [] 5 end;
5
> print let [x := 5] x end;
5
> print let [x := 5 + 4] x * 2 end;
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> print let [x := 5] let [x := 6] x end end ;
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> print let [x := 5] let [x := 6; y := x] y end end;
5
> print let [f := fn [a, b] a + b end] apply f(5,4) end;
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> print let [x := 3] let [g := fn [k] k * x end] apply g(5) end end;
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Statements
The exec :: Stmt -> PEnv -> Env -> Result function takes a statement and
the current procedure and variable environments, and executes that statement in
those environments to get its result. The result of executing a statement consists
of an output string (possibly empty) and the possibly updated procedure and
variable environments. You will need to handle the various kinds of statements.
The rules describing the semantics are in semantics.pdf. You can test the
function by calling exec directly, or by inputting the statements into the REPL.
PrintStmt is done for you, and QuitStmt is already handled in repl.
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Set Statements
Modify exec to handle SetStmtss. The output string should be empty.
*Main> exec (SetStmt "x" (IntExp 5)) H.empty H.empty
("",fromList [],fromList [("x",5)])
*Main> exec (SetStmt "y" (VarExp "x")) H.empty (H.fromList [("x", IntVal 5)])
("",fromList [],fromList [("x",5),("y",5)])
Welcome to your interpreter!
> x := 5;
> print x;
5
> do x := 7 ; print x; od;
7
> print x;
7
> do f := fn [a, b] a + b end; print apply f(5, 4); od;
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Sequencing
Modify exec to handle SeqStmts. The output string should be the concatenation
of all output strings of each individual statement.
*Main> exec (SeqStmt [PrintStmt (IntExp 5)]) H.empty H.empty
("5",fromList [],fromList [])
*Main> exec (SeqStmt [PrintStmt (IntExp 4), PrintStmt (IntExp 2)]) H.empty H.empty
("42",fromList [],fromList [])
Welcome to your interpreter!
> do print 6; od;
6
> do print 4; print 2; od;
42
> do print true + 5; print 7; print 7; od;
exn: Cannot lift77
If Statements
Modify exec to handle IfStmts. The output string depends on which statement
is executed, and may also possibly be an error message.
*Main> exec (IfStmt (BoolExp True) (PrintStmt (IntExp 5)) (PrintStmt (IntExp 10))) H.empty H("5",fromList [],fromList [])
*Main> exec (IfStmt (BoolExp False) (PrintStmt (IntExp 5)) (PrintStmt (IntExp 10))) H.empty12
("10",fromList [],fromList [])
*Main> exec (IfStmt (IntExp 1) (PrintStmt (IntExp 5)) (PrintStmt (IntExp 10))) H.empty H.emp("exn: Condition is not a Bool",fromList [],fromList [])
Welcome to your interpreter!
> if true then print 5; else print 10; fi
5
> if false then print 5; else print 10; fi
10
> if 4 < 3 then print true + 5; else do print 4; print 21; od; fi
421
> if 1 then print 5; else print 10; fi
exn: Condition is not a Bool
Procedure and Call Statements
Modify exec to handle ProcedureStmts and CallStmts. Procedures allows us
to repeat code (much like a function would), but without the restoration of the
old environment.
Note: You can assume for the purposes of this assignment that the number of
arguments passed into a procedure is the same as the number that it needs.
Welcome to your interpreter!
> procedure p() print 5; endproc
> call p();
5
> call q();
Procedure q undefined
> do procedure f(a, b) print a + b; endproc call f(5, 4); od;
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> do procedure s(v) x:= v; endproc call s(10); print x; od;
10
> do procedure e(x) print true; endproc call e(23); print x; od;
True23
> do y := 0; procedure c() if y < 10 then do print y; y := y + 1; call c(); od; else print t0123456789True
> do procedure fog(f, g, x) x := apply f(apply g(x)); endproc call fog(fn [x] x * 2 end, fn14
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