Fundamentals and lambda calculus. Deian Stefan (adopted from my & Edward Yang s CSE242 slides)
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1 Fundamentals and lambda calculus Deian Stefan (adopted from my & Edward Yang s CSE242 slides)
2 Logistics Assignments: Programming assignment 1 is out Homework 1 will be released tomorrow night Podcasting: everything should be set Section: 8-8:50AM default; TA choice to do 3-3:50PM Clickers, sign up: see piazza/course page for link We ll use them today
3 JavaScript functions JavaScript functions are first-class Syntax is a bit ugly/terse when you want to use functions as values; recall block scoping: (function () { //... do something }) (); New version has cleaner syntax called fat arrows Semantics not always the same (this has different meaning), but for this class should always be safe to use
4 fat-arrows.js
5 In this lecture
6 In this lecture
7 In this lecture
8 In this lecture λ
9 What is the lambda calculus? Simplest reasonable programming language Only has one feature: functions
10 Why study it? Captures the idea of first-class functions Good system for studying the concept of variable binding that appears in almost all languages Historically important Competing model of computation introduced by Church as an alternative to Turing machines: substitution (you ll see this today) = symbolic comp Influenced Lisp (thus JS), ML, Haskell, C++, etc.
11 Why else? Base for studying many programming languages You can use lambda calculus and extended it in different ways to study languages and features E.g., we can study the difference between strict languages like JavaScript and lazy ones like Haskell λ + evaluation strategy E.g., we can study different kinds of type systems Simply-typed λ calculus, polymorphic, etc.
12 Why else? Most PL papers describe language models that build on lambda calculus Understanding λ will help you interpret what you are reading in PL research papers Understanding λ will help you get started with other formal/theoretical foundations: Operational semantics Denotational semantics
13 Before we get started, some terminology Syntax (grammar) The symbols used to write a program E.g., (x + y) is a grammatical expression Semantics The actions that occur when a program is executed PL implementation: Syntax -> Semantics
14 Before we get started, some terminology Syntax (grammar) The symbols used to write a program E.g., (x + y) is a grammatical expression Semantics The actions that occur when a program is executed PL implementation: Syntax -> Semantics
15 Before we get started, some terminology Syntax (grammar) The symbols used to write a program E.g., (x + y) is a grammatical expression Semantics The actions that occur when a program is executed PL implementation: Syntax -> Semantics
16 Today Syntax of λ calculus Semantics of λ calculus Free and bound variables Substitution Evaluation order
17 Lambda calculus Language syntax (grammar): Expressions: e ::= x λx.e e 1 e 2 Variables: x Functions or λ abstractions: λx.e This is the same as x => e in JavaScript! Function application: e 1 e 2 This is the same as e 1 (e2) in JavaScript!
18 Example terms λx.(2+x) Same as: x => (2 + x) (λx.(2 + x)) 5 Same as: (x => (2 + x)) (5) (λf.(f 3)) (λx.(x + 1)) Same as: (f => (f (3))) (x => (x+1))
19 Example terms λx.(2+x) LIES! What is this 2 and +? (Sugar.) Same as: x => (2 + x) (λx.(2 + x)) 5 Same as: (x => (2 + x)) (5) (λf.(f 3)) (λx.(x + 1)) Same as: (f => (f (3))) (x => (x+1))
20 Example terms λx.(2+x) Same as: x => (2 + x) (λx.(2 + x)) 5 Same as: (x => (2 + x)) (5) (λf.(f 3)) (λx.(x + 1)) Same as: (f => (f (3))) (x => (x+1))
21 Example terms λx.(2+x) Same as: x => (2 + x) (λx.(2 + x)) 5 Same as: (x => (2 + x)) (5) (λf.(f 3)) (λx.(x + 1)) Same as: (f => (f (3))) (x => (x+1))
22 Example terms λx.(2+x) Same as: x => (2 + x) (λx.(2 + x)) 5 Same as: (x => (2 + x)) (5) (λf.(f 3)) (λx.(x + 1)) Same as: (f => (f (3))) (x => (x+1))
23 JavaScript to λ calculus Let s look at function composition: (f f)(x) In JavaScript: f => (x => f (f (x))) ((f => (x => f (f (x)))) (x => x+1)) (4) In λ: λf.(λx. f (f x)) ((λf.(λx. f (f x)) (λx.x+1)) 4
24 JavaScript to λ calculus Let s look at function composition: (f f)(x) In JavaScript: f => (x => f (f (x))) ((f => (x => f (f (x)))) (x => x+1)) (4) In λ: λf.(λx. f (f x)) ((λf.(λx. f (f x)) (λx.x+1)) 4
25 JavaScript to λ calculus Let s look at function composition: (f f)(x) In JavaScript: f => (x => f (f (x))) ((f => (x => f (f (x)))) (x => x+1)) (4) In λ: λf.(λx. f (f x)) ((λf.(λx. f (f x)) (λx.x+1)) 4
26 Today Syntax of λ calculus Semantics of λ calculus Free and bound variables Substitution Evaluation order
27 Semantics of λ calculus Reduce a term to another as much as we can If we can t reduce it any further, the term is said to be in normal form How? Rewrite terms!
28 Example terms Example: (λx.(2 + x)) 5 (2 + 5) 7 In JavaScript: (x => (2 + x)) (5) (2 + 5) 7 Example: (λf.(f 3)) (λx.(x + 1)) ((λx.(x + 1)) 3) (3 + 1) 4 In JavaScript: (f => (f (3))) (x => (x+1)) ((x => (x+1)) (3) (3+1) 4
29 Example terms Example: (λx.(2 + x)) 5 (2 + 5) 7 In JavaScript: (x => (2 + x)) (5) (2 + 5) 7 Example: (λf.(f 3)) (λx.(x + 1)) ((λx.(x + 1)) 3) (3 + 1) 4 In JavaScript: (f => (f (3))) (x => (x+1)) ((x => (x+1)) (3) (3+1) 4
30 Example terms Example: (λx.(2 + x)) 5 (2 + 5) 7 In JavaScript: (x => (2 + x)) (5) (2 + 5) 7 Example: (λf.(f 3)) (λx.(x + 1)) ((λx.(x + 1)) 3) (3 + 1) 4 In JavaScript: (f => (f (3))) (x => (x+1)) ((x => (x+1)) (3) (3+1) 4
31 Example terms Example: (λx.(2 + x)) 5 (2 + 5) 7 In JavaScript: (x => (2 + x)) (5) (2 + 5) 7 Example: (λf.(f 3)) (λx.(x + 1)) ((λx.(x + 1)) 3) (3 + 1) 4 In JavaScript: (f => (f (3))) (x => (x+1)) ((x => (x+1)) (3) (3+1) 4
32 Example terms Example: (λx.(2 + x)) 5 (2 + 5) 7 In JavaScript: (x => (2 + x)) (5) (2 + 5) 7 Example: (λf.(f 3)) (λx.(x + 1)) ((λx.(x + 1)) 3) (3 + 1) 4 In JavaScript: (f => (f (3))) (x => (x+1)) ((x => (x+1)) (3) (3+1) 4
33 Example terms Example: (λx.(2 + x)) 5 (2 + 5) 7 In JavaScript: (x => (2 + x)) (5) (2 + 5) 7 Example: (λf.(f 3)) (λx.(x + 1)) ((λx.(x + 1)) 3) (3 + 1) 4 In JavaScript: (f => (f (3))) (x => (x+1)) ((x => (x+1)) (3) (3+1) 4
34 Example terms Example: (λx.(2 + x)) 5 (2 + 5) 7 In JavaScript: (x => (2 + x)) (5) (2 + 5) 7 Example: (λf.(f 3)) (λx.(x + 1)) ((λx.(x + 1)) 3) (3 + 1) 4 In JavaScript: (f => (f (3))) (x => (x+1)) ((x => (x+1)) (3) (3+1) 4
35 Example terms Example: (λx.(2 + x)) 5 (2 + 5) 7 In JavaScript: (x => (2 + x)) (5) (2 + 5) 7 Example: (λf.(f 3)) (λx.(x + 1)) ((λx.(x + 1)) 3) (3 + 1) 4 In JavaScript: (f => (f (3))) (x => (x+1)) ((x => (x+1)) (3) (3+1) 4
36 Example terms Example: (λx.(2 + x)) 5 (2 + 5) 7 In JavaScript: (x => (2 + x)) (5) (2 + 5) 7 Example: (λf.(f 3)) (λx.(x + 1)) ((λx.(x + 1)) 3) (3 + 1) 4 In JavaScript: (f => (f (3))) (x => (x+1)) ((x => (x+1)) (3) (3+1) 4
37 Easy! Pattern: for function application substitute the term you are applying the function to for the argument variable
38 Substitution (not right) Substitution: e 1 [x := e 2 ] Replace every occurrence of x in e1 with e 2 General reduction rule for λ calculus: (λx.e 1 ) e 2 e 1 [x := e 2 ] Function application rewritten to e 1 (the function body) with every x in e 1 substituted with e 2 (argument)
39 A more complicated example
40 A more complicated example Compose function that adds 1 to arg & apply it to 4 ((λf.(λx. f (f x)) (λx.x+1)) 4
41 A more complicated example Compose function that adds 1 to arg & apply it to 4 ((λf.(λx. f (f x)) (λx.x+1)) 4 (λx. (λx.x+1) ((λx.x+1) x)) 4
42 A more complicated example Compose function that adds 1 to arg & apply it to 4 ((λf.(λx. f (f x)) (λx.x+1)) 4 (λx. (λx.x+1) ((λx.x+1) x)) 4 (λx.x+1) ((λx.x+1) 4)
43 A more complicated example Compose function that adds 1 to arg & apply it to 4 ((λf.(λx. f (f x)) (λx.x+1)) 4 (λx. (λx.x+1) ((λx.x+1) x)) 4 (λx.x+1) ((λx.x+1) 4) (λx.x+1) (4+1)
44 A more complicated example Compose function that adds 1 to arg & apply it to 4 ((λf.(λx. f (f x)) (λx.x+1)) 4 (λx. (λx.x+1) ((λx.x+1) x)) 4 (λx.x+1) ((λx.x+1) 4) (λx.x+1) (4+1) 4+1+1
45 A more complicated example Compose function that adds 1 to arg & apply it to 4 ((λf.(λx. f (f x)) (λx.x+1)) 4 (λx. (λx.x+1) ((λx.x+1) x)) 4 (λx.x+1) ((λx.x+1) 4) (λx.x+1) (4+1)
46 Let s make this even more fun!
47 Let s make this even more fun! Instead of 1, let s add x to argument (& do it 2 times): (λf.(λx. f (f x)) (λy.y+x)
48 Let s make this even more fun! Instead of 1, let s add x to argument (& do it 2 times): (λf.(λx. f (f x)) (λy.y+x) λx. (λy.y+x) ((λy.y+x) x)
49 Let s make this even more fun! Instead of 1, let s add x to argument (& do it 2 times): (λf.(λx. f (f x)) (λy.y+x) λx. (λy.y+x) ((λy.y+x) x) λx. (λy.y+x) (x+x)
50 Let s make this even more fun! Instead of 1, let s add x to argument (& do it 2 times): (λf.(λx. f (f x)) (λy.y+x) λx. (λy.y+x) ((λy.y+x) x) λx. (λy.y+x) (x+x) λx. (x+x+x)
51 Let s make this even more fun! Instead of 1, let s add x to argument (& do it 2 times): (λf.(λx. f (f x)) (λy.y+x) λx. (λy.y+x) ((λy.y+x) x) λx. (λy.y+x) (x+x) λx. (x+x+x)???? that s not a function that adds x to argument two times
52 Substitution is surprisingly complex Recall our reduction rule for application: (λx.e 1 ) e 2 e 1 [x := e 2 ] This function application reduces to e 1 (the function body) where every x in e 1 is substituted with e 2 (value we re applying func to) Where did we go wrong? When we substituted: (λf.(λx. f (f x)) (λy.y+x) (λx. (λy.y+x) ((λy.y+x) x) the x is captured!
53 Another way to see the problem Syntactic sugar: let x = e 1 in e 2 (λx.e 2 ) e 1 Let syntax makes this easy to see: let x = a+b in let x = a+b in let a = 7 in let a = 7 in x + a (a+b) + a Very obviously wrong! But, guess what: your C macro preprocessor does this!
54 Another way to see the problem Syntactic sugar: let x = e 1 in e 2 (λx.e 2 ) e 1 Let syntax makes this easy to see: let x = a+b in let x = a+b in let a = 7 in let a = 7 in x + a (a+b) + a Very obviously wrong! But, guess what: your C macro preprocessor does this!
55 Fixing the problem How can we fix this? 1.Rename variables! let x = a+b in let x = a+b in let x = a+b in let a = 7 in let a123 = 7 in let a123 = 7 in x + a x + a123 (a+b) + a123 2.Do the dumb substitution!
56 Fixing the problem How can we fix this? 1.Rename variables! let x = a+b in let x = a+b in let x = a+b in let a = 7 in let a123 = 7 in let a123 = 7 in x + a x + a123 (a+b) + a123 2.Do the dumb substitution!
57 Fixing the problem How can we fix this? 1.Rename variables! let x = a+b in let x = a+b in let x = a+b in let a = 7 in let a123 = 7 in let a123 = 7 in x + a x + a123 (a+b) + a123 2.Do the dumb substitution!
58 Why is this the way to go? We can always rename bound variables! Def: variable x is bound in λx.(x+y) Bound variables are just placeholders Above: x is not special, we could have used z We say they are equivalent: λx.(x+y) =α λz.(z+y) Renaming amounts to converting bound variable names to avoid capture: e.g., λx.(x+y) to λz.(z+y)
59 Can we rename everything? Can we rename y in λx.(x+y)? (A: yes, B: no) No! We don t know what y may be, so we must keep it as is! Intuition: Can change the name of your function argument variables but not of variables from the outer scope E.g., x. P(x, y) or i {1,,10} x i + y
60 Can we rename everything? Can we rename y in λx.(x+y)? (A: yes, B: no) No! We don t know what y may be, so we must keep it as is! Intuition: Can change the name of your function argument variables but not of variables from the outer scope E.g., x. P(x, y) or i {1,,10} x i + y
61 Can we rename everything? Can we rename y in λx.(x+y)? (A: yes, B: no) No! We don t know what y may be, so we must keep it as is! Intuition: Can change the name of your function argument variables but not of variables from the outer scope E.g., x. P(x, y) or i {1,,10} x i + y
62 Let s think about this more formally
63 Def: free variables If a variable is not bound by a λ, we say that it is free e.g., y is free in λx.(x+y) is x free? No! We say x is bound in λx.(x+y) We can compute the free variables of any term: FV(x) = {x} FV(λx.e) = FV(e) \ {x} FV(e 1 e 2 ) = FV(e 1 ) FV(e 2 )
64 Def: free variables If a variable is not bound by a λ, we say that it is free e.g., y is free in λx.(x+y) is x free? No! We say x is bound in λx.(x+y) We can compute the free variables of any term: FV(x) = {x} FV(λx.e) = FV(e) \ {x} FV(e 1 e 2 ) = FV(e 1 ) FV(e 2 )
65 Def: free variables If a variable is not bound by a λ, we say that it is free e.g., y is free in λx.(x+y) is x free? No! We say x is bound in λx.(x+y) We can compute the free variables of any term: FV(x) = {x} FV(λx.e) = FV(e) \ {x} FV(e 1 e 2 ) = FV(e 1 ) FV(e 2 )
66 Def: free variables If a variable is not bound by a λ, we say that it is free e.g., y is free in λx.(x+y) is x free? No! We say x is bound in λx.(x+y) We can compute the free variables of any term: FV(x) = {x} FV(λx.e) = FV(e) \ {x} FV(e 1 e 2 ) = FV(e 1 ) FV(e 2 )
67 Def: Capture-avoiding substitution Capture-avoiding substitution: x[x:=e] = e y[x:=e] = y if y x (e1 e 2 )[x := e] = (e 1 [x := e]) (e 2 [ x:= e]) (λx.e 1 )[x := e] = λx.e 1 (λy.e1 )[x := e 2 ] = λy.e 1 [x := e 2 ] if y x and y FV(e 2 ) Why the if? If y is free in e2 this would capture it!
68 Def: Capture-avoiding substitution Capture-avoiding substitution: x[x:=e] = e y[x:=e] = y if y x (e1 e 2 )[x := e] = (e 1 [x := e]) (e 2 [ x:= e]) (λx.e 1 )[x := e] = λx.e 1 (λy.e1 )[x := e 2 ] = λy.e 1 [x := e 2 ] if y x and y FV(e 2 ) Why the if? If y is free in e2 this would capture it!
69 Def: Capture-avoiding substitution Capture-avoiding substitution: x[x:=e] = e y[x:=e] = y if y x (e1 e 2 )[x := e] = (e 1 [x := e]) (e 2 [ x:= e]) (λx.e 1 )[x := e] = λx.e 1 (λy.e1 )[x := e 2 ] = λy.e 1 [x := e 2 ] if y x and y FV(e 2 ) Why the if? If y is free in e2 this would capture it!
70 Def: Capture-avoiding substitution Capture-avoiding substitution: x[x:=e] = e y[x:=e] = y if y x (e1 e 2 )[x := e] = (e 1 [x := e]) (e 2 [ x:= e]) (λx.e 1 )[x := e] = λx.e 1 (λy.e1 )[x := e 2 ] = λy.e 1 [x := e 2 ] if y x and y FV(e 2 ) Why the if? If y is free in e2 this would capture it!
71 Def: Capture-avoiding substitution Capture-avoiding substitution: x[x:=e] = e y[x:=e] = y if y x (e1 e 2 )[x := e] = (e 1 [x := e]) (e 2 [ x:= e]) (λx.e 1 )[x := e] = λx.e 1 (λy.e1 )[x := e 2 ] = λy.e 1 [x := e 2 ] if y x and y FV(e 2 ) Why the if? If y is free in e2 this would capture it!
72 Def: Capture-avoiding substitution Capture-avoiding substitution: x[x:=e] = e y[x:=e] = y if y x (e1 e 2 )[x := e] = (e 1 [x := e]) (e 2 [ x:= e]) (λx.e 1 )[x := e] = λx.e 1 (λy.e1 )[x := e 2 ] = λy.e 1 [x := e 2 ] if y x and y FV(e 2 ) Why the if? If y is free in e2 this would capture it!
73 Def: Capture-avoiding substitution Capture-avoiding substitution: x[x:=e] = e y[x:=e] = y if y x (e1 e 2 )[x := e] = (e 1 [x := e]) (e 2 [ x:= e]) (λx.e 1 )[x := e] = λx.e 1 (λy.e1 )[x := e 2 ] = λy.e 1 [x := e 2 ] if y x and y FV(e 2 ) Why the if? If y is free in e2 this would capture it!
74 Def: Capture-avoiding substitution Capture-avoiding substitution: x[x:=e] = e y[x:=e] = y if y x (e1 e 2 )[x := e] = (e 1 [x := e]) (e 2 [ x:= e]) (λx.e 1 )[x := e] = λx.e 1 (λy.e1 )[x := e 2 ] = λy.e 1 [x := e 2 ] if y x and y FV(e 2 ) Why the if? If y is free in e2 this would capture it!
75 Lambda calculus: equational theory α-renaming or α-conversion λx.e = λy.e[x:=y] where y FV(e) β-reduction (λx.e1 ) e 2 = e 1 [x:=e 2 ] η-conversion λx.(e x) = e where x FV(e) We define our relation using these equations!
76 Back to our example (what we should ve done)
77 Back to our example (what we should ve done) Instead of 1, let s add x to argument (and do it 2x): (λf.(λx. f (f x)) (λy.y+x)
78 Back to our example (what we should ve done) Instead of 1, let s add x to argument (and do it 2x): (λf.(λx. f (f x)) (λy.y+x) =α (λf.(λz. f (f z)) (λy.y+x)
79 Back to our example (what we should ve done) Instead of 1, let s add x to argument (and do it 2x): (λf.(λx. f (f x)) (λy.y+x) =α (λf.(λz. f (f z)) (λy.y+x) =β λz. (λy.y+x) ((λy.y+x) z)
80 Back to our example (what we should ve done) Instead of 1, let s add x to argument (and do it 2x): (λf.(λx. f (f x)) (λy.y+x) =α (λf.(λz. f (f z)) (λy.y+x) =β λz. (λy.y+x) ((λy.y+x) z) =β λz. (λy.y+x) (z+x)
81 Back to our example (what we should ve done) Instead of 1, let s add x to argument (and do it 2x): (λf.(λx. f (f x)) (λy.y+x) =α (λf.(λz. f (f z)) (λy.y+x) =β λz. (λy.y+x) ((λy.y+x) z) =β λz. (λy.y+x) (z+x) =β λz. z+x+x
82 Today Syntax of λ calculus Semantics of λ calculus Free and bound variables Substitution Evaluation order
83 Evaluation order What should we reduce first in (λx.x) ((λy.y) z)? A: The outer term: (λy.y) z B: The inner term: (λx.x) z Does it matter? No! They both reduce to z! Church-Rosser Theorem: If you reduce to a normal form, it doesn t matter what order you do the reductions. This is known as confluence.
84 Evaluation order What should we reduce first in (λx.x) ((λy.y) z)? A: The outer term: (λy.y) z B: The inner term: (λx.x) z Does it matter? No! They both reduce to z! Church-Rosser Theorem: If you reduce to a normal form, it doesn t matter what order you do the reductions. This is known as confluence.
85 Evaluation order What should we reduce first in (λx.x) ((λy.y) z)? A: The outer term: (λy.y) z B: The inner term: (λx.x) z Does it matter? No! They both reduce to z! Church-Rosser Theorem: If you reduce to a normal form, it doesn t matter what order you do the reductions. This is known as confluence.
86 Does evaluation order really not matter?
87 Does evaluation order really not matter? Consider a curious term called Ω Ω (λx.x x) (λx.x x)
88 Does evaluation order really not matter? Consider a curious term called Ω Ω (λx.x x) (λx.x x) =β (x x)[ x:= (λx.x x)]
89 Does evaluation order really not matter? Consider a curious term called Ω Ω (λx.x x) (λx.x x) =β (x x)[ x:= (λx.x x)] =β (λx.x x) (λx.x x)
90 Does evaluation order really not matter? Consider a curious term called Ω Ω (λx.x x) (λx.x x) =β (x x)[ x:= (λx.x x)] =β (λx.x x) (λx.x x) = Ω Deja vu!
91 Ω Ω Ω Ω Ω Ω (Ω has no normal form)
92 Does evaluation order really not matter? Consider a function that ignores its argument: (λx.y) What happens when we call it on Ω? (λx.y) Ω
93 Does evaluation order really not matter? Consider a function that ignores its argument: (λx.y) What happens when we call it on Ω? y (λx.y) Ω
94 Does evaluation order really not matter? Consider a function that ignores its argument: (λx.y) What happens when we call it on Ω? (λx.y) Ω y (λx.y) Ω
95 Does evaluation order really not matter? Consider a function that ignores its argument: (λx.y) What happens when we call it on Ω? (λx.y) Ω y (λx.y) Ω y
96 Does evaluation order really not matter? Consider a function that ignores its argument: (λx.y) What happens when we call it on Ω? (λx.y) Ω y (λx.y) Ω y (λx.y) Ω
97 Does evaluation order really not matter? Consider a function that ignores its argument: (λx.y) What happens when we call it on Ω? y y y y (λx.y) Ω (λx.y) Ω (λx.y) Ω (λx.y) Ω
98 Does evaluation order really not matter? Nope! Evaluation order does matter!
99 Call-by-value Reduce function, then reduce args, then apply e 1 e 2 (λx.e 1 ) e 2 (λx.e 1 ) n e 1 [x:=n] JavaScript s evaluation strategy is call-by-value (ish) What does this program do? (x => 33) ((x => x(x)) (x => x(x))) RangeError: Maximum call stack size exceeded
100 Call-by-value Reduce function, then reduce args, then apply e 1 e 2 (λx.e 1 ) e 2 (λx.e 1 ) n e 1 [x:=n] JavaScript s evaluation strategy is call-by-value (ish) What does this program do? (x => 33) ((x => x(x)) (x => x(x))) RangeError: Maximum call stack size exceeded
101 Call-by-value Reduce function, then reduce args, then apply e 1 e 2 (λx.e 1 ) e 2 (λx.e 1 ) n e 1 [x:=n] JavaScript s evaluation strategy is call-by-value (ish) What does this program do? (x => 33) ((x => x(x)) (x => x(x))) RangeError: Maximum call stack size exceeded
102 Call-by-value Reduce function, then reduce args, then apply e 1 e 2 (λx.e 1 ) e 2 (λx.e 1 ) n e 1 [x:=n] JavaScript s evaluation strategy is call-by-value (ish) What does this program do? (x => 33) ((x => x(x)) (x => x(x))) RangeError: Maximum call stack size exceeded
103 Call-by-value Reduce function, then reduce args, then apply e 1 e 2 (λx.e 1 ) e 2 (λx.e 1 ) n e 1 [x:=n] JavaScript s evaluation strategy is call-by-value (ish) What does this program do? (x => 33) ((x => x(x)) (x => x(x))) RangeError: Maximum call stack size exceeded
104 Call-by-name Reduce function, then reduce args, then apply e 1 e 2 (λx.e 1 ) e 2 e 1 [x:=e 2 ] Haskell s evaluation strategy is call-by-name It only does what is absolutely necessary!
105 Call-by-name Reduce function, then reduce args, then apply e 1 e 2 (λx.e 1 ) e 2 e 1 [x:=e 2 ] Haskell s evaluation strategy is call-by-name It only does what is absolutely necessary!
106 Call-by-name Reduce function, then reduce args, then apply e 1 e 2 (λx.e 1 ) e 2 e 1 [x:=e 2 ] Haskell s evaluation strategy is call-by-name It only does what is absolutely necessary!
107 Call-by-name Reduce function, then reduce args, then apply e 1 e 2 (λx.e 1 ) e 2 e 1 [x:=e 2 ] Haskell s evaluation strategy is call-by-name It only does what is absolutely necessary!
108 Summary A term may have many redexes (subterms can reduce) Evaluation strategy says which redex to evaluate Evaluation not guaranteed to find normal form Call-by-value: evaluate function & args before β reduce Call-by-name: evaluate function, then β-reduce
109 Today Syntax of λ calculus Semantics of λ calculus Free and bound variables Substitution Evaluation order
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