Hindley-Milner Type Checking Automatic Type Inference What can be - - PowerPoint PPT Presentation

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Hindley-Milner Type Checking Automatic Type Inference What can be - - PowerPoint PPT Presentation

Jul 17, 2023 565 likes •1.32k views

CSE340 Principles of Programming Languages Hindley-Milner Type Checking Automatic Type Inference What can be inferred about type of f or x from this definition? f(x) = x Automatic Type Inference f(x) = x f is a function that takes a single

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CSE340 Principles of Programming Languages

Hindley-Milner Type Checking

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f(x) = x

Automatic Type Inference

What can be inferred about type of f or x from this definition?

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SLIDE 3

f(x) = x

Automatic Type Inference

f is a function that takes a single

  • argument. So the type of f can be

described as: T1(*)(T2)

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f(x) = x

Automatic Type Inference

The return value of f is equal to its input, so their types must match: T1 = T2

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f(x) = x

Automatic Type Inference

So f is a function that takes one argument and its return type is the same as its argument’s type. Therefore type of f is: T1(*)(T1)

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f(x) = x

Automatic Type Inference

And we don’t know anything about the type of x So f is a function that takes one argument and its return type is the same as its argument’s type. Therefore type of f is: T1(*)(T1)

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g(x) = x + 1

Automatic Type Inference

How about function g?

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g(x) = x + 1

Automatic Type Inference

What can be inferred from this term?

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g(x) = x + 1

Automatic Type Inference

x is used in an arithmetic expression involving the integer constant 1. So x must be of integer type

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g(x) = x + 1

Automatic Type Inference

So the type of function g should be further restricted to: int(*)(int)

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To perform Hindley-Milner type checking:

  • Start by generating the abstract syntax tree of the function
  • Assume unknown types for arguments: T1, T2, …
  • Examine the tree nodes and apply type constraints to further restrict the

types

  • The type constraints that can be applied depend on the programming

language used. In the following examples we use simple rules similar to those in functional languages like OCaml

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Examples

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f(a,b,c,d) = if b = 1 then if a[1](0) then c(1) else c(d) else if a[b](0) then c(b) else f(a, b - 1, c, d)

Example #1

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f(a,b,c,d) = if b = 1 then if a[1](0) then c(1) else c(d) else if a[b](0) then c(b) else f(a, b - 1, c, d)

Example #1

f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

Condition Then Else

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

Condition Then Else

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

Condition Then Else

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

The function Parameters in order from left to right

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f T1(*)(T2,T3,T4,T5) a T2 b T3 c T4 d T5

f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

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Top-Down order

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

f T1(*)(T2,T3,T4,T5) a T2 b T3 c T4 d T5

T1

The return type of the function should be the same as the type of the if node

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

f T1(*)(T2,T3,T4,T5) a T2 b T3 c T4 d T5

T1 bool

The condition should be of type boolean

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

f T1(*)(T2,T3,T4,T5) a T2 b T3 c T4 d T5

T1 bool T1

The then node should be of the same type as the if node

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

f T1(*)(T2,T3,T4,T5) a T2 b T3 c T4 d T5

T1 bool T1 T1

The else node should be of the same type as the if node

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int

f T1(*)(T2,int,T4,T5) a T2 b int c T4 d T5

The operands of a comparison

  • perator (=) should be of the same
  • type. The right operand is int, so b

should be int too i.e. T3 = int

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool

f T1(*)(T2,int,T4,T5) a T2 b int c T4 d T5

The condition should be of type boolean

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T4,T5) a T2 b int c T4 d T5

The then node should be of the same type as the if node

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T4,T5) a T2 b int c T4 d T5

T1

The else node should be of the same type as the if node

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T4,T5) a T2 b int c T4 d T5

T1 bool

The condition should be of type boolean

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T4,T5) a T2 b int c T4 d T5

T1 bool T1

The then node should be of the same type as the if node

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T4,T5) a T2 b int c T4 d T5

T1 bool T1 T1

The else node should be of the same type as the if node

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T4,T5) a T2 b int c T4 d T5

T1 bool T1 T1 bool(*)(int) int

The left-most child node of an apply node must be a function, the other children are the parameters passed to that

  • function. The return type of the function is

the type of the apply node

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T1(*)(int),T5) a T2 b int c T1(*)(int) d T5

T1 bool T1 T1 bool(*)(int) int int T1(*)(int)

c must be a function that takes one integer argument and returns a value of type T1, i.e. T4 = T1(*)(int)

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T1(*)(int),int) a T2 b int c T1(*)(int) d int

T1 bool T1 T1 bool(*)(int) int int T1(*)(int) int

We know that c is a function that takes an integer as argument and returns

  • T1. The argument passed here is d, so d

must be of type int, i.e. T5 = int

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T1(*)(int),int) a T2 b int c T1(*)(int) d int

T1 bool T1 T1 bool(*)(int) int int T1(*)(int) int int bool(*)(int)

The type of [] node must be a function that takes an integer as argument and returns boolean, i.e. bool(*)(int)

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T1(*)(int),int) a T2 b int c T1(*)(int) d int

T1 bool T1 T1 bool(*)(int) int int T1(*)(int) int int bool(*)(int)

We know that c is a function that takes an integer as argument and returns T1, we also know that b is integer

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(T2,int,T1(*)(int),int) a T2 b int c T1(*)(int) d int

T1 bool T1 T1 bool(*)(int) int int T1(*)(int) int int bool(*)(int) int

We know that f is a function that takes 4 arguments of types T2, int, T1(*)(int) and int and returns a value of type T1. We know that a is of type T2, and c is of type T1(*)(int) and d is of type int. The - node should be of type int

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(array(bool(*)(int)),int,T1(*)(int),int) a array(bool(*)(int)) b int c T1(*)(int) d int

T1 bool T1 T1 bool(*)(int) int int T1(*)(int) int int bool(*)(int) int

Since the indexing operator is applied to a, it must be an array. Each element of the array should be the same type as the [] node, i.e. T2 = array(bool(*)(int))

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(array(bool(*)(int)),int,T1(*)(int),int) a array(bool(*)(int)) b int c T1(*)(int) d int

T1 bool T1 T1 bool(*)(int) int int T1(*)(int) int int bool(*)(int) int

b is used as an index value, it must be int (we already know that —> no conflict). Also a must be an array of bool(*)(int) which is consistent with what we already inferred

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(array(bool(*)(int)),int,T1(*)(int),int) a array(bool(*)(int)) b int c T1(*)(int) d int

T1 bool T1 T1 bool(*)(int) int int T1(*)(int) int int bool(*)(int) int

b is an operand of an arithmetic

  • peration involving an integer (1), so it must

be int which is consistent with what we already

  • inferred. Also the type of - node is consistent

with the types of operands

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f(a,b,c,d) def if = b 1 if if apply [] a 1 apply 1 c apply d c apply [] a b apply b c apply f a

  • b

1 c d

T1 bool T1 T1 int int bool T1

f T1(*)(array(bool(*)(int)),int,T1(*)(int),int) a array(bool(*)(int)) b int c T1(*)(int) d int

T1 bool T1 T1 bool(*)(int) int int T1(*)(int) int int bool(*)(int) int

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g(a,b,c) = if a(1) = 3.5 then b - 1 else if c[5] < a(0) then else g(a, b * 2, c)

Example #2

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g T1(*)(T2,T3,T4) a T2 b T3 c T4

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Bottom-Up order

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g T1(*)(T2,int,T4) a T2 b int c T4

b must be an integer, i.e. T3 = int

int int int

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g T1(*)(T5(*)(int),int,T4) a T5(*)(int) b int c T4

a must be a function that takes an integer as argument and we don’t know its return type yet, i.e. T2 = T5(*)(int)

int int int T5(*)(int) T5

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g T1(*)(T5(*)(int),int,array(T6)) a T5(*)(int) b int c array(T6)

c must be an array since it is the left child of an indexing node. The type

  • f the elements of the array

are not known yet, i.e. T4 = array(T6)

int int int T5(*)(int) T5 array(T6) T6

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g T1(*)(T5(*)(int),int,array(T5)) a T5(*)(int) b int c array(T5)

The operands of a comparison operator must be of the same type, i.e. T6 = T5

int int int T5(*)(int) T5 array(T5) T6 = T5

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g T1(*)(T5(*)(int),int,array(T5)) a T5(*)(int) b int c array(T5)

Recursive call to g with no conflict. The type of the apply node must be the same as return type of g, i.e. T1

int int int T5(*)(int) T5 array(T5) T5 T1

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g T1(*)(T5(*)(int),int,array(T5)) a T5(*)(int) b int c array(T5)

int int int T5(*)(int) T5 array(T5) T5 T1

We know that a is a function that takes a single integer argument and returns a value of type T5

T5

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g T1(*)(float(*)(int),int,array(float)) a float(*)(int) b int c array(float)

int int int float(*)(int) float array(float) float T1

The operands of a comparison operator must be

  • f the same type, so T5

should be float

T5=float float

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g T1(*)(float(*)(int),int,array(float)) a float(*)(int) b int c array(float)

int int int float(*)(int) float array(float) float T1

b is already known to be of type int, no conflict. Also the type of the “-” node must be int

float float int

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g int(*)(float(*)(int),int,array(float)) a float(*)(int) b int c array(float)

int int int float(*)(int) float array(float) float T1 = int float float int bool int int

The condition of an if node must be of type boolean which it is. The then and else nodes must have the same type as the if node. So T1 = int

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g int(*)(float(*)(int),int,array(float)) a float(*)(int) b int c array(float)

int int int float(*)(int) float array(float) float int float float int bool int int int bool

The condition of an if node must be of type boolean which it is. The then and else nodes must have the same type as the if node. No conflict

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g int(*)(float(*)(int),int,array(float)) a float(*)(int) b int c array(float)

int int int float(*)(int) float array(float) float int float float int bool int int int bool

The return type of the function g must be the same as the type of the if node which it is —> No conflict

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g(a,b,c) def if = a 3.5 if apply g a * b 2 c apply 1 b 1

  • <

[] c 5 a apply

g int(*)(float(*)(int),int,array(float)) a float(*)(int) b int c array(float)

int int int float(*)(int) float array(float) float int float float int bool int int int bool

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h(x,y,z) = if x > 1 then y * 2.0 else z + h(y, x, z)

Example #3

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h(x,y,z) def if > x + apply h y x z 1 y 2.0 * z

h T1(*)(T2,T3,T4) x T2 y T3 z T4

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Random order

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h(x,y,z) def if > x + apply h y x z 1 y 2.0 * z

h T1(*)(T2,float,T4) x T2 y float z T4

y must be float since it is used in an arithmetic operation involving 2.0, hence T3 = float

float float float

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h(x,y,z) def if > x + apply h y x z 1 y 2.0 * z

h T1(*)(T2,float,T4) x T2 y float z T4

The condition node must be boolean and the then and else nodes should be of the same type as the if node

float float float float float bool

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h(x,y,z) def if > x + apply h y x z 1 y 2.0 * z

h float(*)(T2,float,T4) x T2 y float z T4

The return type of h must be the same as the type of the if node, i.e. T1 = float

float float float float float bool

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h(x,y,z) def if > x + apply h y x z 1 y 2.0 * z

h float(*)(int,float,T4) x int y float z T4

float float float float float bool

x must be int since it is compared with integer constant 1, i.e. T2 = int

int

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h(x,y,z) def if > x + apply h y x z 1 y 2.0 * z

h float(*)(int,float,float) x int y float z float

float float float float float bool

z must be float since the result of the arithmetic operation is

  • f type float. Also the apply node

must be of type float

int float float

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h(x,y,z) def if > x + apply h y x z 1 y 2.0 * z

h float(*)(int,float,float) x int y float z float

float float float float float bool int float float float float int float(*)(int,float,float)

Call to h with the following argument types: float, int, float —> Type Mismatch

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Type Constraints

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Function Definitions

def function body

T T(*)(…)

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If-Then-Else

if Condition Else Then

bool T T T

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Arithmetic Expressions

  • p

left right

T T T

  • p must be an arithmetic
  • perator: + - * /
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Comparisons

  • p

left right

bool T T

  • p must be a comparison
  • perator: < > = <= etc.
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Function Calls

apply function param1

T1

param2 param3

T2 T3 … T0(*)(T1,T2,T3,…) T0

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SLIDE 73

Array Indexing

[] array index

int array(T) T