Lambda Calculus

This is a simple programming language derived from lambda calculus.

Usage

Run the program using stack run (or run the tests with stack test).

Type in your expression at the prompt: >> . The expression will be evaluated to normal form using the call-by-value evaluation strategy and then printed. Exit the prompt with Ctrl-c (or equivalent).

Syntax

The parser's error messages currently are virtually useless, so be very careful with your syntax.

• Variable names: any sequence of letters.
• Function application: f x y
• Lambda abstraction: \x y z. E or λx y z. E
• Let expressions: let x = E; y = F in G
• Parenthetical expressions: (E)
• Constructors: (), (x, y) (or (,) x y), Left x, Right y, Z, S, [], (x : xs) (or (:) x xs), Char n.
  • The parentheses around the cons constructor are not optional.
  • Char takes a natural number and turns it into a character.
• Pattern matchers: { Left x -> e ; Right y -> f }
  • Pattern matchers can be applied like functions, e.g. { Z -> x, S -> y } 10 reduces to y.
  • Patterns must use the regular form of the constructor, e.g. (x : xs) and not ((:) x xs).
  • There are no nested patterns or default patterns.
  • Incomplete pattern matches will crash the interpreter.
• Literals: 1234, [e, f, g, h], 'a, "abc"
  • Strings are represented as lists of characters.

Call/CC

This interpreter has preliminary support for the call-with-current-continuation control flow operator. However, it has not been thoroughly tested.

To use it, simply apply the variable callcc like you would a function, e.g. (callcc (\k. ...)).

Continuations are printed as λ!. ... ! ..., like a lambda abstraction with an argument named ! which is used exactly once; however, continuations are not the same as lambda abstractions because they perform the side effect of modifying the current continuation, and this is not valid syntax you can input into the REPL.

Example code

The fixpoint function:

(\x. x x) \fix f x. f (fix fix f) x

Create a list by iterating f n times:

fix \iterate f x. { Z -> x ; S n -> iterate f (f x) n }

Create a list whose first element is n - 1, counting down to a last element of 0:

\n. { (n, x) -> x } (iterate { (n, x) -> (S n, (n : x)) } (0, []) n)

Putting it all together to count down from 10:

>> let fix = (\x. x x) \fix f x. f (fix fix f) x; iterate = fix \iterate f x. { Z -> x ; S n -> iterate f (f x) n }; countDownFrom = \n. { (n, x) -> x } (iterate { (n, x) -> (S n, (n : x)) } (0, []) n) in countDownFrom 10
[9, 8, 7, 6, 5, 4, 3, 2, 1, 0]

Append two lists together:

fix \append xs ys. { [] -> ys ; (x : xs) -> (x : append xs ys) } xs

Reverse a list:

fix \reverse. { [] -> [] ; (x : xs) -> append (reverse xs) [x] }

Putting them together so we can reverse "reverse":

>> let fix = (\x. x x) \fix f x. f (fix fix f) x; append = fix \append xs ys. { [] -> ys ; (x : xs) -> (x : append xs ys) } xs; reverse = fix \reverse. { [] -> [] ; (x : xs) -> append (reverse xs) [x] } in reverse "reverse"
"esrever"

Calculating 3 + 2 with the help of Church-encoded numerals:

>> let Sf = \n f x. f (n f x); plus = \x. x Sf in plus (\f x. f (f (f x))) (\f x. f (f x)) S Z
5

This expression would loop forever, but callcc saves the day!

>> y (callcc \k. (\x. (\x. x x) (\x. x x)) (k z))
y z

A few other weird expressions:

>> let D = \x. x x; F = \f. f (f y) in D (F \x. x)
y y
>> let T = \f x. f (f x) in (\f x. T (T (T (T T))) f x) (\x. x) y
y
>> (\x y z. x y) y
λy' z. y y'
>> { Char c -> Char (S c) } 'a
'b