Types and Programming Languages - SJTUbasics.sjtu.edu.cn/~xiaojuan/tapl2016/files/lec1.pdf ·...

Types and Programming Languages

Lecture 1. Introduction & Untyped arithmeticexpressions

Xiaojuan Cai

cxj@sjtu.edu.cn

BASICS Lab, Shanghai Jiao Tong University

Fall, 2016

Why study this course?

I For credits?

I To meet the requirement that every master student shouldtake at least TWO English courses?

I Being curious about Programming Languages.

I For credits?

Programming languages

I How many programming languages in the world?

I https://en.wikipedia.org/wiki/List_of_programming_

languagesI https://en.wikipedia.org/wiki/List_of_programming_

languages_by_typeI http://99-bottles-of-beer.net/

I Which programming language is the best?I Which type of cars is the best?

I Winning an F1 raceI Taking kids to the kindergarten

I How many programming languages in the world?I https://en.wikipedia.org/wiki/List_of_programming_

I Which programming language is the best?

I Which type of cars is the best?I Winning an F1 raceI Taking kids to the kindergarten

Where do we start?

Which programming language should we start with?

A popular one or a classical one?

We will focus mostly on functional programming languagesbecause

I mutation is discouraged which makes programs safe.

I higher-order functions are very convenient.

ML-like languages: SML, OCaml, Haskell, ...

Where do we start?

Which programming language should we start with?A popular one or a classical one?

Where do we start?

Why types?

There are many excellent textbooks on Programming languages,such as:

I Programming Language Pragmatics, by Michael L. Scott

I Practical Foundations of Programming Languages, by RobertHarper

I Programming Languages, Principles and Paradigms, by AllenTucker and Robert Noonan

We will focus on types because

I most language features can be discussed in the framework oftype theory, and

I language design goes hand-in-hand with type system design.

Why types?

There are many excellent textbooks on Programming languages,such as:

I Programming Language Pragmatics, by Michael L. Scott

I Practical Foundations of Programming Languages, by RobertHarper

I Programming Languages, Principles and Paradigms, by AllenTucker and Robert Noonan

We will focus on types because

I most language features can be discussed in the framework oftype theory, and

I language design goes hand-in-hand with type system design.

Course overview

Benjamin C. Pierce

We will discuss in this course:1. Structural induction2. λ-calculus3. simple type systems and safety4. Production types, sum types5. Reference types, exceptions6. Subtyping7. Recursive types8. Type reconstruction9. Universal type and polymorphism10. type operators ...

Course overview

Benjamin C. Pierce

Course overview

Benjamin C. Pierce

After this course, you can ...

I Know how to judge a language.

I Learn to think about software in“PL” way and become abetter programmer

I Confidently pick up new languages for your lifetime.

Course policy

I Final exam: 70%

I Homework: 30%

I TA:Yuwei Wang, wangywgg@sjtu.edu.cn

I Course homepage:http://basics.sjtu.edu.cn/~xiaojuan/tapl2016

Course policy

I Final exam: 70%

I Homework: 30%

I TA:Yuwei Wang, wangywgg@sjtu.edu.cn

I Course homepage:http://basics.sjtu.edu.cn/~xiaojuan/tapl2016

Outline

Introduction

Preliminaries

Untyped arithmetic expressionsAbstract syntaxInduction on termsSemantics

BooleansNumbers and Booleans

Types in Computer Science

Type systems is the most popular and best established lightweightformal methods.

DefinitionA type system is a tractable syntactic method for proving theabsence of certain program behaviors by classifying phrasesaccording to the kinds of values they compute.

Types in Computer Science

Type systems is the most popular and best established lightweightformal methods.

DefinitionA type system is a tractable syntactic method for proving theabsence of certain program behaviors by classifying phrasesaccording to the kinds of values they compute.

Brief history

Types system (type theory) refers to a much broader field.

I 1900. Formalized, Russell’s paradox

I 1925. Simple theory of types, Ramsey

I 1940. Simply typed λ-calculus, Church

I 1973. Constructive type theory, Martin Lof

I 1992. Pure type theory, Barendregt

Some definitions

I Static type system. Type checking during compile-time

I Dynamic type system. Type checking during run-time

I Static ⇒ Conservative ⇒ prove the absence of bad behaviours

I Incapable of finding all undesired program behavirous, e.g.divide by zero

I Type checkersI automatic: no manual interactionI type annotations

Some definitions

What types good for

I Detecting errors early.

I Maintenance tools.

I Abstracting

I Documentation

I Efficiency

Applications: network security, program analysis, theorem prover,database, xml, ...

Language design goes hand-in-hand with type system design.

Outline

Introduction

Preliminaries

Relations

I An n-place relation is a set R ⊆ S1 × S2 × · · · × Sn.

I A two-place relation R on sets S and T is called a binaryrelation. We often write s R t instead of (s, t) ∈ R.

I The ”mixfix” concrete syntax, e.g, Γ ` s : T means “the triple(Γ, s,T ) in the typing relation”.

I P is preserved by R if whenever we have s R t and P(s), wealso have P(t).

Relations

I An n-place relation is a set R ⊆ S1 × S2 × · · · × Sn.

I A two-place relation R on sets S and T is called a binaryrelation. We often write s R t instead of (s, t) ∈ R.

I The ”mixfix” concrete syntax, e.g, Γ ` s : T means “the triple(Γ, s,T ) in the typing relation”.

I P is preserved by R if whenever we have s R t and P(s), wealso have P(t).

Functions

I dom(R): the domain of a relation R on sets S and T is theset of elements s ∈ S such that (s, t) ∈ R for some t.

I A relation R on sets S and T is called a partial function if,whenever (s, t1) ∈ R and (s, t2) ∈ R, we have t1 = t2. Ifdom(R) = S , then R is a total function.

I We write f (x) ↑ to mean “f is undefined on x,” and f (x) ↓ tomean “f is defined on x.”

Ordered setsA binary relation R on a set S is

I Reflexive: ∀x ∈ S .x R x .

I Transitive: x R y ∧ y R z implies x R z .

I Symmetric: x R y implies y R x .

I Antisymmetric: x R y ∧ y R x implies x = y .

1. Preorder (or Quasi order): Reflexive + Transitive

2. Equivalence: Preorder + Symmetric

3. Partial order: Preorder + Antisymmetric

4. Total order: Partial oder + (∀x , y ∈ S .x R y ∨ y R x)

5. Well quasi order: Preorder + (Any infinite sequence containsan increasing pair)

6. Well founded order: Preorder + (No infinite decreasingsequences)

Quiz: 1. Can Transitivity + Symmetry indicate Reflexivity?2. Give examples of these orders.

Inductions

I Ordinary induction on natural numbersIf P(0)and for all i , P(i) implies P(i + 1),then P(n) holds for all n.

I Complete induction on natural numbersIf, for each natural number k,

given P(i) for all i < kwe can show P(k)

then P(n) holds for all n.

Outline

Introduction

Preliminaries

Untyped systems

I Untyped arithmetic expressions

I Untyped λ-calculus

Introduction

t ::= true constant truefalse constant falseif t then t else t conditional0 constant zerosucc t successorpred t predecessoriszero t zero test

I BNF grammar

I t is metavariable.

I For simplicity, we use arabic numbers, e.g. 3 stands for(succ (succ (succ 0)))

I Currently, if (succ 0) then true else (pred 0) is a validterm.

Other ways to give syntax definition

The set of terms is the smallest set T such thatI Inductively.

I {true, false, 0} ⊆ T ;I if t1 ∈ T , then {succ t1, pred t1, iszero t1} ⊆ T ;I if t1, t2, t3 ∈ T , then if t1 then t2 else t3 ∈ T

I By inference rules

true ∈ T false ∈ T 0 ∈ T

t1 ∈ T

succ t1 ∈ T

t1 ∈ T

pred t1 ∈ T

t1 ∈ T

iszero t1 ∈ T

t1, t2, t3 ∈ T

if t1 then t2 else t3 ∈ T

Other ways to give syntax definition, cont’d

I Concretely.For each natural number i , define Si as follows:

S0 = ∅Si+1 = {true, false, 0}

∪{succ t1, pred t1, iszero t1 | t1 ∈ Si}∪{if t1 then t2 else t3 | t1, t2, t3 ∈ Si}

S =⋃

Lemma. S = T.Quiz. What if we change the concrete definition of S to

S0 = {true, false, 0}Si+1 = {succ t1, pred t1, iszero t1 | t1 ∈ Si}

∪{if t1 then t2 else t3 | t1, t2, t3 ∈ Si}

Other ways to give syntax definition, cont’d

I Concretely.For each natural number i , define Si as follows:

S0 = ∅Si+1 = {true, false, 0}

∪{succ t1, pred t1, iszero t1 | t1 ∈ Si}∪{if t1 then t2 else t3 | t1, t2, t3 ∈ Si}

S =⋃

Lemma. S = T.Quiz. What if we change the concrete definition of S to

S0 = {true, false, 0}Si+1 = {succ t1, pred t1, iszero t1 | t1 ∈ Si}

∪{if t1 then t2 else t3 | t1, t2, t3 ∈ Si}

Inductive structure

For any t ∈ T , one of three things must be true about t:

1. t is constant

2. t has form succ t1, pred t1, or iszero t1

3. t has form if t1 then t2 else t3.

Two ways to use this observation: inductive definition andinductive proof.

Inductive defintion

consts(true) = {true}consts(false) = {false}

consts(0) = {0}consts(succ t1) = consts(t1)consts(pred t1) = consts(t1)

consts(iszero t1) = consts(t1)consts(if t1 then t2 else t3) = consts(t1) ∪ consts(t2) ∪ consts(t3)

Quiz. 1. Give an inductive definition of size, which is the size ofthe syntax tree of a term t.2. Give an inductive definition of depth, which is the height of thesyntax tree of a term t.

Inductive defintion

consts(true) = {true}consts(false) = {false}

consts(0) = {0}consts(succ t1) = consts(t1)consts(pred t1) = consts(t1)

consts(iszero t1) = consts(t1)consts(if t1 then t2 else t3) = consts(t1) ∪ consts(t2) ∪ consts(t3)

Quiz. 1. Give an inductive definition of size, which is the size ofthe syntax tree of a term t.2. Give an inductive definition of depth, which is the height of thesyntax tree of a term t.

size(true) = 1size(false) = 1

size(0) = 1size(succ t1) = size(t1) + 1size(pred t1) = size(t1) + 1

size(iszero t1) = size(t1) + 1size(if t1 then t2 else t3) = size(t1) + size(t2) + size(t3) + 1

Lemma. |consts(t)| ≤ size(t).Proof. By induction on the structure of t.

Principles of induction on terms

I Induction on depth:If, for each term s,

given P(r) for all r such that depth(r) < depth(s),we can show P(s),

then P(t) holds for all t.

I Induction on size:If, for each term s,

given P(r) for all r such that size(r) < size(s),we can show P(s),

I Structural Induction:If, for each term s,

given P(r) for all immediate subterms r of s,we can show P(s),

Outline

Introduction

Preliminaries

Semantics of languages

I Operational semantics. It specifies the behavior of PL bydefining an abstract machine.

I Denotational semantics. The meaning of a term is taken to besome mathematical object (a number or a function).

I Axiomatic semantics. It takes the laws themselves as thedefinition of the language.

A toy language – Booleans

Syntax

t ::= termstrue constant truefalse constant falseif t then t else t conditional

v ::= valuestrue true valuefalse false value

Evaluation rules for Booleans

Evaluation

E-IfTrueif true then t2 else t3 −→ t2

E-IfFalseif false then t2 else t3 −→ t3

E-Ift1 −→ t ′1

if t1 then t2 else t3 −→ if t ′1 then t2 else t3

E-IfTrue and E-IfFalse are also called computation rules andE-If is called congruence rule.Quiz. Evaluate the following terms:

I true

if (if false then false else false) then true else false

Evaluation

E-Ift1 −→ t ′1

E-IfTrue and E-IfFalse are also called computation rules andE-If is called congruence rule.

Quiz. Evaluate the following terms:

I true

Evaluation

E-Ift1 −→ t ′1

E-IfTrue and E-IfFalse are also called computation rules andE-If is called congruence rule.Quiz. Evaluate the following terms:

I true

Derivation tree of One-step evaluation

sdef= if true then false else false

tdef= if s then false else false

udef= if false then false else false

E-IfE-If

E-IfTrues −→ false

t −→ uif t then false else false −→ if u then false else false

Theorem 3.5.4 [Determinacy of one-step evaluation]: Ift −→ t ′ and t −→ t ′′, then t ′ = t ′′.Proof. By induction on the depth of the derivation tree.

E-IfE-If

Theorem 3.5.4 [Determinacy of one-step evaluation]: Ift −→ t ′ and t −→ t ′′, then t ′ = t ′′.

Proof. By induction on the depth of the derivation tree.

E-IfE-If

Theorem 3.5.4 [Determinacy of one-step evaluation]: Ift −→ t ′ and t −→ t ′′, then t ′ = t ′′.Proof. By induction on the depth of the derivation tree.

Normal form and multi-step evaluation

I A term t is in normal form if no evaluation rule can appliy toit.

Theorem 3.5.7: Every value is in normal form.Theorem 3.5.8: If t is in normal form, then it is a value.

I The multi-step evaluation relation −→∗ is the reflexive,transitive closure of −→.

Theorem 3.5.11 [Uniqueness of normal forms]: If t −→∗ u andt −→∗ u′ where u, u′ are normal forms, then u = u′.Theorem 3.5.12 [Termination of evaluation]: For every term tthere is some normal form u such that t −→∗ u.

Normal form and multi-step evaluation

I A term t is in normal form if no evaluation rule can appliy toit.

Theorem 3.5.7: Every value is in normal form.Theorem 3.5.8: If t is in normal form, then it is a value.

I The multi-step evaluation relation −→∗ is the reflexive,transitive closure of −→.

Theorem 3.5.11 [Uniqueness of normal forms]: If t −→∗ u andt −→∗ u′ where u, u′ are normal forms, then u = u′.Theorem 3.5.12 [Termination of evaluation]: For every term tthere is some normal form u such that t −→∗ u.

Arithmatic Expression

Syntax

t ::= terms· · ·0 constant zerosucc t successorpred t predecessoriszero t zero test

v ::= values· · ·nv numeric value

nv ::= numeric values0 zero valuesucc nv successor value

Evaluation rulesQuiz. Give the definition of evaluation rules to guarantee

[Determinacy of one-step evaluation]:If t −→ t ′ and t −→ t ′′, then t ′ = t ′′.

Evaluation

E-PredZeropred 0 −→ 0

E-IszeroZeroiszero 0 −→ true

E-PredSuccpred (succ nv) −→ nv

E-IszeroSucciszero (succ nv) −→ false

E-Predt1 −→ t ′1

pred t1 −→ pred t ′1E-Iszero

t1 −→ t ′1iszero t1 −→ iszero t ′1

E-Succt1 −→ t ′1

succ t1 −→ succ t ′1

Evaluation rulesQuiz. Give the definition of evaluation rules to guarantee

[Determinacy of one-step evaluation]:If t −→ t ′ and t −→ t ′′, then t ′ = t ′′.

Evaluation

E-PredZeropred 0 −→ 0

E-IszeroZeroiszero 0 −→ true

E-PredSuccpred (succ nv) −→ nv

E-IszeroSucciszero (succ nv) −→ false

E-Predt1 −→ t ′1

pred t1 −→ pred t ′1E-Iszero

t1 −→ t ′1iszero t1 −→ iszero t ′1

E-Succt1 −→ t ′1

succ t1 −→ succ t ′1

Normal form and stuckness

I What if we change E-PredSucc to pred (succ t) −→ t?Does it still satisfy [Determinacy of one-step transition]?

I Note there are meaningless terms, such asif 0 then (succ true) else (iszeoro false).

I A term t is stuck if it is in normal form but not a value.

Conclusion

I Types are very important for PLs.

I This course will give a full view of type systems from thesimplest one to full-fledged one.

I Fundamental concepts for PLs:I syntax, defined inductively, concretely, ...I inductive proofs are very important for PLs, especially,

structural induction.I operational semantics plays more and more important roles.

We define evaluation rules by using operational transitions.

I Properties such as [Determinacy of one-step evaluation],[Uniqueness of normal forms], and [Termination] areimportant for a good language design.

Homework

I 3.3.4, 3.5.10, 3.5.13, 3.5.18.

I Install OCaml and get familiar with this language.http://ocaml.org/

Types and Programming Languages - SJTUbasics.sjtu.edu.cn/~xiaojuan/tapl2016/files/lec1.pdf ·...

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