CS558 Programming Languages

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1 CS558 Programming Languages Fall 2017 Lecture 7b Andrew Tolmach Portland State University

2 Type Inference Some statically typed languages, like ML (and to a lesser extent Scala), offer alternative ways to regain the flexibility of dynamic typing, via type inference and polymorphism. Type inference works like this: The types of identifiers are automatically inferred from the way they are used The programmer is no longer required to declare the types of identifiers (although this is still permitted) Method requires that the types of operators and literals is known

3 Inference Examples (let f (fun (x) (+ x 2)) f y)) The type of x must be int because it is used as an arg to +. So the type of f must be int int (i.e. the type of functions that expect an int argument and return an int result), and y must be an int. (let f (fun (x) (cons x nil)) f true)) Suppose x has some type t. Then the type of f must be t (list t). Since f is applied to a bool, we must have t = bool. For the moment, we assume that f must be given a unique monomorphic type; we will relax this later

4 Systematic Inference Here s a harder example: (let f (fun (x) (if x p q)) (+ 1 f r))) Can only infer types by looking at both the function s body and its application In general, we can solve the inference task by extracting a collection of typing constraints from the program s AST, and then finding a simultaneous solution for the constraints using unification Extracted constraints tell us how types must be related if we are to be able to find a typing derivation. Each node generates one or more constraints

5 Rules for First-class Functions To handle this example, we ll need some extra typing rules: TE + {x t 1 } e : t 2 TE (fun (x) e) : t 1 t 2 (Fn) TE e 1 : t 1 t 2 TE e 2 : t 1 TE e 1 e 2 ) : t 2 (Appl)

6 Inference Example Solution : t 1 = t 7 = t 8 = t 9 = t 3 = t 5 = t p = t 6 = t q = int t 4 = t x = t 11 = t r = bool t 2 = t f = t 10 = bool int Node Rule Constraints 1 Let t f = t 2 t 1 = t 7 2 Fun t 2 = t x t 3 3 If t 4 = bool t 3 = t 5 = t 6 4 Var t 4 = t x 5 Var t 5 = t p Let(f) 1 6 Var t 5 = t q 7 Add t 7 = t 8 = t 9 = int 8 Int t 8 = int 9 Appl t 10 = t 11 t 9 Fun(x) Var t 10 = t f 11 Var t 11 = t r If 3 1 Appl 8 9 x p q f r 10 11

7 Drawbacks of Inference Consider this variant example: (let f (fun (x) (if x p false) (+ 1 f r))) Now the body of f returns type bool, but it is used in a context expecting an int. The corresponding extracted constraints will be inconsistent; no solution can be found. Can report a type error to the programmer. But which is wrong, the definition of f or the use? No good way to associate the error message with a single program point.

8 Polymorphism Consider (let snd (fun (l) (head (tail l))) snd (cons 1 (cons 2 (cons 3 nil))))) By extracting constraints and solving, we will get snd : (list int) int We could also write Same definition! (let snd (fun (l) (head (tail l))) snd (cons true (cons false (cons true nil))))) And get snd : (list bool) bool

9 Polymorphism (2) So why can t we write something like this? (let snd (fun (l) (head (tail l))) (block snd (cons 1 (cons 2 (cons 3 nil)))) snd (cons true (cons false (cons true nil)))))) We can, by treating the type of snd as polymorphic snd : (list t) t Here t is an unconstrained type variable

10 Inferring Polymorphism In fact, if we extract constraints and solve just for the definition (fun l (head (tail l))) without considering its uses, we will end up with exactly the type (list t) t We can assign this polymorphic type to snd allowing it to be used multiple times, each with a different instance of t (e.g. with t = bool or t = int). By default, languages like ML infer the most polymorphic possible type for every function This is the natural result of the inference process we ve described

11 Parametric Polymorphism We can think of these polymorphic types as being universally quantified over their type variables and instantiated at use sites let snd : t. list t -> t = in snd {bool} (true::false::nil); snd {int} (1::2::nil) This is called parametric polymorphism because the function definition is (implicitly) parameterized by the instantiating type In ML-like languages the quantification and instantiation don t actually appear

12 Explicit Parametric Polymorphism Java generics and Scala type parameterization are also a form a parametric polymorphism, in which type abstraction and instantiation are (mostly) explicit def snd[a](l: List[A]) : A = l.tail.head val a = snd[boolean] (List(true,false)) val b = snd(list(1,2)) In parametric polymorphism, the behavior of the polymorphic function is the same no matter what the instantiating type is Scala In fact, an ML compiler typically generates just one piece of machine code for each polymorphic function, shared by all instances

14 Static vs. Dynamic Overloading In most statically-typed languages, overloading is resolved statically; i.e. the compiler selects the right version of the overloaded definition once and for all at compile time. Dynamically-typed languages also often overload operators (e.g. + on different kinds of numbers, strings, etc.) Here the right version of the overloaded operator is picked at runtime after checking the (runtime) types of the arguments Of course, the operator might fail altogether if there is no version suitable for the types discovered Haskell type classes provide an unusual form of dynamic overloading with a static guarantee that a suitable version exists

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