Interpreting functional languages

The lambda-calculus

At the core of functional languages is the λ-calculus. In pure λ-calculus, everything is a function.

Expressions e of pure λ-calculus are given by this grammar:

    e,f ::= x        -- Variable
          | f e      -- Application of function f to argument e
          | λx → e   -- Function: abstraction of x in e

This is a subset of the Haskell expression syntax. The abstract syntax corresponds to this LBNF grammar:

    EId.   Exp3 ::= Ident;
    EApp.  Exp2 ::= Exp2 Exp3;
    EAbs.  Exp  ::= "λ" Ident "→" Exp;

    coercions Exp 3;

Example: x y z should be read (x y) z.

A small extension of the lambda-calculus

We consider an extension by let, numerals, and primitive operators:

     ...  | let x = e in f
          | let x₁ = e₁; ...; xₙ = eₙ in f
          | n                   -- E.g. 0,1,2..
          | e₁ op e₂            -- op could be +,-,...

     ELet.  Exp  ::= "let" [Bind] "in" Exp;
     EInt.  Exp3 ::= Integer;
     EOp.   Exp1 ::= Exp1 Op Exp2;

     Bind.  Bind ::=  Ident "=" Exp;
     separator Bind ";";

let x = e in f could be regarded just syntactic sugar for (λ x → f) e, and let x₁ = e₁; ...; xₙ = eₙ in f as sugar for (λ x₁ → ... λ xₙ → f) e₁ ... eₙ.

But let is conceptually important.

Convenient syntactic sugar:

• multi-λ: λ x₁ ... xₙ → e sugar for λ x₁ → .... → λ xₙ → e
• let with arguments: let f x₁ ... xₙ = e in ... for let f = λ x₁ ... xₙ → e in ...

Example:

    let
      double x     = x + x
      comp   f g x = f (g x)
    in let
      twice  f     = comp f f
    in
        twice twice double 2

Example:

    never  f = λ x → x
    once   f = λ x → f x
    twice  f = λ x → f (f x)
    thrice f = λ x → f (f (f x))

Functional languages vs. imperative languages

In λ-calculus and functional languages, functions are first class. They can be:

1. called (naturally, also in imperative languages)
2. passed as arguments (in some imperative languages)
3. returned as results (impossible in imperative languages)

E.g. 2. in C:

int comp  (int f(int n), int g(int n), int x) {
  return f(g(x));
}

Trying 3. let twice f = comp f f in twice double in C:

• Attempt 1: illegal type

  int twice (int f(int n)) (int x) {
    return comp(f,f,x);
  }
  ... twice(double) ...

• Attempt 2: Cannot partially apply

  int twice (int f(int n), int x) {
    return comp(f,f,x);
  }
  ... twice(double,??) ...

The essential feature of functional languages is not anonymous functions (via λ) but partial application, forming a new function by giving less arguments than the function arity.

Example: The increment function:

    let plus x y = x + y in plus 1

Can be written as anonymous function λ y → 1 + y.

Free and bound variables

In λ x → x y, variable y is considered free while x is bound by the λ. In let x = y in x, y is free and x is bound by the let. Both let and λ are binders.

Free variable computation:

    FV(x)                        = {x}
    FV(f e)                      = FV(f) ∪ FV(e)
    FV(λ x → e)                  = FV(e) \ {x}
    FV(let x₁=e₁;...;xₙ=eₙ in f) = FV(e₁,...,eₙ) ∪ (FV(f) \ {x₁,...,xₙ})

An expression without free variables is called closed, otherwise open. For closed expression, we are interested in their value.

E.g.:

• let ... in double 2 has value 4
• let ... in twice double 2 has value 8
• let ... in twice twice double 2 has value ??
• let ... in twice double has value λ x → (x + x) + (x + x)

A value v (in our language) can be a numeral (int value) or a λ (function value).

OBS! Some closed expressions do not have a value, e.g.

(λ x → x x) (λ x → x x)

How to compute the value of an expression?

Reduction

Idea: compute the value by substituting function arguments for function parameters, and evaluating operations.

    (λ x → 1 + x) 2
    ↦  1 + 2
    ↦  3

    let x = 2 in 1 + x
    ↦  1 + 2
    ↦  3

This is formally a small-step semantics e ↦ e', called reduction.

Reduction rules:

    (λ x → f) e               ↦  f[x=e]    -- named β by Alonzo Church
    let x = e in f            ↦  f[x=e]
    let x₁=e₁;...;xₙ=eₙ in f  ↦  f[x₁=e₁;...;xₙ=eₙ]

The lhs (left hand side) is called a redex (reducible expression) and the rhs (right hand side) its reduct.

Strategy question: where and under which conditions can these rules be applied?

1. Full reduction: anywhere and unconditional
2. Leftmost-outermost reduction: Reduce the redex which is closest to the root of the expression tree, and if several exist, the leftmost of these.
3. Leftmost-innermost: Reduce the leftmost redex that does not contain any redexes in subexpressions.
4. Call-by-name: same as 2. but never reduce under λ.
5. Call-by-value: same as 3. but never reduce under λ, and thus do not consider anything under a λ as a redex.

Substitution

Substitution f[x=e] of course does not substitute bound occurrences of x in f:

      (λ x → (λ x → x)) 1
    ↦ (λ x → x)[x=1]
    = (λ x → x)

Still substitution has some pitfalls related to shadowing:

Example:

• Reducing the let first:

      let x = 1 in (λ f x → f x) (λ y → x)
    ↦ ((λ f x → f x) (λ y → x))[x=1]
    = (λ f x → f x) (λ y → 1)
    ↦ (λ x → f x)[f = λ y → 1]
    = λ x → (λ y → 1) x
    ↦ λ x → 1[y=x]
    = λ x → 1

• Reducing the λ first:

      let x = 1 in (λ f x → f x) (λ y → x)
    ↦ let x = 1 in (λ x → f x)[f = λ y → x]
    = let x = 1 in λ x → (λ y → x) x
    ↦ let x = 1 in λ x → x[y=x]
    = let x = 1 in λ x → x
    ↦ (λ x → x)[x=1]
    = λ x → x

What goes wrong here?

Variable capture problem in (λ x → f x)[f = λ y → x]: Naive substituting under a binder can produce meaningless results. Here, the free variable x is captured by the binder λ x.

Solutions:

1. Never substitute open expressions under a binder.

   For evaluation of closed expressions, we can use strategies that respect this imperative. E.g., call-by-value, call-by-name.

2. Rename bound variables to avoid capture.

       (λ x → f x)[f = λ y → x]
     = (λ z → f z)[f = λ y → x]
     = (λ z → (λ y → x) z)
     ↦ λ z → x

Consistently renaming bound variables does not change the meaning of an expression.

Nontermination

There are terms that reduce to themselves!

      (λ x → x x) (λ x → x x)
    ↦ (x x)[x = (λ x → x x)]
    = (λ x → x x) (λ x → x x)

Such terms do not have a value.

Big-step semantics

We use the notation let γ in e where γ is a list of bindings xᵢ = eᵢ. This list may be called environment.

Just like for C--, we can give a big step semantics γ ⊢ e ⇓ v for the lambda-calculus. In terms of reduction this should mean:

(let γ in e) ↦* v

Call-by-value

The big-step semantics corresponding to the call-by-value strategy uses an environment of closed values, i.e., γ is of the form x₁=v₁,...,xₙ=vₙ.

A value v is a closed expression which is

• a numeral (integer literal) n
• or a function with an environment let δ in λ x → f.

The latter is called a closure and may be written ⟨λx→f; δ⟩ or (λx→f){δ} (IPL book).

Evaluation rules:

Variable:

     ------------
     γ ⊢ x ⇓ γ(x)

Let:

     γ      ⊢ e₁ ⇓ v₁
     γ,x=v₁ ⊢ e₂ ⇓ v₂
     -------------------------
     γ ⊢ let x = e₁ in e₂ ⇓ v₂

Lambda:

     --------------------------------
     γ ⊢ (λx → f) ⇓ (let γ in λx → f)

Application:

     γ      ⊢ e₁ ⇓ let δ in λx → f
     γ      ⊢ e₂ ⇓ v₂
     δ,x=v₂ ⊢ f ⇓ v
     -----------------------------
     γ ⊢ e₁ e₂ ⇓ v

Exercise: Evaluate (((λ x₁ → λ x₂ → λ x₃ → x₁ + x₃) 1) 2) 3

Rules for integer expressions:

     ---------
     γ ⊢ n ⇓ n

     γ ⊢ e₁ ⇓ n₁
     γ ⊢ e₂ ⇓ n₂
     ---------------- n = n₁ + n₂
     γ ⊢ e₁ + e₂ ⇓ n

Call-by-name

Call by name differs from call-by-value by not evaluating arguments when calling functions, but to form a closure. An environment entry c is now itself a closure let δ in e where environment δ is of the form x₁=c₁,...,xₙ=cₙ.

The evaluation judgement is still of the form γ ⊢ e ⇓ v.

Evaluation rules:

Variable:

     δ ⊢ e ⇓ v
     ---------- γ(x) = let δ in e
     γ ⊢ x ⇓ v

Let:

     γ,x=c ⊢ e₂ ⇓ v₂
     ------------------------- c = let γ in e₁
     γ ⊢ let x = e₁ in e₂ ⇓ v₂

Lambda:

     --------------------------------
     γ ⊢ (λx → f) ⇓ (let γ in λx → f)

Application:

     γ     ⊢ e₁ ⇓ let δ in λx → f
     δ,x=c ⊢ f ⇓ v
     ----------------------------- c = let γ in e₂
     γ     ⊢ e₁ e₂ ⇓ v

Rules for integer expressions: (unchanged).

Comparing cbn (call-by-name) with cbv (call-by-value):

• Arguments are only evaluated when needed. This is an advantage if an argument is unused or only used under rare conditions. E.g.: e₁ && e₂ = if e₁ then e₂ else false

• Arguments are every time evaluated when needed. This is a disadvantage if an argument is used twice or more. E.g.: double x = x + x

call-by-need

Synthesis of call-by-name and call-by-value: call-by-need (Haskell)

• lazy: only evaluate when needed, but then store value
• next time the value is needed, grab stored value
• needs global heap rather than local environment

Call-by-need example:

    twice twice double 2

    double = (\ x -> add x x)
    comp   = (\ f -> \ g -> \ x -> f (g x))
    twice  = (\ f -> comp f f)

Step visualization with https://github.com/well-typed/visualize-cbn:

Literature to study call-by-need:

• big-step semantics:

  John Launchbury: A Natural Semantics for Lazy Evaluation. POPL 1993: 144-154

• small-step semantics:

  Peter Sestoft: Deriving a Lazy Abstract Machine. J. Funct. Program. 7(3): 231-264 (1997)