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Program Analysis and Verification 0368- 4479 http://www.cs.tau.ac.il/~maon/teaching/2013-2014/paav/paav1314b.html. Noam Rinetzky Lecture 9: Abstract Interpretation II. Slides credit: Roman Manevich , Mooly Sagiv , Eran Yahav. From verification to analysis. Manual program verification - PowerPoint PPT Presentation
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Program Analysis and Verification
0368-4479http://www.cs.tau.ac.il/~maon/teaching/2013-2014/paav/paav1314b.html
Noam Rinetzky
Lecture 9: Abstract Interpretation II
Slides credit: Roman Manevich, Mooly Sagiv, Eran Yahav
From verification to analysis
• Manual program verification– Verifier provides assertions• Loop invariants
• Program analysis– Automatic program verification – Tool automatically synthesize assertions • Finds loop invariants
3
Abstract Interpretation [Cousot’77]
• Mathematical foundation of static analysis
4
Abstract Interpretation [Cousot’77]
• Mathematical framework for approximating semantics (aka abstraction)– Allows designing sound static analysis algorithms• Usually compute by iterating to a fixed-point
– Computes (loop) invariants• Can be interpreted as axiomatic verification assertions• Generalizes Hoare Logic & WP / SP calculus
5
Abstract Interpretation [Cousot’77]
• Mathematical foundation of static analysis– Abstract domains• Abstract states ~ Assertions• Join () ~ Weakening
– Transformer functions• Abstract steps ~ Axioms
– Chaotic iteration • Structured Programs ~ Control-flow graphs• Abstract computation ~ Loop invariants
6
Concrete Semantics
set of statesoperational semantics(concrete semantics)
statement Sset of states
7
Conservative Semantics
set of states set of statesoperational semantics(concrete semantics)
statement Sset of states
8
Abstract (conservative) interpretation
set of states set of statesoperational semantics(concrete semantics)
statement S
abstract representation abstract
representation abstract semanticsstatement S abstract
representation
abstraction abstraction
{P} S {Q} sp(S, P)generalizes axiomatic verification
9
Abstract (conservative) interpretation
set of states set of statesoperational semantics(concrete semantics)
statement Sset of states
abstract representation abstract semantics
statement S abstract representation
concretization concretization
10
Abstract (conservative) interpretation
set of states set of statesoperational semantics(concrete semantics)
statement Sset of states
abstract state abstract semantics (transfer function)
statement S abstract state
concretization concretization
11
Abstract Interpretation [Cousot’77]
• Mathematical foundation of static analysis– Abstract domains• Abstract states ~ Assertions• Join () ~ Weakening
– Transformer functions• Abstract steps ~ Axioms
– Chaotic iteration • Abstract computation ~ Loop invariants• Structured Programs ~ Control-flow graphs
Lattices(D, , , , , )
Monotonic functions
Fixpoints
12
A taxonomy of semantic domain typesComplete Lattice(D, , , , , )
Lattice(D, , , , , )
Join semilattice(D, , , )
Meet semilattice(D, , , )
Complete partial order (CPO)(D, , )
Partial order (poset)(D, )
Preorder(D, )
13
Preorder• We say that a binary order relation over a
set D is a preorder if the following conditions hold for every d, d’, d’’ D– Reflexive: d d– Transitive: d d’ and d’ d’’ implies d d’’
14
Preorder
d d’
d’’
• We say that a binary order relation over a set D is a preorder if the following conditions hold for every d, d’, d’’ D– Reflexive: d d– Transitive: d d’ and d’ d’’ implies d d’’
Hasse Diagram
15
Preorder• We say that a binary order relation over a
set D is a preorder if the following conditions hold for every d, d’, d’’ D– Reflexive: d d– Transitive: d d’ and d’ d’’ implies d d’’
d d’
d’’
Hasse Diagram
16
Partial order• We say that a binary order relation over a
set D is a preorder if the following conditions hold for every d, d’, d’’ D– Reflexive: d d– Transitive: d d’ and d’ d’’ implies d d’’– Anti-symmetric: d d’ and d’ d implies d = d’
d d’
d’’
Hasse Diagram
17
Chains
• d d’ means d d’ and d d’
• An ascending chain is a sequencex1 x2 … xk …
• A descending chain is a sequencex1 x2 … xk …
• The height of a poset (D, ) is the length of the maximal ascending chain in D
18
poset Hasse diagram (for CP)
{x=0}
{x=-1}{x=-2} {x=1} {x=2} ……
19
Some posets-related terminology
• If x y (alt y ⊒x) we can say– x is lower than y – x is more precise than y – x is more concrete than y – x under-approximates y
– y is greater than x – y is less precise than x – y is more abstract than x – y over-approximates x
Least upper bound (LUB)
• (D, ) is a poset
• b D is an ∊ upper bound of A D if a ⊆ ∀ A: a b
• b D is ∊ the least upper bound of A D if ⊆– b is an upper bound of A – If b’ is an upper bound of A then b b’– Join: X = LUB of X• x y = {x, y}
May not exist
May not exist
21
Join operator
• Properties of a join operator– Commutative: x y = y x– Associative: (x y) z = x (y z)– Idempotent: x x = x
• A kind of abstract union (disjunction) operator
• Top element of (D, ) is = D
22
Join Example
{x=0}
{x=-1}{x=-2} {x=1} {x=2} ……
23
Join Example
{x=0}
{x=-1}{x=-2} {x=1} {x=2} ……
24
Join Example
{x=0}
{x=-1}{x=-2} {x=1} {x=2} ……
Greatest lower bound (GLB)
• (D, ) is a poset
• b D is an ∊ lower bound of A D if a ⊆ ∀ A: b a
• b D is ∊ the greatest lower bound of A D if ⊆– b is an lower bound of A – If b’ is an lower bound of A then b’ b– Meet: X = GLB of X• x y = {x, y}
May not exist
May not exist
26
Meet operator
• Properties of a meet operator– Commutative: x y = y x– Associative: (x y) z = x (y z)– Idempotent: x x = x
• A kind of abstract intersection (conjunction) operator
• Bottom element of (D, ) is = D
27
Complete partial order (CPO)
• A poset (D , ) is a complete partial if every ascending chain x1 x2 … xk … has a LUB
28
Meet Example
x=0
x0
x<0 x>0
x0
29
Meet Example
x=0
x0
x<0 x>0
x0
30
Meet Example
x=0
x0
x<0 x>0
x0
31
Complete partial order (CPO)
• A poset (D , ) is a complete partial if every ascending chain x1 x2 … xk … has a LUB
32
Join semilattices
• (D, , , ) is a join semilattice – (D, ) is a partial order – ∀X FIN D . X is defined – A top element
33
Meet semilattices
• (D, , , ) is a meet semilattice – (D, ) is a partial order – ∀X FIN D . X is defined – A bottom element
34
Lattices
• (D, , , , , ) is a lattice if– (D, , , ) is a join semilattice– (D, , , ) is a meet semilattice
• A lattice (D, , , , , ) is a complete lattice if– X and Y are defined for arbitrary sets
35
Example: Powerset lattices
• (2X, , , , , X) is the powerset lattice of X– A complete lattice
36
Example: Sign lattice
x=0
x0
x<0 x>0
x0
37
A taxonomy of semantic domain typesComplete Lattice(D, , , , , )
Lattice(D, , , , , )
Join semilattice(D, , , )
Meet semilattice(D, , , )
Join/Meet exist for every subset of DJoin/Meet exist for every finite
subset of D (alternatively, binary join/meet)
Join of the empty set Meet of the empty set
Complete partial order (CPO)(D, , )
Partial order (poset)(D, )
Preorder(D, )
reflexivetransitiveanti-symmetric: d d’ and d’ d implies d = d’
reflexive: d dtransitive: d d’, d’ d’’ implies d d’’
poset with LUB for all ascending chains
38
Towards a recipe for static analysis
39
Collecting semantics
• For a set of program states State, we define the collecting lattice
(2State, , , , , State)
• The collecting semantics accumulates the (possibly infinite) sets of states generated during the execution– Not computable in general
40
Abstract (conservative) interpretation
set of states set of statesoperational semantics(concrete semantics)
statement Sset of states
abstract representation abstract semantics
statement S abstract representation
concretization concretization
41
Abstract (conservative) interpretation
{x 1, x 2, …}↦ ↦ {x 0, x 1, …}↦ ↦operational semantics(concrete semantics)
x=x-1{x 0, x 1, …}↦ ↦
0 < x abstract semanticsx = x -1
0 ≤ x
concretization concretization
42
Abstract (conservative) interpretation
{x 1, x 2, …}↦ ↦ {…, x 0, …}↦operational semantics(concrete semantics)
x=x-1{x 0, x 1, …}↦ ↦
0 < x abstract semanticsx = x -1
concretization concretization
43
Abstract (non-conservative) interpretation
{x 1, x 2, …}↦ ↦ { x 1, …}↦operational semantics(concrete semantics)
x=x-1{x 0, x 1, …}↦ ↦ ⊈
0 < x abstract semanticsx = x -1
0 < x
concretization concretization
44
But …
• what if we have x & y?– Define lattice (semantics) for each variable– Compose lattices• Goal: compositional definition
• What if we have more than 1 statement?– Define semantics for entire program via CFG– Different “abstract states” at every CFG node
45
One lattice per variabletrue
false
x=0
x0
x<0 x>0
x0
true
false
y=0
y0
y<0 y>0
y0
How can we compose them?
46
Cartesian product of complete lattices
• For two complete lattices L1 = (D1, 1, 1, 1, 1, 1) L2 = (D2, 2, 2, 2, 2, 2)
• Define the posetLcart = (D1D2, cart, cart, cart, cart, cart)as follows:– (x1, x2) cart (y1, y2) iff x1 1 y1 x2 2 y2
– cart = ? cart = ? cart = ? cart = ?
• Lemma: L is a complete lattice• Define the Cartesian constructor Lcart = Cart(L1, L2)
47
Cartesian product example=(,)
x<0,y<0 x<0,y=0 x<0,y>0 x=0,y<0 x=0,y=0 x=0,y>0 x>0,y<0 x>0,y=0 x>0,y>0
x0,y<0 x0,y<0 x0,y=0 x0,y=0 x0,y>0 x0,y>0 x>0,y0 x>0,y0……
x0,y0 x0,y0 x0,y0x0,y0
x0 x0 y0 y0
…
(false, false)How does it represent
(x<0y<0) (x>0y>0)?
=(, )
48
Disjunctive completion• For a complete lattice
L = (D, , , , , )• Define the powerset lattice
L = (2D, , , , , ) = ? = ? = ? = ? = ?
• Lemma: L is a complete lattice
• L contains all subsets of D, which can be thought of as disjunctions of the corresponding predicates
• Define the disjunctive completion constructorL = Disj(L)
49
The base lattice CP
{x=0}
{x=-1}{x=-2} {x=1} {x=2} ……
50
The disjunctive completion of CP
{x=0}
true
{x=-1}{x=-2} {x=1} {x=2} ……
false
{x=-2x=-1} {x=-2x=0} {x=-2x=1} {x=1x=2}… … …
{x=0 x=1x=2}{x=-1 x=1x=-2}… ………
What is the height of this lattice?
51
Relational product of lattices
• L1 = (D1, 1, 1, 1, 1, 1)L2 = (D2, 2, 2, 2, 2, 2)
• Lrel = (2D1D2, rel, rel, rel, rel, rel)as follows:– Lrel = ?
52
Relational product of lattices
• L1 = (D1, 1, 1, 1, 1, 1)L2 = (D2, 2, 2, 2, 2, 2)
• Lrel = (2D1D2, rel, rel, rel, rel, rel)as follows:– Lrel = Disj(Cart(L1, L2))
• Lemma: L is a complete lattice• What does it buy us?
53
Cartesian product exampletrue
false
x<0,y<0 x<0,y=0 x<0,y>0 x=0,y<0 x=0,y=0 x=0,y>0 x>0,y<0 x>0,y=0 x>0,y>0
x0,y<0 x0,y<0 x0,y=0 x0,y=0 x0,y>0 x0,y>0 x>0,y0 x>0,y0……
x0,y0 x0,y0 x0,y0x0,y0
x0 x0 y0 y0
…
How does it represent(x<0y<0) (x>0y>0)?
What is the height of this lattice?
54
Relational product exampletrue
false
(x<0y<0)(x>0y>0)
x0 x0 y0 y0
How does it represent(x<0y<0) (x>0y>0)?
(x<0y<0)(x>0y=0) (x<0y0)(x<0y0)
…
What is the height of this lattice?
Collecting semantics1 label0: if x <= 0 goto label1 x := x – 1 goto label0
label1:
234
5
if x > 0
x := x - 1
2
3
entry
exit
[x1]
[x1]
[x1]
[x0]
[x0]
[x-1]
[x-1]
[x2]
[x2]
[x2]
[x2]
[x3]
[x3]
[x3]
…
…
…55
[x-2]…
56
Defining the collecting semantics
• How should we represent the set of states at a given control-flow node by a lattice?
• How should we represent the sets of states at all control-flow nodes by a lattice?
57
Finite maps• For a complete lattice
L = (D, , , , , )and finite set V
• Define the posetLVL = (VD, VL, VL, VL, VL, VL)as follows:– f1 VL f2 iff for all vV
f1(v) f2(v)
– VL = ? VL = ? VL = ? VL = ?
• Lemma: L is a complete lattice• Define the map constructor LVL = Map(V, L)
58
The collecting lattice
• Lattice for a given control-flow node v: ?
• Lattice for entire control-flow graph with nodes V:
?• We will use this lattice as a baseline for static
analysis and define abstractions of its elements
59
The collecting lattice
• Lattice for a given control-flow node v: Lv=(2State, , , , , State)
• Lattice for entire control-flow graph with nodes V:
LCFG = Map(V, Lv)• We will use this lattice as a baseline for static
analysis and define abstractions of its elements
60
Equational definition of the semantics• Define variables of type
set of states for each control-flow node
• Define constraints between them
if x > 0
x := x - 1
2
3
entry
exit
R[entry]
R[2]
R[3]R[exit]
61
Equational definition of the semantics• R[2] = R[entry] x:=x-1 R[3]• R[3] = R[2] {s | s(x) > 0}• R[exit] = R[2] {s | s(x) 0}• A system of recursive equations• How can we approximate it using what
we have learned so far?if x > 0
x := x - 1
2
3
entry
exit
R[entry]
R[2]
R[3]R[exit]
62
An abstract semantics• R[2] = R[entry] x:=x-1# R[3]• R[3] = R[2] {s | s(x) > 0}#
• R[exit] = R[2] {s | s(x) 0}#
• A system of recursive equations
if x > 0
x := x - 1
2
3
entry
exit
R[entry]
R[2]
R[3]R[exit]
Abstract transformer for x:=x-1
Abstract representationof {s | s(x) < 0}
63
Abstract interpretation via abstraction
set of states set of statescollecting semantics
statement S
abstract representationof sets of states
abstract
representationof sets of states abstract semantics
statement Sabstract
representationof sets of states
abstraction abstraction
{P} S {Q} sp(S, P)generalizes axiomatic verification
64
Abstract interpretation via concretization
set of states set of statescollecting semantics
statement Sset of states
abstract representationof sets of states
abstract semanticsstatement S abstract
representationof sets of states
concretization concretization
{P} S {Q}
models(P) models(sp(S, P)) models(Q)
65
Required knowledge
Collecting semanticsAbstract semantics • Connection between collecting semantics and
abstract semantics• Algorithm to compute abstract semantics
66
The collecting lattice (sets of states)
• Lattice for a given control-flow node v: Lv=(2State, , , , , State)
• Lattice for entire control-flow graph with nodes V:
LCFG = Map(V, Lv)• We will use this lattice as a baseline for static
analysis and define abstractions of its elements
Collecting semantics example1234
5
67
label0: if x <= 0 goto label1 x := x – 1 goto label0
label1:
if x > 0
x := x-1
entry
exit
2
3
[x1][x0][x-1][x2][x2][x3]…[x1][x0][x-1][x2][x2][x3]…
[x1][x3][x4]
[x2]
…[x-1][x-2]… [x0]
[x0][x1][x2][x3] …
Shorthand notation1234
5
68
label0: if x <= 0 goto label1 x := x – 1 goto label0
label1:
if x > 0
x := x-1
entry
exit
{xZ}
{xZ}
{x>0}{x0}
{x0}
2
3
69
Equational definition example• A vector of variables R[0, 1, 2, 3, 4]• R[0] = {xZ} // established input
R[1] = R[0] R[4]R[2] = R[1] {s | s(x) > 0}R[3] = R[1] {s | s(x) 0}R[4] = x:=x-1 R[2]
• A (recursive) system of equations
if x > 0
x := x-1
entry
exit
R[0]
R[1]
R[2]R[4]
R[3]
Semantic function for assume x>0
Semantic function for x:=x-1 lifted to sets of states
70
General definition• A vector of variables R[0, …, k] one per input/output of a node
– R[0] is for entry• For node n with multiple predecessors add equation
R[n] = {R[k] | k is a predecessor of n}• For an atomic operation node R[m] S R[n] add equation
R[n] = S R[m]• Treat true/false branches of conditions as atomic statements
assume bexpr and assume bexpr
if x > 0
x := x-1
entry
exit
R[0]
R[1]
R[2]R[4]
R[3]
71
Equation systems in general• Let L be a complete lattice (D, , , , , )• Let R be a vector of variables R[0, …, n] D … D• Let F be a vector of functions of the type
F[i] : R[0, …, n] R[0, …, n]• A system of equations
R[0] = f[0](R[0], …, R[n])…R[n] = f[n](R[0], …, R[n])
• In vector notation R = F(R)• Questions:
1. Does a solution always exist?
2. If so, is it unique?
3. If so, is it computable?
72
Equation systems in general• Let L be a complete lattice (D, , , , , )• Let R be a vector of variables R[0, …, n] D … D• Let F be a vector of functions of the type
F[i] : R[0, …, n] R[0, …, n]• A system of equations
R[0] = f[0](R[0], …, R[n])…R[n] = f[n](R[0], …, R[n])
• In vector notation R = F(R)• Questions:
1. Does a solution always exist?2. If so, is it unique?3. If so, is it computable?
73
Monotone functions
• Let L1=(D1, ) and L2=(D2, ) be two posets
• A function f : D1 D2 is monotone if for every pair x, y D1
x y implies f(x) f(y)• A special case: L1=L2=(D, )
f : D D
74
Important cases of monotonicity
• Join: f(X, Y) = X YProve it!
• For a set X and any function gF(X) = { g(x) | x X }Prove it!
• Notice that the collecting semantics function is defined in terms of– Join (set union)– Semantic function for atomic statements lifted to sets of
states
75
Extensive/reductive functions
• Let L=(D, ) be a poset• A function f : D D is extensive if for every
pair x D, we have that x f(x)• A function f : D D is reductive if for every
pair x D, we have that x f(x)
76
Fixed-points• L = (D, , , , , )• f : D D monotone• Fix(f) = { d | f(d) = d }• Red(f) = { d | f(d) d }• Ext(f) = { d | d f(d) }• Theorem [Tarski 1955]– lfp(f) = Fix(f) = Red(f) Fix(f)– gfp(f) = Fix(f) = Ext(f) Fix(f)
Red(f)
Ext(f)
Fix(f)
lfp
gfp
fn()
fn()
1.Does a solution always exist? Yes2. If so, is it unique? No, but it has least/greatest solutions3. If so, is it computable? Under some conditions…
77
Fixed point example for program• R[0] = {xZ}
R[1] = R[0] R[4]R[2] = R[1] {s | s(x) > 0}R[3] = R[1] {s | s(x) 0}R[4] = x:=x-1 R[2]
if x>0
x := x-1
2
3
entry
exit
xZ
xZ
{x>0}{x<0}
if x>0
x := x-1
2
3
entry
exit
xZ
xZ
{x>0}{x<0}
F(d) : Fixed-point
=
d
78
Fixed point example for program• R[0] = {xZ}
R[1] = R[0] R[4]R[2] = R[1] {s | s(x) > 0}R[3] = R[1] {s | s(x) 0}R[4] = x:=x-1 R[2]
if x>0
x := x-1
2
3
entry
exit
xZ
xZ
{x>0}{x<-5}
if x>0
x := x-1
2
3
entry
exit
xZ
xZ
{x>0}{x<0}
F(d) : pre Fixed-point
d
79
Fixed point example for program• R[0] = {xZ}
R[1] = R[0] R[4]R[2] = R[1] {s | s(x) > 0}R[3] = R[1] {s | s(x) 0}R[4] = x:=x-1 R[2]
if x>0
x := x-1
2
3
entry
exit
xZ
xZ
{x>0}{x<9}
if x>0
x := x-1
2
3
entry
exit
xZ
xZ
{x>0}{x<0}
F(d) : post Fixed-point
d
80
Continuity and ACC condition
• Let L = (D, , , ) be a complete partial order– Every ascending chain has an upper bound
• A function f is continuous if for every increasing chain Y D*,
f(Y) = { f(y) | yY }• L satisfies the ascending chain condition (ACC)
if every ascending chain eventually stabilizes:d0 d1 … dn = dn+1 = …
81
Fixed-point theorem [Kleene]• Let L = (D, , , ) be a complete partial order and a continuous function f: D D
then
lfp(f) = nN fn()
• Lemma: Monotone functions on posets satisfying ACC are continuousProof:
1. Every ascending chain Y eventually stabilizes d0 d1 … dn = dn+1 = … hence dn is the least upper bound of {d0, d1, … , dn},thus f(Y) = f(dn)
2. From monotonicity of f f(d0) f(d1) … f(dn) = f(dn+1) = … Hence f(dn) is the least upper bound of {f(d0), f(d1), … , f(dn)},thus { f(y) | yY } = f(dn)
82
Resulting algorithm • Kleene’s fixed point theorem
gives a constructive method for computing the lfp
lfpfn()
f()f2()
…d := while f(d) d do d := d f(d)return d
Algorithm
lfp(f) = nN fn()Mathematical definition
83
Chaotic iteration• Input:
– A cpo L = (D, , , ) satisfying ACC– Ln = L L … L– A monotone function f : Dn Dn – A system of equations { X[i] | f(X) | 1 i n }
• Output: lfp(f)• A worklist-based algorithm
for i:=1 to n do X[i] := WL = {1,…,n}while WL do j := pop WL // choose index non-deterministically N := F[i](X) if N X[i] then X[i] := N add all the indexes that directly depend on i to WL (X[j] depends on X[i] if F[j] contains X[i])return X
84
Chaotic iteration for static analysis• Specialize chaotic iteration for programs• Create a CFG for program• Choose a cpo of properties for the static analysis to infer: L
= (D, , , )• Define variables R[0,…,n] for input/output of each CFG node
such that R[i] D• For each node v let vout be the variable at the output of that
node:vout = F[v]( u | (u,v) is a CFG edge)– Make sure each F[v] is monotone
• Variable dependence determined by outgoing edges in CFG
85
Constant propagation example
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
x := 4;while (y5) do z := x; x := 4
86
Constant propagation lattice• For each variable x define L as
• For a set of program variables Var=x1,…,xn
Ln = L L … L
0-1-2 1 2 ......
no information
not-a-constant
87
Write down variables
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
x := 4;while (y5) do z := x; x := 4
88
Write down equations
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2R2R2
R3
R4
R6
R1
R5
R0x := 4;while (y5) do z := x; x := 4
89
Collecting semantics equations
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2R2R2
R3
R4
R6
R0 = StateR1 = x:=4 R0
R2 = R1 R5
R3 = assume y5 R2
R4 = z:=x R3
R5 = x:=4 R4
R6 = assume y=5 R2
R1
R5
R0
90
Constant propagation equations
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2R2R2
R3
R4
R6
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R1
R5
R0
abstract transformer
91
Abstract operations for CPR0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
Lattice elements have the form: (vx, vy, vz)x:=4# (vx,vy,vz) = (4, vy, vz)z:=x# (vx,vy,vz) = (vx, vy, vx)assume y5# (vx,vy,vz) = (vx, vy, vx)assume y=5# (vx,vy,vz) = if vy=k5 then (, , ) else (vx, 5, vz)R1 R5 = (a1, b1, c1) (a5, b5, c5) = (a1a5, b1b5, c1c5)
0-1-2 1 2 ......
CP lattice for a single variable
92
Chaotic iteration for CP: initialization
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2=(, , )R2R2
R3=(, , )
R4=(, , )
R6=(, , )
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R1=(, , )
R5=(, , )
R0=(, , )
WL = {R0, R1, R2, R3, R4, R5, R6}
93
Chaotic iteration for CP
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2=(, , )R2R2
R3=(, , )
R4=(, , )
R6=(, , )
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R1=(, , )
R5=(, , )
R0=(, , )
WL = {R1, R2, R3, R4, R5, R6}
94
Chaotic iteration for CP
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2=(, , )R2R2
R3=(, , )
R4=(, , )
R6=(, , )
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R1=(4, , )
R5=(, , )
R0=(, , )
WL = {R2, R3, R4, R5, R6}
95
Chaotic iteration for CP
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2=(, , )R2R2
R3=(, , )
R4=(, , )
R6=(, , )
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R1=(4, , )
R5=(, , )
R0=(, , )
0-1-2 1 2 ......
3 4
WL = {R2, R3, R4, R5, R6}
96
Chaotic iteration for CP
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2=(4, , )R2R2
R3=(, , )
R4=(, , )
R6=(, , )
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R1=(4, , )
R5=(, , )
R0=(, , )
0-1-2 1 2 ......
3 4
WL = {R2, R3, R4, R5, R6}
97
Chaotic iteration for CP
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2=(4, , )R2R2
R3=(, , )
R4=(, , )
R6=(, , )
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R1=(4, , )
R5=(, , )
R0=(, , )
WL = {R3, R4, R5, R6}
98
Chaotic iteration for CP
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2R2
R3=(4, , )
R4=(, , )
R6=(, , )
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R5=(, , )
R1=(4, , )
R0=(, , )
R2=(4, , )
WL = {R4, R5, R6}
99
Chaotic iteration for CP
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2R2
R3=(4, , )
R4=(4, , 4)
R6=(, , )
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R5=(, , )
R1=(4, , )
R0=(, , )
R2=(4, , )
WL = {R5, R6}
100
Chaotic iteration for CP
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2R2
R6=(, , )
R0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
R5=(4, , 4)
R4=(4, , 4)
R3=(4, , )
R1=(4, , )
R0=(, , )
R2=(4, , )R2=(4, , )
WL = {R2, R6}
added R2 back to worklist since it depends on R5
101
Chaotic iteration for CPR0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2R2
R6=(, , )
R5=(4, , 4)
R4=(4, , 4)
R3=(4, , )
R1=(4, , )
R0=(, , )
R2=(4, , )
WL = {R6}
102
Chaotic iteration for CPR0 = R1 = x:=4# R0
R2 = R1 R5
R3 = assume y5# R2
R4 = z:=x# R3
R5 = x:=4# R4
R6 = assume y=5# R2
x := 4
if (*)
assume y5
assume y=5
z := x
x := 4
entry
exit
R2R2
R6=(4, 5, )
R5=(4, , 4)
R4=(4, , 4)
R3=(4, , )
R1=(4, , )
R0=(, , )
R2=(4, , )
Fixed-point
WL = {}
In practice maintaina worklist of nodes
103
Chaotic iteration for static analysis• Specialize chaotic iteration for programs• Create a CFG for program• Choose a cpo of properties for the static analysis to infer: L
= (D, , , )• Define variables R[0,…,n] for input/output of each CFG node
such that R[i] D• For each node v let vout be the variable at the output of that
node:vout = F[v]( u | (u,v) is a CFG edge)– Make sure each F[v] is monotone
• Variable dependence determined by outgoing edges in CFG
104
Complexity of chaotic iteration• Parameters:
– n the number of CFG nodes– k is the maximum in-degree of edges – Height h of lattice L– c is the maximum cost of
• Applying Fv
• • Checking fixed-point condition for lattice L
• Complexity: O(n h c k)• Incremental (worklist) algorithm reduces the n factor in factor
– Implement worklist by priority queue and order nodes by reversed topological order
105
Required knowledge
Collecting semanticsAbstract semantics (over lattices)Algorithm to compute abstract semantics
(chaotic iteration)• Connection between collecting semantics and
abstract semantics• Abstract transformers
106
Recap• We defined a reference semantics –
the collecting semantics• We defined an abstract semantics for a given lattice and
abstract transformers• We defined an algorithm to compute abstract least fixed-point
when transformers are monotone and lattice obeys ACC• Questions:1. What is the connection between the two least fixed-points?2. Transformer monotonicity is required for termination – what
should we require for correctness?
107
Recap• We defined a reference semantics –
the collecting semantics• We defined an abstract semantics for a given lattice and
abstract transformers• We defined an algorithm to compute abstract least fixed-point
when transformers are monotone and lattice obeys ACC• Questions:1. What is the connection between the two least fixed-points?2. Transformer monotonicity is required for termination – what
should we require for correctness?
108
Galois Connection• Given two complete lattices
C = (DC, C, C, C, C, C) – concrete domainA = (DA, A, A, A, A, A) – abstract domain
• A Galois Connection (GC) is quadruple (C, , , A)that relates C and A via the monotone functions– The abstraction function : DC DA
– The concretization function : DA DC
• for every concrete element c DC
and abstract element a DA
( (a)) a and c ( (c))• Alternatively (c) a iff c (a)
109
Galois Connection: c ((c))
1
c2(c)
3((c))
The most precise (least) element in A representing c
C A
110
Galois Connection: ((a)) a
1
3((a))
2(a) a
C AWhat a represents in C(its meaning)
111
Example: lattice of equalities
• Concrete lattice:C = (2State, , , , , State)
• Abstract lattice:EQ = { x=y | x, y Var}A = (2EQ, , , , EQ , )– Treat elements of A as both formulas and sets of constraints
• Useful for copy propagation – a compiler optimization•
(X) = ?(Y) = ?
112
Example: lattice of equalities
• Concrete lattice:C = (2State, , , , , State)
• Abstract lattice:EQ = { x=y | x, y Var}A = (2EQ, , , , EQ , )– Treat elements of A as both formulas and sets of constraints
• Useful for copy propagation – a compiler optimization• (s) = ({s}) = { x=y | s x = s y} that is s x=y
(X) = {(s) | sX} = A {(s) | sX} (Y) = { s | s Y } = models(Y)
113
Galois Connection: c ((c))
1
[x5, y5, z5]
2
x=x, y=y, z=z,x=y, y=x,x=z, z=x,y=z, z=y
3
…[x6, y6, z6][x5, y5, z5][x4, y4, z4]
…
4
x=x, y=y, z=z
The most precise (least) element in A representing [x5, y5, z5]
C A
114
Most precise abstract representation
1c
5
C A
46
2
7 3 8
9
(c)
(c) = {c’ | c (c’)}
115
Most precise abstract representation
1c
5
C A
46
2
7 3 8
9
(c)= x=x, y=y, z=z,x=y, y=x,x=z, z=x,y=z, z=y
(c) = {c’ | c (c’)}
[x5, y5, z5]
x=y, y=zx=y, z=y
x=y
116
Galois Connection: ((a)) a
1
3x=x, y=y, z=z,x=y, y=x,x=z, z=x,y=z, z=y
2
…[x6, y6, z6][x5, y5, z5][x4, y4, z4]
…
x=y, y=z
What a represents in C(its meaning)
is called a semanticreduction
C A
117
Galois Insertion a: ((a))=a
1x=x, y=y, z=z,x=y, y=x,x=z, z=x,y=z, z=y
2
…[x6, y6, z6][x5, y5, z5][x4, y4, z4]
…
C AHow can we obtain a Galois Insertion
from a Galois Connection?
All elementsare reduced
118
Properties of a Galois Connection
• The abstraction and concretization functions uniquely determine each other:
(a) = {c | (c) a}(c) = {a | c (a)}
119
Abstracting (disjunctive) sets
• It is usually convenient to first define the abstraction of single elements
(s) = ({s}) • Then lift the abstraction to sets of elements
(X) = A {(s) | sX}
120
The case of symbolic domains• An important class of abstract domains are symbolic domains –
domains of formulas• C = (2State, , , , , State)
A = (DA, A, A, A, A, A)• If DA is a set of formulas then the abstraction of a state is defined
as (s) = ({s}) = A{ | s }
the least formula from DA that s satisfies• The abstraction of a set of states is
(X) = A { (s) | s X}• The concretization is
( ) = { s | s } = models( )
121
Inducing along the connections
• Assume the complete latticesC = (DC, C, C, C, C, C) A = (DA, A, A, A, A, A)M = (DM, M, M, M, M, M)andGalois connectionsGCC,A=(C, C,A, A,C, A) and GCA,M=(A, A,M, M,A, M)
• Lemma: both connections induce theGCC,M= (C, C,M, M,C, M)
defined by C,M = C,A A,M and M,C = M,A A,C
122
Inducing along the connections
1 C,A
A,C
c 2C,A(c)
5
C A
3
M
A,M
4
M,A
c’a’
=A,M(C,A(c))
123
Sound abstract transformer
• Given two latticesC = (DC, C, C, C, C, C)A = (DA, A, A, A, A, A)and GCC,A=(C, , , A) with
• A concrete transformer f : DC DC
an abstract transformer f# : DA DA
• We say that f# is a sound transformer (w.r.t. f) if• c: f(c)=c’ (f#(c)) (c’)• For every a and a’ such that
(f((a))) A f#(a)
124
Transformer soundness condition 1
1 2
C A
f 34f#
5
c: f(c)=c’ (f#(c)) (c’)
125
Transformer soundness condition 2
C A
1 2
f#35f
4
a: f#(a)=a’ f((a)) (a’)
126
Best (induced) transformer
C A
23f
f#(a)= (f((a)))
1 f#
4
Problem: incomputable directly
127
Best abstract transformer [CC’77]
• Best in terms of precision– Most precise abstract transformer– May be too expensive to compute
• Constructively defined asf# = f
– Induced by the GC• Not directly computable because first step is
concretization• We often compromise for a “good enough” transformer– Useful tool: partial concretization
128
Transformer example
• C = (2State, , , , , State)• EQ = { x=y | x, y Var}
A = (2EQ, , , , EQ , )• (s) = ({s}) = { x=y | s x = s y} that is s x=y
(X) = {(s) | sX} = A {(s) | sX} () = { s | s } = models()
• Concrete: x:=y X = { s[xs y] | sX}• Abstract: x:=y# X = ?
129
Developing a transformer for EQ - 1
• Input has the form X = {a=b}• sp(x:=expr, ) = v. x=expr[v/x] [v/x]
• sp(x:=y, X) = v. x=y[v/x] {a=b}[v/x] = …• Let’s define helper notations:– EQ(X, y) = {y=a, b=y X}• Subset of equalities containing y
– EQc(X, y) = X \ EQ(X, y)• Subset of equalities not containing y
130
Developing a transformer for EQ - 2
• sp(x:=y, X) = v. x=y[v/x] {a=b}[v/x] = …• Two cases– x is y: sp(x:=y, X) = X – x is different from y:
sp(x:=y, X) = v. x=y EQ)X, x)[v/x] EQc(X, x)[v/x] = x=y EQc(X, x) v. EQ)X, x)[v/x] x=y EQc(X, x)
• Vanilla transformer: x:=y#1 X = x=y EQc(X, x)
• Example: x:=y#1 {x=p, q=x, m=n} = {x=y, m=n}Is this the most precise result?
131
Developing a transformer for EQ - 3
• x:=y#1 {x=p, x=q, m=n} = {x=y, m=n} {x=y, m=n, p=q}– Where does the information p=q come from?
• sp(x:=y, X) =x=y EQc(X, x) v. EQ)X, x)[v/x]
• v. EQ)X, x)[v/x] holds possible equalities between different a’s and b’s – how can we account for that?
132
Developing a transformer for EQ - 4
• Define a reduction operator:Explicate(X) = if exist {a=b, b=c}X
but not {a=c} X then Explicate(X{a=c})elseX
• Define x:=y#2 = x:=y#1 Explicate• x:=y#2){x=p, x=q, m=n}) = {x=y, m=n, p=q}
is this the best transformer?
133
Developing a transformer for EQ - 5
• x:=y#2 ){y=z}) = {x=y, y=z} {x=y, y=z, x=z}• Idea: apply reduction operator again after the vanilla
transformer• x:=y#3 = Explicate x:=y#1 Explicate • Observation : after the first time we apply Explicate, all
subsequent values will be in the image of the abstraction so really we only need to apply it once to the input
• Finally: x:=y#(X) = Explicate x:=y#1
– Best transformer for reduced elements (elements in the image of the abstraction)
134
Negative property of best transformers
• Let f# = f • Best transformer does not compose
(f(f((a)))) f#(f#(a))
135
(f(f((a)))) f#(f#(a))
C A
23f
1 f#
6
54
f
7
f#
8
9
f
136
Soundness theorem 1
1. Given two complete latticesC = (DC, C, C, C, C, C)A = (DA, A, A, A, A, A)and GCC,A=(C, , , A) with
2. Monotone concrete transformer f : DC DC
3. Monotone abstract transformer f# : DA DA
4. a DA : f( (a)) (f#(a))Then
lfp(f) (lfp(f#)) (lfp(f)) lfp(f#)
137
Soundness theorem 1
C A
f
fn …
lpf(f)
f2 f3
f#n
…
lpf(f#)
f#2 f#3
f#
aDA : f((a)) (f#(a)) aDA : fn((a)) (f#n(a)) aDA : lfp(fn)((a)) (lfp(f#n)(a)) lfp(f) lfp(f#)
138
Soundness theorem 2
1. Given two complete latticesC = (DC, C, C, C, C, C)A = (DA, A, A, A, A, A)and GCC,A=(C, , , A) with
2. Monotone concrete transformer f : DC DC
3. Monotone abstract transformer f# : DA DA
4. c DC : (f(c)) f#( (c))Then
(lfp(f)) lfp(f#)lfp(f) (lfp(f#))
139
Soundness theorem 2
C A
f
fn …
lpf(f)
f2 f3
f#n
…
lpf(f#)
f#2 f#3
f#
c DC : (f(c)) f#((c)) c DC : (fn(c)) f#n((c)) c DC : (lfp(f)(c)) lfp(f#)((c)) lfp(f) lfp(f#)
140
A recipe for a sound static analysis
• Define an “appropriate” operational semantics• Define “collecting” structural operational semantics • Establish a Galois connection between collecting
states and abstract states• Local correctness: show that the abstract
interpretation of every atomic statement is soundw.r.t. the collecting semantics
• Global correctness: conclude that the analysis is sound
141
Completeness
• Local property:– forward complete: c: (f#(c)) = (f(c))– backward complete: a: f((a)) = (f#(a))
• A property of domain and the (best) transformer• Global property:– (lfp(f)) = lfp(f#)– lfp(f) = (lfp(f#))
• Very ideal but usually not possible unless we change the program model (apply strong abstraction) and/or aim for very simple properties
142
Forward complete transformer
1 2
C A
f 34
f#
c: (f#(c)) = (f(c))
143
Backward complete transformer
C A
1 2
f#35f
a: f((a)) = (f#(a))
144
Global (backward) completeness
C A
f
fn …
lpf(f)
f2 f3
f#n
…
lpf(f#)
f#2 f#3
f#
a: f((a)) = (f#(a)) a: fn((a)) = (f#n(a)) aDA : lfp(fn)((a)) = (lfp(f#n)(a)) lfp(f) = lfp(f#)
145
Global (forward) completeness
C A
f
fn …
lpf(f)
f2 f3
f#n
…
lpf(f#)
f#2 f#3
f#
c DC : (f(c)) = f#((c)) c DC : (fn(c)) = f#n((c)) c DC : (lfp(f)(c)) = lfp(f#)((c)) lfp(f) = lfp(f#)
146
Three example analyses
• Abstract states are conjunctions of constraints• Variable Equalities– VE-factoids = { x=y | x, y Var} false
VE = (2VE-factoids, , , , false, )• Constant Propagation– CP-factoids = { x=c | x Var, c Z} false
CP = (2CP-factoids, , , , false, )• Available Expressions– AE-factoids = { x=y+z | x Var, y,z VarZ} false
A = (2AE-factoids, , , , false, )
147
Lattice combinators reminder
• Cartesian Product– L1 = (D1, 1, 1, 1, 1, 1)
L2 = (D2, 2, 2, 2, 2, 2)– Cart(L1, L2) = (D1 D2, cart, cart, cart, cart, cart)
• Disjunctive completion– L = (D, , , , , )– Disj(L) = (2D, , , , , )
• Relational Product– Rel(L1, L2) = Disj(Cart(L1, L2))
148
Cartesian product of complete lattices
• For two complete lattices L1 = (D1, 1, 1, 1, 1, 1) L2 = (D2, 2, 2, 2, 2, 2)
• Define the posetLcart = (D1 D2, cart, cart, cart, cart, cart)as follows:– (x1, x2) cart (y1, y2) iff
x1 1 y1 andx2 2 y2
– cart = ? cart = ? cart = ? cart = ?
• Lemma: L is a complete lattice• Define the Cartesian constructor Lcart = Cart(L1, L2)
149
Cartesian product of GCs
• GCC,A=(C, C,A, A,C, A)GCC,B=(C, C,B, B,C, B)
• Cartesian ProductGCC,AB = (C, C,AB, AB,C, AB)
– C,AB(X)= ?– AB,C(Y) = ?
150
Cartesian product of GCs
• GCC,A=(C, C,A, A,C, A)GCC,B=(C, C,B, B,C, B)
• Cartesian ProductGCC,AB = (C, C,AB, AB,C, AB)
– C,AB(X) = (C,A(X), C,B(X))– AB,C(Y) = A,C(X) B,C(X)
• What about transformers?
151
Cartesian product transformers
• GCC,A=(C, C,A, A,C, A) FA[st] : A AGCC,B=(C, C,B, B,C, B) FB[st] : B B
• Cartesian ProductGCC,AB = (C, C,AB, AB,C, AB)
– C,AB(X) = (C,A(X), C,B(X))– AB,C(Y) = A,C(X) B,C(X)
• How should we define FAB[st] : AB AB
152
Cartesian product transformers
• GCC,A=(C, C,A, A,C, A) FA[st] : A AGCC,B=(C, C,B, B,C, B) FB[st] : B B
• Cartesian ProductGCC,AB = (C, C,AB, AB,C, AB)
– C,AB(X) = (C,A(X), C,B(X))– AB,C(Y) = A,C(X) B,C(X)
• How should we define FAB[st] : AB AB• Idea: FAB[st](a, b) = (FA[st] a, FB[st] b)• Are component-wise transformers precise?
153
Cartesian product analysis example• Abstract interpreter 1: Constant Propagation• Abstract interpreter 2: Variable Equalities• Let’s compare
– Running them separately and combining results– Running the analysis with their Cartesian product
a := 9;b := 9;c := a;
a := 9;b := 9;c := a;
CP analysis VE analysis{a=9}{a=9, b=9}{a=9, b=9, c=9}
{}{}{c=a}
154
Cartesian product analysis example• Abstract interpreter 1: Constant Propagation• Abstract interpreter 2: Variable Equalities• Let’s compare
– Running them separately and combining results– Running the analysis with their Cartesian product
CP analysis + VE analysisa := 9;b := 9;c := a;
{a=9}{a=9, b=9}{a=9, b=9, c=9, c=a}
155
Cartesian product analysis example• Abstract interpreter 1: Constant Propagation• Abstract interpreter 2: Variable Equalities• Let’s compare
– Running them separately and combining results– Running the analysis with their Cartesian product
CPVE analysisMissing
{a=b, b=c}
a := 9;b := 9;c := a;
{a=9}{a=9, b=9}{a=9, b=9, c=9, c=a}
156
Transformers for Cartesian product
• Naïve (component-wise) transformers do not utilize information from both components– Same as running analyses separately and then
combining results• Can we treat transformers from each analysis
as black box and obtain best transformer for their combination?
157
Can we combine transformer modularly?
• No generic method for any abstract interpretations
158
Reducing values for CPVE
• X = set of CP constraints of the form x=c(e.g., a=9)
• Y = set of VE constraints of the form x=y• ReduceCPVE(X, Y) = (X’, Y’) such that
(X’, Y’) (X’, Y’)• Ideas?
159
Reducing values for CPVE• X = set of CP constraints of the form x=c
(e.g., a=9)• Y = set of VE constraints of the form x=y• ReduceCPVE(X, Y) = (X’, Y’) such that
(X’, Y’) (X’, Y’)• ReduceRight:
– if a=b X and a=c Y then add b=c to Y• ReduceLeft:
– If a=c and b=c Y then add a=b to X• Keep applying ReduceLeft and ReduceRight and reductions
on each domain separately until reaching a fixed-point
160
Transformers for Cartesian product
• Do we get the best transformer by applying component-wise transformer followed by reduction?– Unfortunately, no (what’s the intuition?)– Can we do better?– Logical Product [Gulwani and Tiwari, PLDI 2006]
161
Product vs. reduced productCPVE lattice
{a=9}{c=a} {c=9}{c=a}
{a=9, c=9}{c=a}{[a9, c 9]}
collecting lattice
{}
162
Reduced product
• For two complete lattices L1 = (D1, 1, 1, 1, 1, 1) L2 = (D2, 2, 2, 2, 2, 2)
• Define the reduced posetD1D2 = {(d1,d2)D1D2 | (d1,d2) = (d1,d2) } L1L2 = (D1D2, cart, cart, cart, cart, cart)
163
Transformers for Cartesian product
• Do we get the best transformer by applying component-wise transformer followed by reduction?– Unfortunately, no (what’s the intuition?)– Can we do better?– Logical Product [Gulwani and Tiwari, PLDI 2006]
164
165
Logical product--
• Assume A=(D,…) is an abstract domain that supports two operations: for xD– inferEqualities(x) = { a=b | (x) a=b }
returns a set of equalities between variables that are satisfied in all states given by x
– refineFromEqualities(x, {a=b}) = ysuch that• (x)=(y)• y x
166
Developing a transformer for EQ - 1
• Input has the form X = {a=b}• sp(x:=expr, ) = v. x=expr[v/x] [v/x]
• sp(x:=y, X) = v. x=y[v/x] {a=b}[v/x] = …• Let’s define helper notations:– EQ(X, y) = {y=a, b=y X}• Subset of equalities containing y
– EQc(X, y) = X \ EQ(X, y)• Subset of equalities not containing y
167
Developing a transformer for EQ - 2
• sp(x:=y, X) = v. x=y[v/x] {a=b}[v/x] = …• Two cases
– x is y: sp(x:=y, X) = X – x is different from y:
sp(x:=y, X) = v. x=y EQ)X, x)[v/x] EQc(X, x)[v/x] = x=y EQc(X, x) v. EQ)X, x)[v/x] x=y EQc(X, x)
• Vanilla transformer: x:=y#1 X = x=y EQc(X, x)
• Example: x:=y#1 {x=p, q=x, m=n} = {x=y, m=n}Is this the most precise result?
168
Developing a transformer for EQ - 3
• x:=y#1 {x=p, x=q, m=n} = {x=y, m=n} {x=y, m=n, p=q}– Where does the information p=q come from?
• sp(x:=y, X) =x=y EQc(X, x) v. EQ)X, x)[v/x]
• v. EQ)X, x)[v/x] holds possible equalities between different a’s and b’s – how can we account for that?
169
Developing a transformer for EQ - 4
• Define a reduction operator:Explicate(X) = if exist {a=b, b=c}X
but not {a=c} X then Explicate(X{a=c})elseX
• Define x:=y#2 = x:=y#1 Explicate• x:=y#2){x=p, x=q, m=n}) = {x=y, m=n, p=q}
is this the best transformer?
170
Developing a transformer for EQ - 5
• x:=y#2 ){y=z}) = {x=y, y=z} {x=y, y=z, x=z}• Idea: apply reduction operator again after the
vanilla transformer• x:=y#3 = Explicate x:=y#1 Explicate
171
Logical Product-
basically the strongest postcondition
safely abstracting the existential quantifier
172
Abstracting the existentialReduce the pair
Abstract away existential quantifier for each domain
173
Example
174
Information loss exampleif (…) b := 5else b := -5
if (b>0) b := b-5else b := b+5assert b==0
{}{b=5}
{b=-5}{b=}
{b=}
{b=}can’t prove
175
Disjunctive completion of a lattice• For a complete lattice
L = (D, , , , , )• Define the powerset lattice
L = (2D, , , , , ) = ? = ? = ? = ? = ?
• Lemma: L is a complete lattice
• L contains all subsets of D, which can be thought of as disjunctions of the corresponding predicates
• Define the disjunctive completion constructorL = Disj(L)
176
Disjunctive completion for GCs
• GCC,A=(C, C,A, A,C, A)GCC,B=(C, C,B, B,C, B)
• Disjunctive completionGCC,P(A) = (C, P(A), P(A), P(A))
– C,P(A)(X) = ?– P(A),C(Y) = ?
177
Disjunctive completion for GCs
• GCC,A=(C, C,A, A,C, A)GCC,B=(C, C,B, B,C, B)
• Disjunctive completionGCC,P(A) = (C, P(A), P(A), P(A))
– C,P(A)(X) = {C,A({x}) | xX}– P(A),C(Y) = {P(A)(y) | yY}
• What about transformers?
178
Information loss exampleif (…) b := 5else b := -5
if (b>0) b := b-5else b := b+5assert b==0
{}{b=5}
{b=-5}{b=5 b=-5}
{b=0}
{b=0}proved
179
The base lattice CP
{x=0}
true
{x=-1}{x=-2} {x=1} {x=2} ……
false
180
The disjunctive completion of CP
{x=0}
true
{x=-1}{x=-2} {x=1} {x=2} ……
false
{x=-2x=-1} {x=-2x=0} {x=-2x=1} {x=1x=2}… … …
{x=0 x=1x=2}{x=-1 x=1x=-2}… ………
What is the height of this lattice?
181
Taming disjunctive completion• Disjunctive completion is very precise
– Maintains correlations between states of different analyses– Helps handle conditions precisely– But very expensive – number of abstract states grows
exponentially– May lead to non-termination
• Base analysis (usually product) is less precise– Analysis terminates if the analyses of each component
terminates• How can we combine them to get more precision yet
ensure termination and state explosion?
182
Taming disjunctive completion
• Use different abstractions for different program locations– At loop heads use coarse abstraction (base)– At other points use disjunctive completion
• Termination is guaranteed (by base domain)• Precision increased inside loop body
183
With Disj(CP)while (…) { if (…) b := 5 else b := -5
if (b>0) b := b-5 else b := b+5 assert b==0}
Doesn’t terminate
184
With tamed Disj(CP)while (…) { if (…) b := 5 else b := -5
if (b>0) b := b-5 else b := b+5 assert b==0}
terminates
CP
Disj(CP)
What MultiCartDomain implements
185
Reducing disjunctive elements
• A disjunctive set X may contain within it an ascending chain Y=a b c…
• We only need max(Y) – remove all elements below
186
Relational product of lattices
• L1 = (D1, 1, 1, 1, 1, 1)L2 = (D2, 2, 2, 2, 2, 2)
• Lrel = (2D1D2, rel, rel, rel, rel, rel)as follows:– Lrel = ?
187
Relational product of lattices
• L1 = (D1, 1, 1, 1, 1, 1)L2 = (D2, 2, 2, 2, 2, 2)
• Lrel = (2D1D2, rel, rel, rel, rel, rel)as follows:– Lrel = Disj(Cart(L1, L2))
• Lemma: L is a complete lattice• What does it buy us?– How is it relative to Cart(Disj(L1), Disj(L2))?
• What about transformers?
188
Relational product of GCs
• GCC,A=(C, C,A, A,C, A)GCC,B=(C, C,B, B,C, B)
• Relational ProductGCC,P(AB) = (C, C,P(AB), P(AB),C, P(AB))
– C,P(AB)(X) = ?– P(AB),C(Y) = ?
189
Relational product of GCs
• GCC,A=(C, C,A, A,C, A)GCC,B=(C, C,B, B,C, B)
• Relational ProductGCC,P(AB) = (C, C,P(AB), P(AB),C, P(AB))
– C,P(AB)(X) = {(C,A({x}), C,B({x})) | xX}– P(AB),C(Y) = {A,C(yA) B,C(yB) | (yA,yB)Y}
190
Cartesian product example
Correlations preserved
191
Function space• GCC,A=(C, C,A, A,C, A)
GCC,B=(C, C,B, B,C, B)• Denote the set of monotone functions from A to B by AB• Define for elements of AB as follows
(a1, b1) (a2, b2) = if a1=a2 then {(a1, b1B b1)} else {(a1, b1), (a2, b2)}
• Reduced cardinal powerGCC,AB = (C, C,AB, AB,C, AB)
– C,AB(X) = {(C,A({x}), C,B({x})) | xX}– AB,C(Y) = {A,C(yA) B,C(yB) | (yA,yB)Y}
• Useful when A is small and B is much larger– E.g., typestate verification
192
Widening/Narrowing
193
How can we prove this automatically?
RelProd(CP, VE)
194
Intervals domain
• One of the simplest numerical domains• Maintain for each variable x an interval [L,H]– L is either an integer of -– H is either an integer of +
• A (non-relational) numeric domain
195
Intervals lattice for variable x
[0,0][-1,-1][-2,-2] [1,1] [2,2] ......
[-,+]
[0,1] [1,2] [2,3][-1,0][-2,-1]
[-10,10]
[1,+][-,0]
... ...
[2,+][0,+][-,-1][-,-1]... ...
[-20,10]
196
Intervals lattice for variable x
• Dint[x] = { (L,H) | L-,Z and HZ,+ and LH}• • =[-,+]• = ?– [1,2] [3,4] ?– [1,4] [1,3] ?– [1,3] [1,4] ?– [1,3] [-,+] ?
• What is the lattice height?
197
Intervals lattice for variable x
• Dint[x] = { (L,H) | L-,Z and HZ,+ and LH}• • =[-,+]• = ?– [1,2] [3,4] no– [1,4] [1,3] no– [1,3] [1,4] yes– [1,3] [-,+] yes
• What is the lattice height? Infinite
198
Joining/meeting intervals
• [a,b] [c,d] = ?– [1,1] [2,2] = ?– [1,1] [2, +] = ?
• [a,b] [c,d] = ?– [1,2] [3,4] = ?– [1,4] [3,4] = ?– [1,1] [1,+] = ?
• Check that indeed xy if and only if xy=y
199
Joining/meeting intervals
• [a,b] [c,d] = [min(a,c), max(b,d)]– [1,1] [2,2] = [1,2]– [1,1] [2,+] = [1,+]
• [a,b] [c,d] = [max(a,c), min(b,d)] if a proper interval and otherwise – [1,2] [3,4] = – [1,4] [3,4] = [3,4]– [1,1] [1,+] = [1,1]
• Check that indeed xy if and only if xy=y
200
Interval domain for programs
• Dint[x] = { (L,H) | L-,Z and HZ,+ and LH}• For a program with variables Var={x1,…,xk}• Dint[Var] = ?
201
Interval domain for programs
• Dint[x] = { (L,H) | L-,Z and HZ,+ and LH}• For a program with variables Var={x1,…,xk}
• Dint[Var] = Dint[x1] … Dint[xk]• How can we represent it in terms of formulas?
202
Interval domain for programs
• Dint[x] = { (L,H) | L-,Z and HZ,+ and LH}• For a program with variables Var={x1,…,xk}
• Dint[Var] = Dint[x1] … Dint[xk]• How can we represent it in terms of formulas?– Two types of factoids xc and xc– Example: S = {x9, y5, y10}– Helper operations• c + + = +• remove(S, x) = S without any x-constraints• lb(S, x) =
203
Assignment transformers• x := c# S = ?• x := y# S = ?• x := y+c# S = ?• x := y+z# S = ?• x := y*c# S = ?• x := y*z# S = ?
204
Assignment transformers• x := c# S = remove(S,x) {xc, xc}• x := y# S = remove(S,x) {xlb(S,y), xub(S,y)}• x := y+c# S = remove(S,x) {xlb(S,y)+c, xub(S,y)+c}• x := y+z# S = remove(S,x) {xlb(S,y)+lb(S,z),
xub(S,y)+ub(S,z)}• x := y*c# S = remove(S,x) if c>0 {xlb(S,y)*c,
xub(S,y)*c} else {xub(S,y)*-c, xlb(S,y)*-c}
• x := y*z# S = remove(S,x) ?
205
assume transformers• assume x=c# S = ?• assume x<c# S = ?• assume x=y# S = ?• assume xc# S = ?
206
assume transformers• assume x=c# S = S {xc, xc}• assume x<c# S = S {xc-1}• assume x=y# S = S {xlb(S,y), xub(S,y)}• assume xc# S = ?
207
assume transformers• assume x=c# S = S {xc, xc}• assume x<c# S = S {xc-1}• assume x=y# S = S {xlb(S,y), xub(S,y)}• assume xc# S = (S {xc-1}) (S {xc+1})
208
Effect of function f on lattice elements• L = (D, , , , , )• f : D D monotone• Fix(f) = { d | f(d) = d }• Red(f) = { d | f(d) d }• Ext(f) = { d | d f(d) }• Theorem [Tarski 1955]– lfp(f) = Fix(f) = Red(f) Fix(f)– gfp(f) = Fix(f) = Ext(f) Fix(f)
Red(f)
Ext(f)
Fix(f)
lfp
gfp
fn()
fn()
209
Effect of function f on lattice elements• L = (D, , , , , )• f : D D monotone• Fix(f) = { d | f(d) = d }• Red(f) = { d | f(d) d }• Ext(f) = { d | d f(d) }• Theorem [Tarski 1955]– lfp(f) = Fix(f) = Red(f) Fix(f)– gfp(f) = Fix(f) = Ext(f) Fix(f)
Red(f)
Ext(f)
Fix(f)
lfp
gfp
fn()
fn()
210
Continuity and ACC condition
• Let L = (D, , , ) be a complete partial order– Every ascending chain has an upper bound
• A function f is continuous if for every increasing chain Y D*,
f(Y) = { f(y) | yY }• L satisfies the ascending chain condition (ACC)
if every ascending chain eventually stabilizes:d0 d1 … dn = dn+1 = …
211
Fixed-point theorem [Kleene]
• Let L = (D, , , ) be a complete partial order and a continuous function f: D D then
lfp(f) = nN fn()
212
Resulting algorithm • Kleene’s fixed point theorem
gives a constructive method for computing the lfp
lfpfn()
f()f2()
…d := while f(d) d do d := d f(d)return d
Algorithm
lfp(f) = nN fn()Mathematical definition
213
Chaotic iteration• Input:
– A cpo L = (D, , , ) satisfying ACC– Ln = L L … L– A monotone function f : Dn Dn – A system of equations { X[i] | f(X) | 1 i n }
• Output: lfp(f)• A worklist-based algorithm
for i:=1 to n do X[i] := WL = {1,…,n}while WL do j := pop WL // choose index non-deterministically N := F[i](X) if N X[i] then X[i] := N add all the indexes that directly depend on i to WL (X[j] depends on X[i] if F[j] contains X[i])return X
214
Concrete semantics equations
• R[0] = {xZ} R[1] = x:=7R[2] = R[1] R[4]R[3] = R[2] {s | s(x) < 1000}R[4] = x:=x+1 R[3]R[5] = R[2] {s | s(x) 1000} R[6] = R[5] {s | s(x) 1001}
R[0]R[2]R[3] R[4]
R[1]
R[5]R[6]
215
Abstract semantics equations
• R[0] = ({xZ}) R[1] = x:=7#
R[2] = R[1] R[4]R[3] = R[2] ({s | s(x) < 1000})R[4] = x:=x+1# R[3]R[5] = R[2] ({s | s(x) 1000})R[6] = R[5] ({s | s(x) 1001}) R[5] ({s | s(x) 999})
R[0]R[2]R[3] R[4]
R[1]
R[5]R[6]
216
Abstract semantics equations
• R[0] = R[1] = [7,7] R[2] = R[1] R[4]R[3] = R[2] [-,999]R[4] = R[3] + [1,1]R[5] = R[2] [1000,+] R[6] = R[5] [999,+] R[5] [1001,+]
R[0]R[2]R[3] R[4]
R[1]
R[5]R[6]
217
Too many iterations to converge
218
How many iterations for this one?
219
Widening
• Introduce a new binary operator to ensure termination– A kind of extrapolation
• Enables static analysis to use infinite height lattices– Dynamically adapts to given program
• Tricky to design• Precision less predictable then with finite-
height domains (widening non-monotone)
220
Formal definition• For all elements d1 d2 d1 d2 • For all ascending chains d0 d1 d2 …
the following sequence is finite– y0 = d0 – yi+1 = yi di+1
• For a monotone function f : DD define– x0 = – xi+1 = xi f(xi )
• Theorem:– There exits k such that xk+1 = xk
– xkRed(f) = { d | dD and f(d) d }
221
Analysis with finite-height lattice
A
f#n = lpf(f#) …
f#2 f#3
f#
Red(f)
Fix(f)
222
Analysis with widening
A
f#2 f#3
f#2 f#3
f#
Red(f)
Fix(f) lpf(f#)
223
Widening for Intervals Analysis
• [c, d] = [c, d]• [a, b] [c, d] = [
if a cthen aelse -,
if b dthen belse
224
Semantic equations with widening
• R[0] = R[1] = [7,7] R[2] = R[1] R[4]R[2.1] = R[2.1] R[2]R[3] = R[2.1] [-,999]R[4] = R[3] + [1,1]R[5] = R[2] [1001,+] R[6] = R[5] [999,+] R[5] [1001,+]
R[0]R[2]R[3] R[4]
R[1]
R[5]R[6]
225
Choosing analysis with widening
Enable widening
Non monotonicity of widening
• [0,1] [0,2] = ?• [0,2] [0,2] = ?
Non monotonicity of widening
• [0,1] [0,2] = [0, ]• [0,2] [0,2] = [0,2]
228
Analysis results with widening
Did we prove it?
229
Analysis with narrowing
A
f#2 f#3
f#2 f#3
f#
Red(f)
Fix(f) lpf(f#)
Formal definition of narrowing• Improves the result of widening• y x y (x y) x• For all decreasing chains x0 x1 …
the following sequence is finite– y0 = x0
– yi+1 = yi xi+1
• For a monotone function f: DDand xkRed(f) = { d | dD and f(d) d }define– y0 = x– yi+1 = yi f(yi )
• Theorem:– There exits k such that yk+1 =yk
– ykRed(f) = { d | dD and f(d) d }
Narrowing for Interval Analysis • [a, b] = [a, b]• [a, b] [c, d] = [
if a = - then celse a,
if b = then delse b
]
232
Semantic equations with narrowing
• R[0] = R[1] = [7,7] R[2] = R[1] R[4]R[2.1] = R[2.1] R[2]R[3] = R[2.1] [-,999]R[4] = R[3]+[1,1]R[5] = R[2]# [1000,+] R[6] = R[5] [999,+] R[5] [1001,+]
R[0]R[2]R[3] R[4]
R[1]
R[5]R[6]
233
Analysis with widening/narrowing• Two phases– Phase 1: analyze with
widening until converging
– Phase 2: use values to analyze with narrowing
Phase 2:R[0] = R[1] = [7,7] R[2] = R[1] R[4]R[2.1] = R[2.1] R[2]R[3] = R[2.1] [-,999]R[4] = R[3]+[1,1]R[5] = R[2]# [1000,+] R[6] = R[5] [999,+] R[5] [1001,+]
Phase 1:R[0] = R[1] = [7,7] R[2] = R[1] R[4]R[2.1] = R[2.1] R[2]R[3] = R[2.1] [-,999]R[4] = R[3] + [1,1]R[5] = R[2] [1001,+] R[6] = R[5] [999,+] R[5] [1001,+]
234
Analysis with widening/narrowing
235
Analysis results widening/narrowing
Precise invariant