역사의 흐름을 따라! 우리 언어에 추가할 것dreameye/pl/slide/pl4.pdf ·...

우리 언어에 추가할 것

•환경 추가 (유효 범위를 지원하기 위해)

•프로시져 (procedure) 추가 •Static & Dynamic Scoping

•Call-by-value & Call-by-reference

•Return Value

•Recursive Call

•레코드 (record) 추가

•포인터 (pointer) 추가

역사의 흐름을 따라!

기계 중심의

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ ∈ Env = Id → AddrM ∈ Mem = Addr → Val

let x := Ein S

{int x = E;S

}

Ids1−1←→ Addr ←→ Val

for x := E1 to E2 do Sx := E1;high := E2;while not(high < x) do

S;x := x+ 1

The variable high should not appear in S.

1

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ,M � S ⇓ M

σ,M � E ⇓ v

σ,M � x := E ⇓ M [v/σ(x)]

σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S ⇓ M1

σ,M � let x := E in S ⇓ M1


let x := Ein S

{int x = E;S

}



S;x := x+ 1


1

환경과 메모리Env Mem

xabc

1T3F

환경과 메모리

let x := 0in

let x := 10in

x := x + 1;x := x + 1

let x := 0in

let y := malloc(1)in

*y = x; x := &y;write **x

dom(M �)

Env {x �→ �1, y �→ �2}Memory {�2 �→ �3, �3 �→ 0, �1 �→ �2}

Env Memory{x �→ �1} {�1 �→ 1}

{x �→ �1, f �→ �y, x := x+ y, {x �→ �1}�} {�1 �→ 1}{f �→ �y, x := x+ y, {x �→ �1}�, x �→ �2} {�1 �→ 1, �2 �→ 2}{f �→ �y, x := x+ y, {x �→ �1}�, x �→ �2} {�1 �→ 4, �2 �→ 2}{x �→ �2, f �→ �y, x := x+ y, {x �→ �1}�} {�1 �→ 6, �2 �→ 2}{x �→ �1, f �→ �y, x := x+ y, {x �→ �1}�} {�1 �→ 7, �2 �→ 2}

{x �→ �1}{�1 �→ 7, �2 �→ 2, �3 �→ 3, �4 �→ 2}

Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}

r ∈ Record = Field → Addrn ∈ Zb ∈ B = {T, F}v ∈ Val = Z + B + {·}+ Recordσ ∈ Env = Id → Addr + ProcedureM ∈ Mem = Addr → Val

Procedure = Id × E × Env� ∈ Addr = Nf ∈ Field = Id

1

C SEnv Mem{} {}{x �→ �} {� �→ 0}{x �→ ��} {� �→ 0, �� → 10}{x �→ ��} {� �→ 0, �� → 11}{x �→ �} {� �→ 0, �� → 11}{x �→ �} {� �→ 1, �� → 11}

let x := 0inlet x := 10in

x := x + 1;x := x + 1

let x := 0in



dom(M �)

Env {x �→ �1, y �→ �2}Memory {�2 �→ �3, �3 �→ 0, �1 �→ �2}

Env Memory{x �→ �1} {�1 �→ 1}


{x �→ �1}{�1 �→ 7, �2 �→ 2, �3 �→ 3, �4 �→ 2}

Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}

1

σ,M � S1 ⇓ M1 σ,M1 � S2 ⇓ M2

σ,M � S1;S2 ⇓ M2

Env Mem{} {}{x �→ �} {� �→ 0}{x �→ ��} {� �→ 0, �� → 10}{x �→ ��} {� �→ 0, �� → 11}{x �→ �} {� �→ 0, �� → 11}{x �→ �} {� �→ 1, �� → 11}


x := x + 1;x := x + 1

let x := 0in



dom(M �)

Env {x �→ �1, y �→ �2}Memory {�2 �→ �3, �3 �→ 0, �1 �→ �2}

Env Memory{x �→ �1} {�1 �→ 1}


{x �→ �1}{�1 �→ 7, �2 �→ 2, �3 �→ 3, �4 �→ 2}

1

Procedure

•기계 중심의 프로그래밍 언어에서 procedure란 명령문들에 이름을 붙여 놓은 것

•인자를 가지고 어느 정도 일반화 시킨 것

procedure f(x) = S f(E)

Procedure 선언은 프로그램의 어디서 하도록 할까?

Procedure에 있는 자유로운 변수들은 어떻게 하나?

재사용가독성

문제 쪼개기

Syntax

ProgramP → S

StatementS → x := E

| S;S| if E then S else S| while E do S| for x := E to E do S| read x| write E| skip

StatementS → ...

| let procedure f(x) = S in S| call f(E)

ExpressionE → n | true | false | x

| E + E | E − E | E ∗ E | E/E| E = E | E < E | not E

let x := 1 inlet y := 2

in x + y

let x := 1 inx + y

x := 0;y := 0;while x < 11 do

y := y + x;x := x + 1;

write y

for x:= 0 to 10 doy := y + x;

write y

Γ ∈ TypeEnv = Id → Type

Type → τ | Int | Bool | Type → Type

4

어떤 x가 증가할까?

let x := 1 inlet procedure addx(y) = x := x + y in...addx(3)

let x := 1 inlet procedure addx(y) = x := x + y in

let x := 2in

addx(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ,M � E ⇓ v

σ,M � x := E ⇓ M [v/σ(x)]

σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S ⇓ M1



let x := Ein S

{int x = E;S

}



S;x := x+ 1


1

let x := 1 inlet procedure addx(y) = x := x + yin

addx(3)


let x := 2in

addx(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ,M � E ⇓ v

σ,M � x := E ⇓ M [v/σ(x)]

σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S ⇓ M1



let x := Ein S

{int x = E;S

}



S;x := x+ 1


1

Static Scoping

•프로그램에서 이름은 이제 메모리 주소나 procedure를 지칭한다.

σ ∈ Env = Id → Addr + Procedure


addx(3)


let x := 2in

addx(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ,M � E ⇓ v

σ,M � x := E ⇓ M [v/σ(x)]

σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S ⇓ M1



let x := Ein S

{int x = E;S

}



S;x := x+ 1

1

int x = 1;void f(int y){

x = x + y;}

{int x = 2;f(3);

};x = x + 1;

let x := 1in call x(2)

let procedure f(x) = x + 10in f + 3

let x := 1in

let y := truein x + y

while 1 do E

while (while x < 10 do x := x + 1) do E

if {name := 1 } then 2 else skip

σ,M � S1 ⇓ M1 σ,M1 � S2 ⇓ M2

σ,M � S1;S2 ⇓ M2

Env Mem{} {}{x �→ �} {� �→ 0}{x �→ ��} {� �→ 0, �� → 10}{x �→ ��} {� �→ 0, �� → 11}{x �→ �} {� �→ 0, �� → 11}{x �→ �} {� �→ 1, �� → 11}


x := x + 1;

1

Procedure = Id × S × Env

σ,M � E ⇓ v,M �

σ,M � x := E ⇓ ·,M �[v/σ(x)]

σ,M � E ⇓1 v,M1 � /∈ dom(M1) σ[�/x],M1[v/�] � E2 ⇓ v2,M2

σ,M � let x := E1 in E2 ⇓ v2,M2

σ[�x,E1,σ�/f ],M � E2 ⇓ v,M1

σ,M � let procedure f(x) = E1 in E2 ⇓ v,M1

σ(f) = �x,E1,σ1� σ,M � E ⇓ v,M1 � /∈ dom(M) σ1[�/x],M1[v/�] � E1 ⇓ v�,M2

σ,M � call f(E) ⇓ v�,M2

σ(f) = �x,E1,σ1� σ,M � E ⇓ v,M1 � /∈ dom(M)σ1[�x,E1,σ1�/f ][�/x],M1[v/�] � E1 ⇓ v�,M2


σ(f) = �y, E1,σ1� σ1[σ(x)/y],M � E1 ⇓ v,M �

σ,M � call f�x� ⇓ v,M �


Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1

σ,M � let procedure f(x) = S1 in S2 ⇓ M1

σ(f) = �x, S1� σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S1 ⇓ M1

σ,M � call f(E) ⇓ M1

let x := 1 inlet procedure f(y) = x := x + y in

let x := 2in

f(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

5

C style

Static Scopingσ ∈ Env = Id → Addr + Procedure


σ[�x, S1,σ�/f ],M � S2 ⇓ M1


σ(f) = �x, S1,σ1� σ,M � E ⇓ v � /∈ dom(M) σ1[�/x],M [v/�] � S1 ⇓ M1



Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)


let x := 2in

addx(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ,M � E ⇓ v

σ,M � x := E ⇓ M [v/σ(x)]

σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S ⇓ M1



1

Procedure가 정의될 때의 환경을 같이 저장한다!

이름의 실체가 실행 전에 결정됨!

Env Memory{x �→ �1} , {�1 �→ 1}

{x �→ �1, f �→ �y, x := x+ y, {x �→ �1}�} , {�1 �→ 1}{f �→ �y, x := x+ y, {x �→ �1}�, x �→ �2} , {�1 �→ 1, �2 �→ 2}



Record = Field → Addr

age name

{left �→ �1, content �→ �2, right �→ �3}

let x := 1in

let procedure f(y) =x := x + y

inlet x := 2in

f(3);f(2);

x := x + 1

let x := 0in

let procedure sum(n) =for i := 1 to n do

x := x + iin

call sum(10);write x+1;

let x := {height:= 176, weight = 89}in

x.height + x.weight

let tree := {left := {}, v := 0, right := {}}

1

S

C

Env Memory{x �→ �1} {�1 �→ 1}

{x �→ �1, f �→ �y, x := x+ y, {x �→ �1}�} {�1 �→ 1}{f �→ �y, x := x+ y, {x �→ �1}�, x �→ �2} {�1 �→ 1, �2 �→ 2}{f �→ �y, x := x+ y, {x �→ �1}�, x �→ �2} {�1 �→ 4, �2 �→ 2}{f �→ �y, x := x+ y, {x �→ �1}�, x �→ �2} {�1 �→ 6, �2 �→ 2}{x �→ �1, f �→ �y, x := x+ y, {x �→ �1}�} {�1 �→ 7, �2 �→ 2}

Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}




age name


let x := 1in


inlet x := 2in

f(3);f(2);

x := x + 1

let x := 0in


x := x + i

1

Env Memory{x �→ �1} {�1 �→ 1}


Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}




age name


let x := 1in


inlet x := 2in

f(3);f(2);

x := x + 1

let x := 0in


x := x + i

1

메모리 생존 기간Memory Cell’s Lifetime

}}}x

xf

생존기간은 scope와 일치

우리 언어가 커지면 일치하지 않을수도!

Env Memory{x �→ �1} , {�1 �→ 1}

{x �→ �1, f �→ �y, x := x+ y, {x �→ �1}�} , {�1 �→ 1}{f �→ �y, x := x+ y, {x �→ �1}�, x �→ �2} , {�1 �→ 1, �2 �→ 2}




age name


let x := 1in


inlet x := 2in

f(3);f(2);

x := x + 1

let x := 0in


x := x + iin



x.height + x.weight


1

Env Memory{x �→ �1} {�1 �→ 1}


{x �→ �1}{�1 �→ 7, �2 �→ 2, �3 �→ 3, �4 �→ 2}

Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}




age name


let x := 1in


inlet x := 2in

f(3);f(2);

x := x + 1

let x := 0in

let procedure sum(n) =

1

}

버려 버려!

Env Memory{x �→ �1} {�1 �→ 1}


{x �→ �1}{�1 �→ 7, �2 �→ 2, �3 �→ 3, �4 �→ 2}

Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}




age name


let x := 1in


inlet x := 2in

f(3);f(2);

x := x + 1

let x := 0in

let procedure sum(n) =

1

환경

메모리

프로시져에 붙어 있는 환경도 고려

Env Memory{x �→ �1} {�1 �→ 1}


Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}




age name


let x := 1in


inlet x := 2in

f(3);f(2);

x := x + 1

let x := 0in


x := x + i

1

Env Memory{x �→ �1} , {�1 �→ 1}

{x �→ �1, f �→ �y, x := x+ y, {x �→ �1}�} , {�1 �→ 1}{f �→ �y, x := x+ y, {x �→ �1}�, x �→ �2} , {�1 �→ 1, �2 �→ 2}




age name


let x := 1in


inlet x := 2in

f(3);f(2);

x := x + 1

let x := 0in


x := x + iin



x.height + x.weight


1

S

C

인자 넘기는 방법Parameter Passing



σ[�x, S1,σ�/f ],M � S2 ⇓ M1





Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)


let x := 2in

addx(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ,M � E ⇓ v

σ,M � x := E ⇓ M [v/σ(x)]

σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S ⇓ M1



1

call-by-value값을 넘기고 있다!

인자 넘기는 방법Parameter Passing

call-by-reference어떨 땐 주소를 넘겨주고 싶은데?

let procedure reset(x) = x := 0in

let y := 1 in...; call reset(y);

let z := 2 in...; call reset(z);

...

let x := 0 inlet procedure inc(n) = x:= x + n in

call inc(1);let x := truein

call inc(2)



σ[�x, S1,σ�/f ],M � S2 ⇓ M1





Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)


let x := 2in

addx(3);x := x + 1;

1

좋아! 우리 언어에 추가하자!

인자의 주소를 넘기자call-by-reference

ProgramP → S



StatementS → ...

| let procedure f(x) = S in S| call f(E)| call f�x�




in x + y

let x := 1 inx + y

x := 0;y := 0;while x < 11 do

y := y + x;x := x + 1;

write y

for x:= 0 to 10 doy := y + x;

write y

Γ ∈ TypeEnv = Id → Type

Type → τ | Int | Bool | Type → Type

5


let y := 1 in...; call reset <y>;

let z := 2 in...; call reset <z>;

...



call inc(2)



σ[�x, S1,σ�/f ],M � S2 ⇓ M1




σ(f) = �y, S1,σ1� σ1[σ(x)/y],M � S1 ⇓ M1

σ,M � call f�x� ⇓ M1


Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)

1




...



call inc(2)



σ[�x, S1,σ�/f ],M � S2 ⇓ M1




σ(f) = �y, S1,σ1� σ1[σ(x)/y],M � S1 ⇓ M1



Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)

1




...



call inc(2)



σ[�x, S1,σ�/f ],M � S2 ⇓ M1




σ(f) = �y, S1,σ1� σ1[σ(x)/y],M � S1 ⇓ M1



Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)

1

별칭alias

함수가 돌려주는 값은?

let x := 0in


x := x + iin


let procedure sum(n) =let x := 0 in

for i := 1 to n dox := x + i;

return xin

write (call sum(10)) + 1




...



call inc(2)



σ[�x, S1,σ�/f ],M � S2 ⇓ M1




σ(f) = �y, S1,σ1� σ1[σ(x)/y],M � S1 ⇓ M1


1

let x := 0in


x := x + iin




return xin





...



call inc(2)



σ[�x, S1,σ�/f ],M � S2 ⇓ M1




σ(f) = �y, S1,σ1� σ1[σ(x)/y],M � S1 ⇓ M1


1

call f(E)나 call f<x>가 값을 돌려주도록 하자!

함수가 계산하는 값을 저장할 변수

명령문이 값을 돌려주면?


Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)


let x := 2in

addx(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ,M � E ⇓ v

σ,M � x := E ⇓ M [v/σ(x)]

σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S ⇓ M1



let x := Ein S

{int x = E;S

}


2

let x := 0in


x := x + iin




return xin





...



call inc(2)

σ,M � S ⇓ M �



σ[�x, S1,σ�/f ],M � S2 ⇓ M1




σ(f) = �y, S1,σ1� σ1[σ(x)/y],M � S1 ⇓ M1


1

에잇 합체! fusion!

let x := 0in


x := x + iin




return xin





...



call inc(2)

σ,M � E ⇓ v,M �



σ[�x, S1,σ�/f ],M � S2 ⇓ M1




σ(f) = �y, S1,σ1� σ1[σ(x)/y],M � S1 ⇓ M1


1

통일된 의미 구조

let x := 0in


x := x + iin




return xin





...



call inc(2)

σ,M � E ⇓ v,M �



σ[�x, S1,σ�/f ],M � S2 ⇓ M1




σ(f) = �y, S1,σ1� σ1[σ(x)/y],M � S1 ⇓ M1


1

•값을 돌려주던 E는 받은 메모리를 그대로 값과 같이 돌려주면 되겠다.

•메모리를 돌려주던 S는 어떤 값을 계산해 낸다고 정의 할까?

void/텅 빈 값으로 하자.


Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)


let x := 2in

addx(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ,M � E ⇓ v

σ,M � x := E ⇓ M [v/σ(x)]

σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S ⇓ M1



v ∈ Val = Z + B + {·}

let x := Ein S

{int x = E;S

}

2

혹시 텅 빈 값을 만들고 싶으면 ()를 쓰자.

통일된 SyntaxProgram

P → E

ExpressionE → x := E

| E;E| if E then E else E| while E do E| for x := E to E do E| read x| write E| skip| let procedure f(x) = E in E| call f(E)| call f�x�| n | true | false | x| E + E | E − E | E ∗ E | E/E| E = E | E < E | not E

ProgramP → S



StatementS → ...





in x + y

6

통일된 Semantics의미 공간

semantic domain

n ∈ Zb ∈ B = {T, F}v ∈ Val = Z + B + {·}σ ∈ Env = Id → Addr + ProcedureM ∈ Mem = Addr → Val

Procedure = Id × E × Env� ∈ Addr = N

let x := 0in


x := x + iin




return xin





...



call inc(2)

σ,M � E ⇓ v,M �


1

주소는 인데스로 쓸 수만 있으면 어떤 것이든 상관없다.

통일된 SemanticsProcedure = Id × S × Env

σ,M � E ⇓ v,M �

σ,M � x := E ⇓ ·,M �[v/σ(x)]


σ,M � let x := E1 in E2 ⇓ v2,M2

σ[�x,E1,σ�/f ],M � E2 ⇓ v,M1




σ(f) = �y, E1,σ1� σ1[σ(x)/y],M � E1 ⇓ v,M �



Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)


let x := 2in

addx(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

2


σ,M � E ⇓ v,M �

σ,M � x := E ⇓ ·,M �[v/σ(x)]

σ,M � E1 ⇓ v,M1 � /∈ dom(M1) σ[�/x],M1[v/�] � E2 ⇓ v2,M2

σ,M � let x := E1 in E2 ⇓ v2,M2

σ[�x,E1,σ�/f ],M � E2 ⇓ v,M1






σ(f) = �y,E1,σ1� σ1[σ(x)/y],M � E1 ⇓ v,M �



Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1




let x := 1 inlet procedure f(y) = x := x + y in

let x := 2in

f(3);x := x + 1;

σ,M � E ⇓ v

M(σ(x)) = v

σ,M � x ⇓ v

σ,M � E ⇓ v

σ,M � x := E ⇓ M [v/σ(x)]

σ,M � E ⇓ v � /∈ dom(M) σ[�/x],M [v/�] � S ⇓ M1



v ∈ Val = Z + B + {·}+

5

재귀 호출Recursive Call



let x := 0in


x := x + iin




return xin


let procedure fac(n) =if n <= 0 then 1else n * call fac (n-1)

incall fac(2)




...


call inc(1);let x := true

1

incall inc(2)

σ,M � E ⇓ v,M �



σ,M � E ⇓ v,M �

σ,M � x := E ⇓ ·,M �[v/σ(x)]


σ,M � let x := E1 in E2 ⇓ v2,M2

σ[�x,E1,σ�/f ],M � E2 ⇓ v,M1




σ(f) = �y, E1,σ1� σ1[σ(x)/y],M � E1 ⇓ v,M �



Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)


let x := 2in

addx(3);x := x + 1;

2

incall inc(2)

σ,M � E ⇓ v,M �



σ,M � E ⇓ v,M �

σ,M � x := E ⇓ ·,M �[v/σ(x)]


σ,M � let x := E1 in E2 ⇓ v2,M2

σ[�x,E1,σ�/f ],M � E2 ⇓ v,M1






σ(f) = �y, E1,σ1� σ1[σ(x)/y],M � E1 ⇓ v,M �



Procedure = Id × S

σ[�x, S1�/f ],M � S2 ⇓ M1





addx(3)


2

동작하지 않는다. 이유는?어디에 넣지?

우리가 만든 언어Program

P → E



ProgramP → S



StatementS → ...





in x + y

6

지원하는 feature들• for loop and while loop

•name memory cells

•name program codes

•names with scopes

• recursive calls

•call-by-value, call-by-reference

• integer I/O

하지만, 자료 구조는?tree, sets, lists ...

Record•기계 중심의 프로그래밍 언어에서 “레코

드”란 메모리 뭉치 + 뭉치에 있는 주소에 이름을 붙인 것




let x := 0in


x := x + iin




return xin



incall fac(2)




...


call inc(1);

1




{age �→ �1, name �→ �2}

let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1




age name


let x := 0in


x := x + iin




return xin



incall fac(2)




...

1




age name


let x := 0in


x := x + iin




return xin



incall fac(2)




...

1

메모리

Record 추가

ProgramP → E



ExpressionE → ...

| {xi := E1, ..., xn := En}| E.x| E.x := E

6

만들기

사용하기




age name


let x := 0in


x := x + iin



x.height + x.weight

let tree := {left := {}, v := 0, right := {}}in

tree.left := 1;tree.right := {left := {}, v := 2, right := 3}



return xin



incall fac(2)

1

Record 의미

let x := 0in


x := x + iin




return xin





...



call inc(2)

σ,M � E ⇓ v,M �



σ[�x, S1,σ�/f ],M � S2 ⇓ M1




σ(f) = �y, S1,σ1� σ1[σ(x)/y],M � S1 ⇓ M1


1



call inc(2)

σ,M � E ⇓ v,M �

σ,M � E1 ⇓ v1,M1

σ,M1 � E2 ⇓ v2,M2...

σ,Mn � En ⇓ vn,Mn

�1, ..., �n /∈ dom(Mn) ∀i, j ≤ n. i �= j ⇒ �i �= �j

σ,M � {x1 := E1, ..., xn := En} ⇓ {x1 �→ �1, ..., xn �→ �n},Mn[v1/�1]...[vn/�n]



σ,M � E ⇓ v,M �

σ,M � x := E ⇓ ·,M �[v/σ(x)]


σ,M � let x := E1 in E2 ⇓ v2,M2

σ[�x,E1,σ�/f ],M � E2 ⇓ v,M1






σ(f) = �y, E1,σ1� σ1[σ(x)/y],M � E1 ⇓ v,M �



2



call inc(2)

σ,M � E ⇓ v,M �

σ,M � E1 ⇓ v1,M1

σ,M1 � E2 ⇓ v2,M2...


�1, ..., �n /∈ dom(Mn) ∀i, j ≤ n. i �= j ⇒ �i �= �j


σ,M � E ⇓ r,M �

σ,M � E.x ⇓ M �(r(x)),M �



σ,M � E ⇓ v,M �

σ,M � x := E ⇓ ·,M �[v/σ(x)]


σ,M � let x := E1 in E2 ⇓ v2,M2

σ[�x,E1,σ�/f ],M � E2 ⇓ v,M1






σ(f) = �y, E1,σ1� σ1[σ(x)/y],M � E1 ⇓ v,M �



2



call inc(2)

σ,M � E ⇓ v,M �

σ,M � E1 ⇓ v1,M1

σ,M1 � E2 ⇓ v2,M2...


�1, ..., �n /∈ dom(Mn) ∀i, j ≤ n. i �= j ⇒ �i �= �j


σ,M � E ⇓ r,M �

σ,M � E.x ⇓ M �(r(x)),M �

σ,M � E1 ⇓ r,M1 σ,M1 � E2 ⇓ v,M2

σ,M � E1.x := E2 ⇓ ·,M2[v/r(x)]


2

Record 예

r ∈ Record = Field → Addrn ∈ Zb ∈ B = {T, F}v ∈ Val = Z + B + {·}σ ∈ Env = Id → Addr + ProcedureM ∈ Mem = Addr → Val



age name


let x := 0in


x := x + iin



x.height + x.weight

let tree := {left := {}, v := 0, right := {}}in

tree.left := 1;tree.right := {left := {}, v := 2, right := 3}



return xin



incall fac(2)

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1

176

89

x heightweight





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1

tree left

rightv

1

0

구현에 휘둘리지는 말자.





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1





let x := 0in


x := x + iin




return xin



incall fac(2)




...

1

{}

2

3

left

rightv

우리가 만든 언어지원하는 feature들• for loop and while loop•name memory cells•name program codes•names with scopes• recursive calls•call-by-value, call-by-reference• integer I/O•primitive values: integers, booleans•compound values: records

하지만, 메모리 주소에 꼭 이름을 주어야 하나요?

ProgramP → E


| E;E| if E then E else E| while E do E| for x := E to E do E| read x| write E| skip| let procedure f(x) = E in E| call f(E)| call f�x�| n | true | false | x| E + E | E − E | E ∗ E | E/E| E = E | E < E | not E| {xi := E1, ..., xn := En}| E.x| E.x := E

ProgramP → S



StatementS → ...




let x := 1 in

7

주소도 값이다!

주소를 만드는 방법주소를 사용하는 방법

let x := 0in


*y = x; x := &y;write *x

Env {x �→ �1, y �→ �2}Memory {�2 �→ �3, �3 �→ 0, �1 �→ �2}

Env Memory{x �→ �1} {�1 �→ 1}


{x �→ �1}{�1 �→ 7, �2 �→ 2, �3 �→ 3, �4 �→ 2}

Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}




age name


let x := 1in


1

Final Statelet x := 0in



Env {x �→ �1, y �→ �2}Memory {�2 �→ �3, �3 �→ 0, �1 �→ �2}

Env Memory{x �→ �1} {�1 �→ 1}


{x �→ �1}{�1 �→ 7, �2 �→ 2, �3 �→ 3, �4 �→ 2}

Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}




age name


let x := 1in


1

ProgramP → E


| E;E| if E then E else E| while E do E| for x := E to E do E| read x| write E| skip| let procedure f(x) = E in E| call f(E)| call f�x�| n | true | false | x| E + E | E − E | E ∗ E | E/E| E = E | E < E | not E| {xi := E1, ..., xn := En}| E.x| E.x := E

ExpressionE → ...

| malloc(E) | &x| ∗E | E := E

7

ProgramP → E


| E;E| if E then E else E| while E do E| for x := E to E do E| read x| write E| skip| let procedure f(x) = E in E| call f(E)| call f�x�| n | true | false | x| E + E | E − E | E ∗ E | E/E| E = E | E < E | not E| {xi := E1, ..., xn := En}| E.x| E.x := E| malloc(E) | &E| ∗E | E := E

8

포인터 관련식의 의미

4.4 !" !!! II: K- 89

Exercise 1 !!!! !"! !! !!!!, !!!! "!"", !!!" !

"!" !"! !!# """ #$!! ""!!# # ! ! $". "!%?

%# "%! "&& '".

!,M ! E " n,M !

!,M ! malloc E " ",M ! n > 0, {", " + 1, · · · , " + n # 1} $% Dom M !

!,M ! &x " !(x),M

!,M ! E " r,M !

!,M ! &E.x " r(x),M !

!,M ! E " ",M !

!,M ! *E " M !("),M !

!,M ! E1 " ",M1 !,M1 ! E2 " v,M2

!,M ! *E1 := E2 " v,M2{" &' v}

*E! ! '!# "( (!# !"! !"$% $!$ $# (!, %##

"#". !! x# !"$% $!!! %$!!!$ "" %## #" #"!.

*E # !"$% $!$ $! (!! E# #$! !"! !!"# #%". &

! !"! !!#% %". *E# !"$% %$!$ $! (!! E# #$!

!"! !!$ ""#" $! "# #%". (C!"# !#".)

!$" &$ ""!!# #'$# x# ##!! !!% "&# !!$!

3! ""$".

letx := malloc(2)

in*x := 1;*(x+1) := *x + 2

SNU 4190.310 Programming Languages c!Kwangkeun Yi, Seoul National Univ., 2006, 2007, 2008

let x := 0in



dom(M �)

Env {x �→ �1, y �→ �2}Memory {�2 �→ �3, �3 �→ 0, �1 �→ �2}

Env Memory{x �→ �1} {�1 �→ 1}


{x �→ �1}{�1 �→ 7, �2 �→ 2, �3 �→ 3, �4 �→ 2}

Env Memory{x �→ �1, y �→ �3} {�1 �→ 1, �2 �→ 2, �3 �→ 3}

{x �→ �1, y �→ �4} {�1 �→ 4, �2 �→ 2, �4 �→ 2}




age name


1

우리가 만든 언어지원하는 feature들• for loop and while loop•name memory cells•name program codes•names with scopes• recursive calls•call-by-value, call-by-reference• integer I/O•primitive values: integers, booleans•compound values: records• locations as values: pointers

ProgramP → E


| E;E| if E then E else E| while E do E| for x := E to E do E| read x| write E| skip| let procedure f(x) = E in E| call f(E)| call f�x�| n | true | false | x| E + E | E − E | E ∗ E | E/E| E = E | E < E | not E| {xi := E1, ..., xn := En}| E.x| E.x := E| malloc(E) | &E| ∗E | E := E

8

훗!그래야

C언어 답지!

훗!!

훗!!

훗!이래야

C언어 답지!

타입 검사Type Checking

메모리 관리Memory Management

역사의 흐름을 따라! 우리 언어에 추가할 것dreameye/pl/slide/pl4.pdf ·...

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