CSEP505: Programming Languages Lecture 5: continuations, types€¦ · And where we were Ł Go back...
66
CSEP505: Programming Languages Lecture 5: continuations, types Dan Grossman Spring 2006
Transcript of CSEP505: Programming Languages Lecture 5: continuations, types€¦ · And where we were Ł Go back...
CS
EP
505:
Pro
gram
min
g La
ngua
ges
Lect
ure
5: c
ontin
uatio
ns, t
ypes
Dan
Gro
ssm
anS
prin
g 20
06
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an2
Rem
embe
r our
sym
bol-p
ile
Exp
ress
ions
:e::= x|λx.e|e e
Val
ues:
v::= λx.e
e1!λx
.e3
e2 !
v2e3
{v2/
x} !
v��
����
����
��[la
m]
���
����
����
����
����
����
����
��[a
pp]
λx.e
!λx
.e
e
1 e2
!v
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an3
And
whe
re w
e w
ere
�G
o ba
ck to
mat
h m
etal
angu
age
�N
otes
on
conc
rete
syn
tax
(rel
ates
to C
aml)
�D
efin
e se
man
tics
with
infe
renc
e ru
les
�La
mbd
a en
codi
ngs
(sho
w o
ur la
ngua
ge is
mig
hty)
�D
efin
e su
bstit
utio
n pr
ecis
ely
�An
d re
visi
t fun
ctio
n eq
uiva
lenc
es�
Envi
ronm
ents
�
Smal
l-ste
p�
Def
ine
and
mot
ivat
e co
ntin
uatio
ns�
(ver
y fa
ncy
lang
uage
feat
ure)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an4
Sm
all-s
tep
CB
V
�Le
ft-to
-rig
ht s
mal
l-ste
p ju
dgm
ente
→e�
e1→
e1�
e2→
e2�
����
����
����
����
����
����
����
����
����
�e1
e2→
e1� e
2
v
e2→
ve2
�
(λ
x.e)
v →
e{v/
x}
�N
eed
an �o
uter
loop
� as
usua
l: e →
*e�
�*
mea
ns �0
or m
ore
step
s��
Don
�t us
ually
bot
her w
ritin
g ru
les,
but
they
�re e
asy:
e1→
e2
e
2→
* e3
����
����
����
����
����
����
����
����
e→
* e
e
1→
* e3
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an5
In C
aml
type exp=
Vof string| Lof string*exp| Aof exp * exp
letsubste1_with e2_for s= …
let recinterp_onee=
matche with
V _-> failwith“interp_one”(*unbound var*)
|L _ -> failwith“interp_one”(*already done*)
|A(L(s1,e1),L(s2,e2)) -> subste1 L(s2,e2)s1
| A(L(s1,e1),e2) -> A(L(s1,e1),interp_one e2)
| A(e1,e2) -> A(interp_onee1, e2)
let recinterp_smalle=
matche with
V _-> failwith“interp_small”(*unbound var*)
|L _ -> e
| A(e1,e2) -> interp_small(interp_onee)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an6
Unr
ealis
tic, b
ut�
�C
an d
istin
guis
h in
finite
-loop
s fro
m s
tuck
pro
gram
s
�It�
s cl
oser
to a
con
text
ual s
eman
tics
that
can
def
ine
cont
inua
tions
�An
d ca
n be
mad
e ef
ficie
nt b
y �k
eepi
ng tr
ack
of w
here
yo
u ar
e� a
nd u
sing
env
ironm
ents
�Ba
sic
idea
firs
t in
the
SE
CD
mac
hine
[Lan
din
1960
]!�
Triv
ial t
o im
plem
ent i
n as
sem
bly
plus
mal
loc!
�Ev
en w
ith c
ontin
uatio
ns
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an7
Red
ivis
ion
of la
bor
typeectxt=Hole
|Leftof ectxt* exp
|Rightof exp * ectxt(*exp a value*)
let recsplit e=
matche with
A(L(s1,e1),L(s2,e2)) -> (Hole,e)
| A(L(s1,e1),e2) -> let (ctx2,e3) =split e2in
(Right(L(s1,e1),ctx2), e3)
| A(e1,e2) -> let (ctx2,e3) =split e1in
(Left(ctx2,e2), e3)
|_ ->failwith“bad argsto split”
let recfill (ctx,e) = (* plug the hole *)
matchctxwith
Hole
-> e
|Left(ctx2,e2) -> A(fill(ctx2,e), e2)
|Right(e2,ctx2) -> A(e2, fill (ctx2,e))
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an8
So
wha
t?
�H
aven
�t do
ne m
uch
yet: e = fill(splite)
�Bu
t we
can
writ
e in
terp
_sm
allw
ith th
em�
Show
s a
step
has
thre
e pa
rts: s
plit,
sub
st, f
ill
let recinterp_smalle=
matche with
V _-> failwith“interp_small”(*unbound var*)
|L _ -> e
| _ ->
matchsplit e with
(ctx, A(L(s3,e3),v))->
interp_small(fill(ctx, subste3 v s3))
| _-> failwith“bad split”
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an9
Aga
in, s
o w
hat?
�W
ell,
now
we
�hav
e ou
r han
ds� o
n a
cont
ext
�C
ould
sav
e an
d re
stor
e th
em
�(li
ke h
w2
with
hea
ps, b
ut th
is is
the
cont
rols
tack
)�
It�s
easy
giv
en th
is s
eman
tics!
�Su
ffici
ent f
or:
�Ex
cept
ions
�C
oope
rativ
e th
read
s�
Cor
outin
es�
�Tim
e tra
vel�
with
sta
cks
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an10
Lang
uage
w/ c
ontin
uatio
ns
�N
ow 2
kin
ds o
f val
ues,
but
use
L fo
r bot
h�
Cou
ld in
stea
d ha
ve 2
kin
ds o
f app
licat
ion
+ er
rors
�N
ew k
ind
stor
es a
con
text
(tha
t can
be
rest
ored
)�
Letc
cge
ts th
e cu
rrent
con
text
type exp= (* change: 2 kinds of L + Letcc*)
Vof string| Lof string*body| Aof exp * exp
|Letccofstring * exp
andbody=Expofexp |Ctxtofectxt
andectxt=Hole(* no change *)
|Leftof ectxt* exp
|Rightof exp * ectxt
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an11
Spl
it w
ith L
etcc
�O
ld: a
ctiv
e ex
pres
sion
(thi
ng in
the
hole
) alw
ays
som
eA(L(s1,e1),L(s2,e2))
�N
ew: c
ould
als
o be
som
e Letcc(s1,e1)
let recsplit e= (* change: one new case *)
matche with
Letcc(s1,e1)-> (Hole,e) (* new *)
|A(L(s1,e1),L(s2,e2)) -> (Hole,e)
| A(L(s1,e1),e2) -> let (ctx2,e3) =split e2in
(Right(L(s1,e1),ctx2), e3)
| A(e1,e2) -> let (ctx2,e3) =split e1in
(Left(ctx2,e2), e3)
|_ ->failwith“bad argsto split”
let recfill (ctx,e) = … (* no change *)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an12
All
the
actio
n•Letcc
beco
mes
an
L th
at �g
rabs
the
curre
nt c
onte
xt�
•A
whe
re b
ody
is a
Ctxt
�igno
res
curr
ent c
onte
xt�
let recinterp_smalle=
matche with
V _-> failwith“interp_small”(*unbound var*)
|L _ -> e
| _ -> matchsplit e with
(ctx, A(L(s3,Exp e3),v))->
interp_small(fill(ctx, subste3 v s3))
|(ctx, Letcc(s3,e3)) ->
interp_small(fill(ctx,
subste3 (L("",Ctxtctx)) s3))(*woah!!!*)
|(ctx, A(L(s3,Ctxt c3),v)) ->
interp_small(fill(c3, v)) (*woah!!!*)
| _-> failwith“bad split”
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an13
Exa
mpl
es
�C
ontin
uatio
ns fo
r exc
eptio
ns is
�eas
y��
Letc
cfo
r try
, App
ly fo
r rai
se�
Cor
outin
esca
n yi
eld
to e
ach
othe
r (ex
ampl
e: C
GI!)
�Pa
ss a
roun
d a
yiel
d fu
nctio
n th
at ta
kes
an
argu
men
t ��h
ow to
rest
art m
e��
Body
of y
ield
app
lies
the
�old
how
to re
star
t me�
pa
ssin
g th
e �n
ew h
ow to
rest
art m
e��
Can
gen
eral
ize
to c
oope
rativ
e th
read
-sch
edul
ing
�W
ith m
utat
ion
can
real
ly d
o st
rang
e st
uff
�Th
e �g
oto
of fu
nctio
nal p
rogr
amm
ing�
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an14
A lo
wer
-leve
l vie
w
�If
you�
re c
onfu
sed,
thin
k ca
ll-st
acks
�W
hat i
f YFL
had
thes
e op
erat
ions
:�
Stor
e cu
rrent
sta
ck in
x (c
f. Le
tcc)
�R
epla
ce c
urre
nt s
tack
with
sta
ck in
x�
You
need
to �f
ill th
e st
ack�
s ho
le� w
ith s
omet
hing
di
ffere
nt o
r you
�ll h
ave
an in
finite
loop
�C
ompi
ling
Letc
c�
Can
act
ually
cop
y st
acks
(exp
ensi
ve)
�O
r can
avo
id s
tack
s (p
ut fr
ames
in h
eap)
�Ju
st s
hare
and
rely
on
garb
age
colle
ctio
n
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an15
Whe
re a
re w
e
Fini
shed
maj
or p
arts
of t
he c
ours
e�
Func
tiona
l pro
gram
min
g (o
ngoi
ng)
�IM
P, lo
ops,
mod
elin
g m
utat
ion
�La
mbd
a-ca
lcul
us, m
odel
ing
func
tions
�Fo
rmal
sem
antic
s�
Con
text
s, c
ontin
uatio
nsM
oral
? P
reci
se d
efin
ition
s of
rich
lang
uage
s is
diff
icul
t bu
t ele
gant
Maj
or n
ew to
pic:
Typ
es!
�C
ontin
ue u
sing
lam
bda-
calc
ulus
as
our m
odel
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an16
Type
s In
tro
Naï
ve th
ough
t: M
ore
pow
erfu
l PL
is b
ette
r�
Be T
urin
g C
ompl
ete
�H
ave
real
ly fl
exib
le th
ings
(lam
bda,
con
tinua
tions
, �)
�H
ave
conv
enie
nces
to k
eep
prog
ram
s sh
ort
By
this
met
ric, t
ypes
are
a s
tep
back
war
d�
Who
le p
oint
is to
allo
w fe
wer
pro
gram
s�
A �fi
lter�
bet
wee
n pa
rse
and
com
pile
/inte
rp�
Why
a g
reat
idea
?
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an17
Why
type
s
1.C
atch
�stu
pid
mis
take
s� e
arly
�3
+ �h
ello
��
prin
t_st
ring
(Stri
ng.a
ppen
d�h
i�)�
But m
ay b
e to
o ea
rly (c
ode
not u
sed,
�)
2.P
reve
nt g
ettin
g st
uck
/ goi
ng h
ayw
ire�
Kno
wev
alua
tion
cann
ot e
verg
et to
the
poin
t w
here
the
next
ste
p �m
akes
no
sens
e��
Alte
rnat
e: la
ngua
ge m
akes
eve
ryth
ing
mak
e se
nse
(e.g
., C
lass
Cas
tExc
eptio
n)�
Alte
rnat
e: la
ngua
ge c
an d
o w
hate
ver ?
!
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an18
Dig
ress
ion/
serm
on
Uns
afe
lang
uage
s ha
ve o
pera
tions
whe
re u
nder
som
e si
tuat
ions
the
impl
emen
tatio
n �c
an d
o an
ythi
ng�
IMP
with
uns
afe
C a
rray
s ha
s th
is ru
le (a
ny H
�;s�!)
:
Abs
tract
ion,
mod
ular
ity, e
ncap
sula
tion
are
impo
ssib
le
beca
use
one
bad
line
can
have
arb
itrar
y gl
obal
effe
ctA
n en
gine
erin
g di
sast
er (c
f. ci
vil e
ngin
eerin
g)
H;e
1!
{v1,
�,v
n}
H;e
2!
ii>
n��
����
����
����
����
����
����
����
����
���
H; e
1[i]=
e2!
H�;s
�
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an19
Why
type
s, c
ontin
ued
3.E
nfor
ce a
stro
ng in
terfa
ce (v
ia a
n ab
stra
ct ty
pe)
�C
lient
s ca
n�t b
reak
inva
riant
s�
Clie
nts
can�
t ass
ume
an im
plem
enta
tion
�As
sum
es s
afet
y4.
Allo
w fa
ster
impl
emen
tatio
ns�
Com
pile
r kno
ws
run-
time
type
-che
cks
unne
eded
�C
ompi
ler k
now
s pr
ogra
m c
anno
t det
ect
spec
ializ
atio
n/op
timiz
atio
n5.
Sta
tic o
verlo
adin
g (e
.g.,
with
+)
�N
ot s
o in
tere
stin
g�
Late
-bin
ding
ver
y in
tere
stin
g (c
ome
back
to th
is)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an20
Why
type
s, c
ontin
ued
6.N
ovel
use
s�
A po
wer
ful w
ay to
thin
k ab
out m
any
cons
erva
tive
prog
ram
ana
lyse
s/re
stric
tions
�Ex
ampl
es: r
ace-
cond
ition
s, m
anua
l mem
ory
man
agem
ent,
secu
rity
leak
s, �
�I d
o so
me
of th
is; �
a ty
pes
pers
on�
We�
ll fo
cus
on s
afet
y an
d st
rong
inte
rface
s�
And
late
r dis
cuss
the
�sta
tic ty
pes
or n
ot� d
ebat
e (it
�s re
ally
a c
ontin
uum
)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an21
Our
pla
n
�Si
mpl
y-ty
ped
Lam
bda-
Cal
culu
s�
Safe
ty =
(pre
serv
atio
n +
prog
ress
)�
Exte
nsio
ns (p
airs
, dat
atyp
es, r
ecur
sion
, etc
.)�
Dig
ress
ion:
sta
tic v
s. d
ynam
ic ty
ping
�D
igre
ssio
n: C
urry
-How
ard
Isom
orph
ism
�Su
btyp
ing
�Ty
pe V
aria
bles
: �
Gen
eric
s (∀
), A
bstra
ct ty
pes
(∃),
Rec
ursi
ve ty
pes
�Ty
pe in
fere
nce
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an22
Add
ing
inte
gers
Add
ing
inte
gers
to th
e la
mbd
a-ca
lcul
us:
Exp
ress
ions
:e
::= x
|λx.
e|e
e |
cV
alue
s:
v::=
λx.
e |c
Cou
ld a
dd +
and
oth
er p
rimiti
ves
or ju
st p
aram
eter
ize
�pro
gram
s� b
y th
em: λ
plus
.λm
inus
.� e
�Li
ke P
erva
sive
sin
Cam
l�
A gr
eat i
dea
for k
eepi
ng la
ngua
ge d
efin
ition
s sm
all!
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an23
Stu
ck
�Ke
y is
sue:
can
a p
rogr
am e
�get
stu
ck� (
smal
l-ste
p):
�e →
* e1
�e1
is n
ot a
val
ue�
Ther
e is
no
e2 s
uch
that
e1 →
e2�
�Wha
t is
stuc
k� d
epen
ds o
n th
e se
man
tics:
e1→
e1�
e2→
e2�
����
����
����
����
����
����
����
����
����
�e1
e2→
e1� e
2
v
e2→
ve2
�
(λ
x.e)
v →
e{v/
x}
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an24
STL
C S
tuck
•S::=c e|x e|(λx.e) x |S e |(λx.e) S
�It�
s no
t nor
mal
to d
efin
e th
ese
expl
icitl
y, b
ut a
gre
at
way
to th
ink
abou
t it.
�M
ost p
eopl
e do
n�t r
ealiz
e �s
afet
y�de
pend
s on
the
sem
antic
s:
�W
e ca
n ad
d �c
heat
�rul
es to
�avo
id�b
eing
stu
ck.
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an25
Sou
nd a
nd c
ompl
ete
�D
efin
ition
: A ty
pe s
yste
m is
sou
nd if
it n
ever
acc
epts
a
prog
ram
that
can
get
stu
ck
�D
efin
ition
: A ty
pe s
yste
m is
com
plet
e if
it al
way
s ac
cept
s a
prog
ram
that
can
not g
et s
tuck
�So
undn
ess
and
com
plet
enes
s ar
e de
sira
ble
�Bu
t im
poss
ible
(und
ecid
able
) for
lam
bda-
calc
ulus
�If
e ha
s no
con
stan
ts o
r fre
e va
riabl
es th
en e
(3 4
)ge
ts s
tuck
iffe
term
inat
es�
As is
any
non
-triv
ial p
rope
rty fo
r a T
urin
g-co
mpl
ete
PL
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an26
Wha
t to
do
�O
ld c
oncl
usio
n: �s
trong
type
s fo
r wea
k m
inds
��
Nee
d an
unc
heck
ed c
ast (
a ba
ck-d
oor)
�M
oder
n co
nclu
sion
: �
Mak
e fa
lse
posi
tives
rare
and
fals
e ne
gativ
es
impo
ssib
le (b
e so
und
and
expr
essi
ve)
�M
ake
wor
karo
unds
reas
onab
le�
Just
ifica
tion:
fals
e ne
gativ
es to
o ex
pens
ive,
hav
e co
mpi
le-ti
me
reso
urce
s fo
r �fa
ncy�
type
-che
ckin
g
�O
kay,
let�s
act
ually
try
to d
o it�
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an27
Wro
ng a
ttem
pt
τ::=int|function
A ju
dgm
ent: ├e :τ
(for w
hich
we
hope
ther
e�s
an e
ffici
ent a
lgor
ithm
)
����
����
����
����
����
����
����
��├c :int
├(λx.e):function
├e1
: function
├e2: int
����
����
����
����
����
����
����
��├e1 e2: int
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an28
So
very
wro
ng
1.U
nsou
nd: (λx.y)
32.
Dis
allo
ws
func
tion
argu
men
ts: (λx.
x 3)(λy.y)
3.Ty
pes
not p
rese
rved
: (λx.(λy.y)) 3
�R
esul
t is
not a
n in
tege
r
����
����
����
����
����
����
����
��├c :int
├(λx.e):function
├e1: function
├e2: int
����
����
����
����
����
����
����
��├e1 e2: int
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an29
Get
ting
it rig
ht
1.N
eed
to ty
pe-c
heck
func
tion
bodi
es, w
hich
hav
e fre
e va
riabl
es2.
Nee
d to
dis
tingu
ish
func
tions
acc
ordi
ng to
arg
umen
t an
d re
sult
type
s
For (
1):
Γ::=. | Γ,
x : τ
and Γ├e :τ
�A
type
-che
ckin
g en
viro
nmen
t (ca
lled
a co
ntex
t)
For (
2):
τ::=int| τ→
τ�
Arro
w is
par
t of t
he (t
ype)
lang
uage
(not
met
a)�
An in
finite
num
ber o
f typ
es�
Just
like
Cam
l
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an30
Exa
mpl
es a
nd s
ynta
x
�Ex
ampl
es o
f typ
esin
t→in
t(in
t→in
t) →
int
int→
(int→
int)
�C
oncr
etel
y →
is ri
ght-a
ssoc
iativ
e, i.
e.,
�i.e
., τ1→
τ2→
τ3is
τ1→
(τ2→
τ3)
�Ju
st li
ke C
aml
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an31
STL
C in
one
slid
eE
xpre
ssio
ns:e::= x|λx.e|e e |c
Val
ues:
v::= λx.e |e e
Type
s:
τ::=int
| τ→
τC
onte
xts:
Γ::=. | Γ,
x : τ
e1→
e1�
e2→
e2�
����
����
����
����
����
����
����
����
����
�e1
e2→
e1� e
2
v
e2→
ve2
�
(λ
x.e)
v →
e{v/
x}
����
����
���
����
����
����
Γ├c :int
Γ├x : Γ(x)
Γ,x:τ1├e:τ2
Γ├e1:τ1→
τ2 Γ├e2:τ1
����
����
����
����
����
����
����
����
����
����
��Γ├(λx.e):τ1→
τ2 Γ├e1 e2:τ2
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an32
Rul
e-by
-rul
e
�C
onst
ant r
ule:
con
text
irre
leva
nt�
Varia
ble
rule
: loo
kup
(no
inst
antia
tion
if x
not i
n Γ)
�Ap
plic
atio
n ru
le: �
yeah
, tha
t mak
es s
ense
��
Func
tion
rule
the
inte
rest
ing
one�
����
����
���
����
����
����
Γ├c :int
Γ├x : Γ(x)
Γ,x:τ1├e:τ2
Γ├e1:τ1→
τ2 Γ├e2:τ1
����
����
����
����
����
����
����
����
����
����
��Γ├(λx.e):τ1→
τ2 Γ├e1 e2:τ2
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an33
The
func
tion
rule
�W
here
did
τ1co
me
from
?�
Our
rule
�inf
erre
d� o
r �gu
esse
d� it
�To
be
synt
ax-d
irect
ed, c
hang
e λx.e
toλx: τ.
e
and
use
that
τ�
If w
e th
ink
of Γ
as a
par
tial f
unct
ion,
we
need
x n
ot
alre
ady
in it
(alp
ha-c
onve
rsio
n al
low
s)
Γ,x:τ1├e:τ2
����
����
����
����
��Γ├(λx.e):τ1→
τ2
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an34
Our
pla
n
�Si
mpl
y-ty
ped
Lam
bda-
Cal
culu
s�
Safe
ty =
(pre
serv
atio
n +
prog
ress
)�
Exte
nsio
ns (p
airs
, dat
atyp
es, r
ecur
sion
, etc
.)�
Dig
ress
ion:
sta
tic v
s. d
ynam
ic ty
ping
�D
igre
ssio
n: C
urry
-How
ard
Isom
orph
ism
�Su
btyp
ing
�Ty
pe V
aria
bles
: �
Gen
eric
s (∀
), A
bstra
ct ty
pes
(∃),
Rec
ursi
ve ty
pes
�Ty
pe in
fere
nce
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an35
Is it
�rig
ht�?
�C
an d
efin
e an
y ty
pe s
yste
m w
e w
ant
�W
hat w
e de
fined
is s
ound
and
inco
mpl
ete
�C
an p
rove
inco
mpl
ete
with
one
exa
mpl
e�
Ever
y va
riabl
e ha
s ex
actly
one
sim
ple
type
�Ex
ampl
e (d
oesn
�t ge
t stu
ck, d
oesn
�t ty
pech
eck)
(λx. (x(λy.y)) (x 3)) (λz.z)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an36
Sou
nd
�St
atem
ent o
f sou
ndne
ss th
eore
m:
If .├
e:τ
and
e →
*e2,
then
e2
is a
val
ue o
r the
re
exis
ts a
n e3
suc
h th
at e
2 →
e3�
Proo
f is
toug
h �
Mus
t hol
d fo
r all
e an
d an
y nu
mbe
r of s
teps
�Bu
t eas
y if
thes
e tw
o th
eore
ms
hold
1.P
rogr
ess:
If
.├e:τ
then
e is
a v
alue
or t
here
ex
ists
an
e�su
ch th
at e→
e�2.
Pre
serv
atio
n: If
.├e:τ
and
e→e�
then
.├e:τ
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an37
Let�s
pro
ve it
Pro
ve: I
f .├
e:τ
and
e →
*e2,
then
e2
is a
val
ue o
r ∃e
3 su
ch th
at e
2 →
e3, a
ssum
ing:
1.If
.├e:τ
then
e is
a v
alue
or ∃
e�su
ch th
at e→
e�2.
If .├
e:τ
and
e→e�
then
.├e:τ
Pro
ve s
omet
hing
stro
nger
: Als
o sh
ow .├e2:τ
Pro
of: B
y in
duct
ion
on n
whe
re e
→*e
2 in
n s
teps
�C
ase
n=0:
imm
edia
te fr
om p
rogr
ess
(e=e
2)�
Cas
e n>
0: th
en ∃
e2�s
uch
that
�
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an38
Wha
t�s th
e po
int
�Pr
ogre
ss is
wha
t we
care
abo
ut�
But P
rese
rvat
ion
is th
e in
varia
ntth
at h
olds
no
long
er
how
long
we
have
bee
n ru
nnin
g�
(Pro
gres
s an
d P
rese
rvat
ion)
impl
ies
Sou
ndne
ss
�Th
is is
a v
ery
gene
ral/p
ower
ful r
ecip
e fo
r sho
win
g yo
u �d
on�t
get t
o a
bad
plac
e��
If in
varia
nt h
olds
, you
�re in
a g
ood
plac
e (p
rogr
ess)
and
you
go
to a
goo
d pl
ace
(pre
serv
atio
n)
�D
etai
ls o
n ne
xt 2
slid
es le
ss im
porta
nt�
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an39
Forg
et a
cou
ple
thin
gs?
Pro
gres
s: If
.├
e:τ
then
e is
a v
alue
or t
here
exi
sts
an e
�suc
h th
at e→
e�
Pro
of: I
nduc
tion
on (h
eigh
t of)
deriv
atio
n tre
e fo
r .├e:τ
Rou
gh id
ea:
�Tr
ivia
l unl
ess
e is
an
appl
icat
ion
�Fo
r e =
e1
e2,
�If
left
or ri
ght n
ot a
val
ue, i
nduc
tion
�If
both
val
ues,
e1
mus
t be
a la
mbd
a�
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an40
Forg
et a
cou
ple
thin
gs?
Pre
serv
atio
n: If
.├e:τ
and
e→e�
then
.├e:τ
Als
o by
indu
ctio
n on
ass
umed
typi
ng d
eriv
atio
n.
The
troub
le is
whe
n e→
e�in
volv
es s
ubst
itutio
n �
requ
ires
anot
her t
heor
em
Sub
stitu
tion:
If Γ
,x:τ1├e:τa
ndΓ├e1:τ1
, the
nΓ├e{e1/x}:τ
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an41
Our
pla
n
�Si
mpl
y-ty
ped
Lam
bda-
Cal
culu
s�
Safe
ty =
(pre
serv
atio
n +
prog
ress
)�
Exte
nsio
ns (p
airs
, dat
atyp
es, r
ecur
sion
, etc
.)�
Dig
ress
ion:
sta
tic v
s. d
ynam
ic ty
ping
�D
igre
ssio
n: C
urry
-How
ard
Isom
orph
ism
�Su
btyp
ing
�Ty
pe V
aria
bles
: �
Gen
eric
s (∀
), A
bstra
ct ty
pes
(∃),
Rec
ursi
ve ty
pes
�Ty
pe in
fere
nce
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an42
Hav
ing
laid
the
grou
ndw
ork�
�So
far:
�O
ur la
ngua
ge (S
TLC
) is
tiny
�W
e us
ed h
eavy
-dut
y to
ols
to d
efin
e it
�N
ow:
�Ad
d lo
ts o
f thi
ngs
quic
kly
�Be
caus
e ou
r too
ls a
re a
ll w
e ne
ed
�An
d ea
ch a
dditi
on w
ill ha
ve th
e sa
me
form
�
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an43
A m
etho
d to
our
mad
ness
�Th
e pl
an�
Add
synt
ax�
Add
new
sem
antic
rule
s (in
clud
ing
subs
titut
ion)
�Ad
d ne
w ty
ping
rule
s
�If
our a
dditi
on e
xten
ds th
e sy
ntax
of t
ypes
, the
n�
We
will
have
new
val
ues
(of t
hat t
ype)
�An
d w
ays
to m
ake
the
new
val
ues
�(c
alle
d in
trodu
ctio
n fo
rms)
�An
d w
ays
to u
se th
e ne
w v
alue
s�
(cal
led
elim
inat
ion
form
s)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an44
Let b
indi
ngs
(CB
V)
e ::=
� |
letx
=e1
ine2
(no
new
val
ues
or ty
pes)
e1→
e1�
����
����
����
����
����
����
���
letx
=e1
ine2→
letx
=e1
�in
e2��
����
����
����
����
�le
tx =
v in
e2→
e2{v
/x}
Γ├e1:τ1
Γ,x:τ1├e2:τ2
����
����
����
����
����
����
��Γ├
letx
=e1
ine2
: τ2
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an45
Let a
s su
gar?
Let i
s ac
tual
ly s
o m
uch
like
lam
bda,
we
coul
d us
e 2
othe
r diff
eren
t but
equ
ival
ent s
eman
tics
2.le
t x=e
1 in
e2
is s
ugar
(a d
iffer
ent c
oncr
ete
way
to
writ
e th
e sa
me
abst
ract
syn
tax)
for (λx
.e2)
e1
3.In
stea
d of
sem
antic
rule
s on
last
slid
e, u
se ju
st��
����
����
����
����
����
����
�le
tx =
e1 in
e2→
(λx.
e2) e
1
Not
e: In
Cam
l, le
t is
nots
ugar
for a
pplic
atio
n be
caus
e le
t is
type
-che
cked
diff
eren
tly (t
ype
varia
bles
)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an46
Boo
lean
s
e ::
= �
|tru
|fls
|e ?
e :e
v ::
= �
|tru
|fls
τ::=
� |
bool
e1→
e1�
����
����
����
����
����
����
����
����
�e1
?e2
:e3→
e1� ?
e2 :
e3
Γ├
tru:bool
����
����
����
����
����
����
�tru
?e2
:e3→
e2
Γ├
fls:bool
����
����
����
��Γ├
e1:boolΓ├
e2:τΓ├
e3:τ
fls?
e2 :
e3→
e3
�
����
����
����
����
����
����
���
Γ├e1
?e2
:e3
: τ
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an47
Cam
l? L
arge
-ste
p?
�In
hom
ewor
k 3,
you
�ll ad
d co
nditi
onal
s, p
airs
, etc
. to
our e
nviro
nmen
t-bas
ed la
rge-
step
inte
rpre
ter
�C
ompa
red
to la
st s
lide
�D
iffer
ent m
eta-
lang
uage
(cas
es re
arra
nged
)�
Larg
e-st
ep in
stea
d of
sm
all
�If
test
s an
inte
ger f
or 0
(lik
e C
)�
Larg
e-st
ep b
oole
ans
with
infe
renc
e ru
les
e1!
true2
!v
e1!
flse3
!v
����
����
����
����
���
����
����
����
���
e1 ?
e2 :
e3!
v
e
1 ?
e2 :
e3!
v��
����
����
����
��tru
!tru
fls!
fls
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an48
Pai
rs (C
BV
, lef
t-to-
right
)
e ::
= �
|(e
,e)|
e.1
|e.2
v ::
= �
|(v
,v)
τ::=
� | τ*τ
e1→
e1�
e
2→
e2�
e→
e�
e→
e���
����
����
����
����
����
����
����
����
����
����
(e1,
e2)→
(e1�
,e2)
(v
,e2)→
(v,e
2�)
e.1→
e�.1
e
.2→
e�.2
����
����
����
����
����
����
(v1,
v2).1→
v1
(v
1,v2
).2→
v2
Γ├e1:τ1 Γ├
e2:τ2 Γ├
e:τ1*τ2 Γ├
e:τ1*τ2
����
����
����
����
���
����
����
���
����
����
���
Γ├(e
1,e2
): τ1
*τ2 Γ├
e.1:τ1 Γ├
e.2:τ2
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an49
Bes
t gue
ss o
f whe
re le
ctur
e 5
will
end
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an50
Tow
ard
Sum
s
�N
ext a
dditi
on: s
ums
(muc
h lik
e M
L da
taty
pes)
�In
form
al re
view
of M
L da
taty
peba
sics
type t = A of t1 | B of t2 | C of t3
�In
trodu
ctio
n fo
rms:
con
stru
ctor
-app
lied-
to-e
xp�
Elim
inat
ion
form
s: m
atch
e1
with
pat
->
exp
��
Typi
ng: I
f e h
as ty
pe t1
, the
n A
e h
as ty
pe t
�
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an51
Unl
ike
ML,
par
t 1
�M
L da
taty
pes
do a
lot a
t onc
e�
Allo
w re
curs
ive
type
s�
Intro
duce
a n
ew n
ame
for a
type
�Al
low
type
par
amet
ers
�Al
low
fanc
y pa
ttern
mat
chin
g�
Wha
t we
do w
ill be
sim
pler
�Ad
d re
curs
ive
type
s se
para
tely
late
r�
Avoi
d na
mes
(a b
it si
mpl
er in
theo
ry)
�Av
oid
type
par
amet
ers
(for s
impl
icity
)�
Onl
y pa
ttern
s of
form
A x
(res
t is
suga
r)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an52
Unl
ike
ML,
par
t 2
�W
hat w
e ad
d w
ill a
lso
be d
iffer
ent
�O
nly
two
cons
truct
ors
A a
nd B
�Al
l sum
type
s us
e th
ese
cons
truct
ors
�So
A e
can
hav
e an
y su
m ty
pe a
llow
ed b
y e�
sty
pe�
No
need
to d
ecla
re s
um ty
pes
in a
dvan
ce�
Like
func
tions
, will
�gue
ss ty
pes�
in o
ur ru
les
�Th
is s
houl
d st
ill h
elp
expl
ain
wha
t dat
atyp
esar
e
�Af
ter f
orm
alis
m, w
ill c
ompa
re to
C u
nion
s an
d O
OP
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an53
The
mat
h (w
ith ty
pe ru
les
to c
ome)
e ::
= �
|A
e |B
e |m
atch
e w
ithA
x ->
e |B
y ->
ev
::=
� |
Av
|Bv
τ::=
� | τ+τ
e→
e�e→
e�
e1→
e1�
����
����
���
����
���
����
����
����
����
����
����
���
A e→
A e�
B e →
B e�
mat
che1
with
Ax->
e2 |B
y ->
e3→
mat
che1
� with
Ax->
e2 |B
y ->
e3
����
����
����
����
����
����
����
����
����
����
mat
chA
v w
ithA
x->
e2 |
By ->
e3 →
e2{v
/x}
����
����
����
����
����
����
����
����
����
����
mat
chB
v w
ithA
x->
e2 |
By ->
e3 →
e3{y
/x}
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an54
Low
-leve
l vie
w
You
can
thin
k of
dat
atyp
eva
lues
as
�pai
rs�
�Fi
rst c
ompo
nent
is A
or B
(or 0
or 1
if y
ou p
refe
r)�
Seco
nd c
ompo
nent
is �t
he d
ata�
�e2
or e
3 ev
alua
ted
with
�the
dat
a� in
pla
ce o
f the
va
riabl
e�
This
is a
ll lik
e C
amla
s in
lect
ure
1�
Exam
ple
valu
es o
f typ
e in
t+ (i
nt->
int):
017
1
λx. λ
y.x+
y[(�
y�,6
)]
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an55
Typi
ng ru
les
�Ke
y id
ea fo
r dat
atyp
eex
p: �o
ther
can
be
anyt
hing
��
Key
idea
for m
atch
es: �
bran
ches
nee
d sa
me
type
��
Just
like
con
ditio
nals
Γ├e:τ1
Γ├e:τ2
����
����
����
����
����
����
���
Γ├A
e: τ1+τ2 Γ├
B e
: τ1+τ2
Γ├e1
: τ1+τ2 Γ,
x:τ1├
e2: τ
Γ,x:τ2├
e3: τ
����
����
����
����
����
����
����
����
����
����
Γ├m
atch
e1 w
ithA
x->
e2 |
By ->
e3 : τ
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an56
Com
pare
to p
airs
, par
t 1
��p
airs
and
sum
s� is
a b
ig id
ea�
Lang
uage
s sh
ould
hav
e bo
th (i
n so
me
form
)�
Som
ehow
pai
rs c
ome
acro
ss a
s si
mpl
er, b
ut
they
�re re
ally
�dua
l� (s
ee C
urry
-How
ard
soon
)�
Intro
duct
ion
form
s:
�pa
irs �n
eed
both
�, su
ms
�nee
d on
e�
Γ├e1:τ1 Γ├
e2:τ2
Γ├e:τ1
Γ├e:τ2
����
����
����
����
����
����
����
����
����
����
���
Γ├(e
1,e2
): τ1
*τ2
Γ├A
e: τ1+τ2 Γ├
B e
: τ1+τ2
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an57
Com
pare
to p
airs
, par
t 2
�El
imin
atio
n fo
rms
�Pa
irs g
et e
ither
, sum
s m
ust b
e pr
epar
ed fo
r eith
er
Γ├e:τ1*τ2 Γ├
e:τ1*τ2
����
����
���
����
����
���
Γ├e.1:τ1 Γ├
e.2:τ2
Γ├e1
: τ1+τ2
Γ,
x:τ1├
e2: τ
Γ,x:τ2├
e3: τ
����
����
����
����
����
����
����
����
����
����
Γ├m
atch
e1 w
ithA
x->
e2 |
By ->
e3 : τ
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an58
Livi
ng w
ith ju
st p
airs
�If
stub
born
you
can
cra
m s
ums
into
pai
rs (d
on�t!
)�
Rou
nd-p
eg, s
quar
e-ho
le�
Less
effi
cien
t (du
mm
y va
lues
)�
Flat
tene
d pa
irs d
on�t
chan
ge th
at�
Mor
e er
ror-
pron
e (m
ay u
se d
umm
y va
lues
)�
Exam
ple:
int+ (int-> int)
beco
mes
int* (int
* (int
-> int))
1λx
. λy.
x+y
[(�y�
,6]
0
0λx
. λy.
x+y
[ ]17
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an59
Sum
s in
oth
er g
uise
s
typet =A oft1 |B oft2 |C oft3
match
e with
A x ->
…
Mee
ts C
:structt {
enum
{A,B,C}
tag;
union{t1 a;t2 b;t3 c;}data;
};
… switch(e->tag){caseA:t1 x=e->data.a;…
�N
o st
atic
che
ckin
g th
at ta
g is
obe
yed
�As
fat a
s th
e fa
ttest
var
iant
(avo
idab
le w
ith c
asts
)�
Mut
atio
n bi
tes
agai
n!�
Sham
eles
s pl
ug: C
yclo
neha
s M
L-st
yle
data
type
s
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an60
Sum
s in
oth
er g
uise
s
typet =A oft1 |B oft2 |C oft3
match
e with
A x ->
…
Mee
ts J
ava:
abstract classt {abstractObject m();}
classA extendst {t1 x;Object m(){…}}
classB extendst {t2 x;Object m(){…}}
classC extendst {t3 x;Object m(){…}}
… e.m() …
�A
new
met
hod
for e
ach
mat
ch e
xpre
ssio
n�
Supp
orts
orth
ogon
al fo
rms
of e
xten
sibi
lity
(will
com
e ba
ck to
this
)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an61
Whe
re a
re w
e
�H
ave
adde
d le
t, bo
ols,
pai
rs, s
ums
�C
ould
hav
e do
ne s
tring
, flo
ats,
reco
rds,
��
Amaz
ing
fact
:�
Even
with
eve
ryth
ing
we
have
add
ed s
o fa
r, ev
ery
prog
ram
term
inat
es!
�I.e
., if
.├e:τ
then
ther
e ex
ists
a v
alue
vsu
ch
that
e →
* v�
Cor
olla
ry: O
ur e
ncod
ing
of fi
x w
on�t
type
-che
ck�
To re
gain
Tur
ing-
com
plet
enes
s, w
e ne
ed e
xplic
it su
ppor
t for
recu
rsio
n
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an62
Rec
ursi
on
�W
e co
uld
add
�fix
e� (a
sk m
e if
you�
re c
urio
us),
but
mos
t peo
ple
find
�letre
cf x
e� m
ore
intu
itive
e ::
= �
|le
trec
f x e
v ::
= �
|le
trec
f x e
(no
new
type
s)�S
ubst
itute
arg
umen
t lik
e la
mbd
a &
who
le fu
nctio
n fo
r f�
����
����
����
����
����
����
����
����
��(le
trec
f x e
) v →
(e{v
/x})
{(le
trec
f x e
)/f}
Γ, f:τ1→
τ2,x
:τ1 ├e:τ2
����
����
����
����
����
���
Γ├le
trec
f x e
: τ1→
τ2
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an63
Our
pla
n
�Si
mpl
y-ty
ped
Lam
bda-
Cal
culu
s�
Safe
ty =
(pre
serv
atio
n +
prog
ress
)�
Exte
nsio
ns (p
airs
, dat
atyp
es, r
ecur
sion
, etc
.)�
Dig
ress
ion:
sta
tic v
s. d
ynam
ic ty
ping
�D
igre
ssio
n: C
urry
-How
ard
Isom
orph
ism
�Su
btyp
ing
�Ty
pe V
aria
bles
: �
Gen
eric
s (∀
), A
bstra
ct ty
pes
(∃),
Rec
ursi
ve ty
pes
�Ty
pe in
fere
nce
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an64
A c
oupl
e sl
ides
for c
onte
xt e
xam
ples
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an65
Red
ivis
ion
of la
bor
typeectxt=Hole
|Leftof ectxt* exp
|Rightof exp * ectxt(*exp a value*)
let recsplit e=
matche with
A(L(s1,e1),L(s2,e2)) -> (Hole,e)
| A(L(s1,e1),e2) -> let (ctx2,e3) =split e2in
(Right(L(s1,e1),ctx2), e3)
| A(e1,e2) -> let (ctx2,e3) =split e1in
(Left(ctx2,e2), e3)
18 A
pril
2006
CSE
P50
5 Sp
ring
2006
Dan
Gro
ssm
an66
Red
ivis
ion
of la
bor
typeectxt=Hole
|Leftof ectxt* exp
|Rightof exp * ectxt(*exp a value*)
let recsplit e=
matche with
A(L(s1,e1),L(s2,e2)) -> (Hole,e)
| A(L(s1,e1),e2) -> let (ctx2,e3) =split e2in
(Right(L(s1,e1),ctx2), e3)
| A(e1,e2) -> let (ctx2,e3) =split e1in
(Left(ctx2,e2), e3)
