# Type Inference * * * <i> Topics: * Type inference and reconstruction * Constraint collection * Constraint solving (unification) </i> * * * Java and OCaml are *statically typed* languages, meaning every binding has a type that is determined at *compile time*&mdash;that is, before any part of the program is executed. The type-checker is a compile-time procedure that either accepts or rejects a program. By contrast, JavaScript and Ruby are dynamically-typed languages; the type of a binding is not determined ahead of time and computations like binding 42 to x and then treating x as a string result in run-time errors. Unlike Java, OCaml is *implicitly typed*, meaning programmers rarely need to write down the types of bindings. This is often convenient, especially with higher-order functions. (Although some people disagree as to whether it makes code easier or harder to read). But implicit typing in no way changes the fact that OCaml is statically typed. Rather, the type-checker has to be more sophisticated because it must infer what the *type annotations* "would have been" had the programmers written all of them. In principle, type inference and type checking could be separate procedures (the inferencer could figure out the types then the checker could determine whether the program is well-typed), but in practice they are often merged into a single procedure called *type reconstruction*. ## OCaml type reconstruction OCaml was rather cleverly designed so that type reconstruction is a straightforward algorithm. At a very high level, that algorithm works as follows: - Determine the types of definitions in order, using the types of earlier definitions to infer the types of later ones. (Which is one reason you may not use a name before it is bound in an OCaml program.) - For each let definition, analyze the definition to determine *constraints* about its type. For example, if the inferencer sees x+1, it concludes that x must have type int. It gathers similar constraints for function applications, pattern matches, etc. Think of these constraints as a system of equations like you might have in algebra. - Use that system of equations to solve for the type of the name begin defined. The OCaml type reconstruction algorithm attempts to never reject a program that could type-check, if the programmer had written down types. It also attempts never to accept a program that cannot possibly type check. Some more obscure parts of the language can sometimes make type annotations either necessary or at least helpful (see RWO chapter 22, "Type inference", for examples). But for most code you write, type annotations really are completely optional. Since it would be verbose to keep writing "the OCaml type reconstruction algorithm," we'll call the algorithm HM. That name is used throughout the programming languages literature, because the algorithm was independently invented by Roger <u>H</u>indley and Robin <u>M</u>ilner. ## Collecting and solving constraints: Examples To gather the constraints for a definition, HM does the following: - Assign a preliminary type to every subexpression in the definition. For known operations and constants, such as + and 3, use the type that is already known for it. For anything else, use a new type variable that hasn't been used anywhere else. - Use the "shape" of the expressions to generate constraints. For example, if an expression involves applying a function to an argument, then generate a constraint requiring the type of the argument to be the same as the function's input type. We'll give some examples of this first, then we'll give the algorithms for doing it. #### Example 1. Here's an example utop interaction:  # let g x = 5 + x;; val g : int -> int = <fun>  How did OCaml infer the type of g here? Let's work it out. First, let's rewrite g syntactically to make our work a little easier:  let g = fun x -> ((+) 5) x  We've made the anonymous function explicit, and we've made the binary infix operator a prefix function application. **1. Assign preliminary types.** For each subexpression of fun x -> (+) 5 x, including the entire expression itself, we assign a preliminary type. We already know the types of (+) and 5, because those are baked into the language itself, but for everything else we "play dumb" and just invent a new type variable for it. For now we will use uppercase letters to represent those type variables, rather than the OCaml syntax for type variables (e.g., 'a).  Subexpression Preliminary type ------------------ -------------------- fun x -> ((+) 5) x R x U ((+) 5) x S ((+) 5) T (+) int -> (int -> int) 5 int x V  **2. Collect constraints.** Here are some observations we could make about the "shape" of subexpressions and some relationships among them: * Since function argument x has type U and function body ((+) 5) x has type S, it must be the case that R, the type of the anonymous function expression, satisfies the constraint R = U -> S. That is, *the type of the anonymous function* is *the type of its argument* arrow *the type of its body*. * Since function ((+) 5) has type T and function application ((+) 5) x has type S, and since the argument x has type V, it must be the case that T = V -> S. That is, *the type of the function being applied* is *the type of the argument it's being applied to* arrow *the type of the function application expression*. * Since function (+) has type int -> (int -> int) and function application (+) 5 has type T, and since the argument 5 has type int, it must be the case that int -> (int->int) = int -> T. Once again, *the type of the function being applied* is *the type of the argument it's being applied to* arrow *the type of the function application expression*. * Since x occurs with both type U and V, it must be the case that U = V. The set of constraints thus generated is:  U = V R = U -> S T = V -> S int -> (int -> int) = int -> T  **3. Solve constraints.** You can solve that system of equations easily. Starting from the last constraint, we know T must be int -> int. Substituting that into the second constraint, we get that int -> int must equal V -> S, hence V = S = int. Since U=V, U must also be int. Substituting for S and U in the first constraint, we get that R = int -> int. So the inferred type of g is int -> int. #### Example 2.  # let apply f x = f x;; val apply : ('a -> 'b) -> 'a -> 'b = <fun>  Again we rewrite:  let apply = fun f -> (fun x -> f x)  **1. Assign preliminary types.**  Subexpression Preliminary type ----------------------- ------------------ fun f -> (fun x -> f x) R f S (fun x -> f x) T x U f x V f S x U  **2. Collect constraints.** - R = S -> T, because of the anonymous function expression. - T = U -> V, because of the nested anonymous function expression. - S = U -> V, because of the function application. **3. Solve constraints.** Using the third constraint, and substituting for S in the first constraint, we have that R = (U -> V) -> T. Using the second constraint, and substituting for T in the first constraint, we have that R = (U -> V) -> (U -> V). There are no further substitutions that can be made, so we're done solving the constraints. If we now replace the preliminary type variables with actual OCaml type variables, specifically U with 'a and V with 'b, we get that the type of apply is ('a -> 'b) -> ('a -> 'b), which is the same as ('a -> 'b) -> 'a -> 'b. #### Example 3.  # apply g 3;; - : int = 8  We rewrite that as (apply g) 3. **1. Assign preliminary types.** In this running example, the inference for g and apply has already been done, so we can fill in their types as known, much like the type of + is already known.  Subexpression Preliminary type ------------- ------------------------------------------ (apply g) 3 R (apply g) S apply (U -> V) -> (U -> V) g int -> int 3 int  **2. Collect constraints.** - S = int -> R - (U -> V) -> (U -> V) = (int -> int) -> S **3. Solve constraints.** Breaking down the last constraint, we have that U = V = int, and that S = U -> V, hence S = int -> int. Substituting that into the first constraint, we have that int -> int = int -> R. Therefore R = int, so the type of apply g 3 is int. #### Example 4.  # apply not false;; - : bool = true  By essentially the same reasoning as in example 3, HM can infer that the type of this expression is bool. This illustrates the polymorphism of apply: because the type (U -> V) -> (U -> V) of apply contains type variables, the function can be applied to any arguments, so long as those arguments' types can be consistently substituted for the type variables. ## Collecting constraints: Algorithm We now present an algorithm that generates constraints. This algorithm is a precise description of how constraint gathering works in the examples we discussed above. The algorithm is not exactly what HM does, because HM actually performs type checking at the same time as type inference. However, the resulting types are the same, and separating inference from checking hopefully will give you a clearer idea of how inference itself works. The algorithm takes as input an expression e. We'll assume that every function fun x -> e' in that expression has an argument with a different name. (If not, our algorithm could make a pre-pass to rename variables. This is feasible because of lexical scope.) The output of the algorithm is a set of constraints. The first thing the algorithm does is to assign unique preliminary type variables, e.g. R or S, - one to each *defining* occurrence of a variable, which could be as a function argument or a let binding, and - one to each occurrence of each subexpression of e. Call the type variable assigned to x in the former clause D(x), and call the type variable assigned to occurrence of a subexpression e' in the latter clause U(e'). The names of these are mnemonics: U stands for the <u>u</u>se of an expression, and D stands for the <u>d</u>efinition of a variable name. Next, the algorithm generates the following constraints: - For integer constants n: U(n) = int. This constraints follows from the type checking rule for integers, which says that every integer constant has type int. Constraints for other types of constants are generated in a similar way. - For variables x: D(x) = U(x). This constraint follows from the type checking rule for variables, which says the type of a variable use (in this case, U(x)) must be the same as the type at which that variable was defined (here, D(x)). - For function application e1 e2: U(e1) = U(e2) -> U(e1 e2), as well as any constraints resulting from e1 and e2. This constraint follows from the type checking rule for function application. - For anonymous functions fun x -> e: U(fun x -> e) = D(x) -> U(e), as well as any constraints resulting from e. This constraint follows from the type checking rule for anonymous functions. - For let expressions let x=e1 in e2: D(x)=U(e1), U(let x=e1 in e2) = U(e2), as well as any constraints resulting from e1 and e2. This constraint follows from the type checking rule for let expressions. - Other expression forms: similar kinds of constraints likewise derived from the type checking rule for the expression form. The result is a set of constraints, which is the output of the algorithm. It's not too hard to implement this algorithm as a recursive function over a tree representing the syntax of e. **Example.** Given expression fun x -> (fun y -> x), a type variable R is associated with argument x, and S with argument y. For subexpressions, T is associated with the occurrence of fun x -> (fun y -> x), and X with the occurrence of (fun y -> x), and Y with the occurrence of x. (Note that the names we've chosen for the type variables are completely arbitrary.) The constraints generated are T = R -> X, and X = S -> Y, and Y = R. ## Solving constraints: Algorithm What does it mean to solve a set of constraints? To answer this question, we define *type substitutions*. A type substitution is a map from a type variable to a type. We'll write {t/X} for the substitution that maps type variable X to type t. The way a substitution S operates on a type can be defined recursively:  S(X) = if S = {t/X} then t else X S(t1 -> t2) = S(t1) -> S(t2)  A substitution S can be applied to a constraint t = t'; the result S(t = t') is defined to be S(t) = S(t'). And a substitution can be applied to a set C of constraints; the result S(C) is the result of applying S to each of the individual constraints in C. Given two substitutions S and S', we write S;S' for their composition: (S;S')(t) = S'(S(t)). A substitution *unifies* a constraint t_1 = t_2 if S(t_1) = S(t_2). A substitution S unifies a set C of constraints if S unifies every constraint in C. For example, substitution S = {int->int/Y};{int/X} unifies constraint X -> (X -> int) = int -> Y. To solve a set of constraints C, we need to find a substitution that unifies C. If there are no substitutions that unify C, where C is the constraints generated from expression e, then e is not typeable. To find a substitution that unifies C, we use an algorithm appropriately called the *unification* algorithm. It is defined as follows: - if C is the empty set, then unify(C) is the empty substitution. - if C is the union of a constraint t = t' with other constraints C', then unify(C) is defined as follows, based on that constraint: - if t and t' are both the same type variable, e.g. X, then return unify(C'). - if t = X for some type variable X, and X does not occur in t', then let S = {t'/X}, and return unify(S(C'));S. - if t' = X for some type variable X, and X does not occur in t, then let S = {t/X}, and return unify(S(C'));S. - if t = t0 -> t1 and t' = t'0 -> t'1, then let C'' be the union of C' with the constraints t0 = t'0 and t1 = t'1, and return unify(C''). - if t = t0 * t1 and t' = t'0 * t'1, then let C'' be the union of C' with the constraints t0 = t'0 and t1 = t'1, and return unify(C''). - if t = (t0, ..., tn) tc and t' = (t'0, ..., t'n) tc for some type constructor tc, then let C'' be the union of C' with the constraints ti = t'i, and return unify(C''). - otherwise, fail. There is no possible unifier. In the second and third subcases, the check that X should not occur in t ensures that the algorithm doesn't produce a cyclic substitution&mdash;for example, {(X -> X) / X}. It's possible to prove that the unification algorithm always terminates, and that it produces a result if and only a unifier actually exists&mdash;that is, if and only if the set of constraints has a solution. Moreover, the solution the algorithm produces is the *most general unifier*, in the sense that if S = unify(C) and S' unifies C, then there must exist some S'' such that S' = S;S''. If R is the type variable assigned to represent the type of the entire expression e, and if S is the substitution produced by the algorithm, then S(R) is the type inferred for e by HM type inference. Call that type t. It's possible to prove t is the *principal* type for the expression, meaning that if e also has type t' for any other t', then there exists a substitution S such that t' = S(t). So HM actually infers the most lenient type that is possible for any expression. ## Let expressions Consider the following code:  let double f z = f (f z) in (double (fun x -> x+1) 1, double (fun x -> not x) false)  The inferred type for f in double would be X -> X. In the algorithm we've described so far, the use of double in the first component of the pair would produce the constraint X = int, and the use of double in the definition of b would produce the constraint X = bool. Those constraints would be contradictory, causing unification to fail! There is a very nice solution to this called *let-polymorphism*, which is what OCaml actually uses. Let-polymorphism enables a polymorphic function bound by a let expression behave as though it has multiple types. The essential idea is to allow each usage of a polymorphic function to have its own instantiation of the type variables, so that contradictions like the one above can't happen. We won't cover let-polymorphism here, but you can learn more about it in the reading given below. ## Efficiency of HM HM is usually a very efficient algorithm&mdash;you've probably never had to wait for the REPL to print the inferred types of your programs. In practice, it runs in approximately linear time. But in theory, there are some very strange programs that can cause its running-time to blow up. (Technically, it's DEXPTIME-complete.) For fun, try typing the following code in utop:  let b = true;; let f0 = fun x -> x+1;; let f = fun x -> if b then f0 else fun y -> x y;; let f = fun x -> if b then f else fun y -> x y;; (* keep repeating that last line *) ` You'll see the types get longer and longer, and eventually type inference will cause a notable delay. ## The history of HM HM has been rediscovered many times by many people. Curry used it informally in the 1950's (perhaps even the 1930's). He wrote it up formally in 1967 (published 1969). Hindley discovered it independently in 1969; Morris in 1968; and Milner in 1978. In the realm of logic, similar ideas go back perhaps as far as Tarski in the 1920's. Commenting on this history, Hindley wrote, > There must be a moral to this story of continual re-discovery; > perhaps someone along the line should have learned to read. Or someone > else learn to write. ## Summary Hindley&ndash;Milner type inference is one of the core algorithms that makes the OCaml language, and many other functional languages, possible. It is fundamentally based on traversing the source code to collect a system of equations, then solving that system to determine the types. ## Terms and concepts * constraint * Hindley&ndash;Milner (HM) type inference algorithm * implicit typing * let polymorphism * preliminary type variable * static typing * substitution * type annotation * type inference * type reconstruction * unification * unifier ## Further reading * [*Types and Programming Languages*][tapl], chapter 22, by Benjamin C. Pierce. [tapl]: https://newcatalog.library.cornell.edu/catalog/8324012