Symbolic Execution & Constraint Solving

greenpepperwhinnySecurity

Nov 3, 2013 (4 years and 8 months ago)

101 views

Finding bugs: Analysis Techniques & Tools

Symbolic Execution

& Constraint Solving


CS161 Computer
Security

Cho
, Chia Yuan

Lab

•
Q1: Manual reasoning on code

–
Mergesort

implementation published in
Wikibooks


•
Q2: Constraint Solving

–
‘Solve’ for collisions in
ELFHash

function


•
Q3:
Whitebox

&
b
lackbox

fuzzing

–
Use a dynamic symbolic execution tool to find bugs automatically


•
Start early!

Big Picture

Attacks

&

Defenses

Mobile
Security

(Android)

Web
Security

Network

Security

Crypto

Program Analysis & Verification

Symbolic Execution & Constraint Solving

Why?

A little history …

Can we build a machine that can
automatically reason and prove
mathematical facts about programs?


1967


1967


1976

“From one simple view, it is an enhanced

testing technique. Instead of executing a program

on a set of sample inputs, a program is "symbolically"

executed for a set of
classes of inputs.
”

Why now?


Advances in SAT Solvers

Source:
Sanji t

Seshi a

Advances in SAT Solvers


Source:
Sanji t

Seshi a

Significance

How do we know our program is
“correct”?

•
In general, we don’t know.

•
Test it

•
Let users test it for us

•
Fuzz it

•
Try to prove it’s correct

•
Static analysis

Symbolic
Execution

&

Constraint

Solving

Precision

Coverage

Dynamic
Sym

Exec is Directed Testing

•
Path
-
by
-
path exploration

buf
=
malloc

(s);

read(
fd
,
buf
,
len
);

s =
len

s =
len

+ 2

len

= input + 3;

if
len

< 10

if
len

% 2 == 0

s =
len

F

T

T

F

(
len

== input + 3
)



&&

!(
len

< 10)


&&
!(len%2==0)

Dynamic
Sym

Exec is Directed Testing

•
Path
-
by
-
path exploration

buf
=
malloc

(s);

read(
fd
,
buf
,
len
);

s =
len

s =
len

+ 2

len

= input + 3;

if
len

< 10

if
len

% 2 == 0

s =
len

F

T

T

F

(
len

== input + 3
)



&&

!(
len

< 10)


&& (
len%2==0)

Can we combine
all paths
into
1 single formula
?


Bounded Model Checking

How

do we
construct the
formula & use a
solver?

Q2 Goal: ‘Solve’ for Hash Collisions

Constructing Logic Formulas from Code

•
Convert statements into Static Single
Assignment (SSA) form


•
Encode SSA into target solver input format

Static Single Assignment Equations

•
Unroll

loops
to form loop
-
free
program

–
for(i=0
;
i<2
;
i++){a=a+1;}


a=a+1; a=a+1;

•
Rename LHS
of each assignment into a
new

local variable


a1
=a+1;
a2
=a+1;

•
Whenever a variable is
read

(e.g., at RHS),

replace it with
last assigned

variable name


a1=
a0
+1
;
a2=
a1
+1
;


Conditional (if) statements

•
Dynamic Symbolic Execution:

–
2
separate

path formulas


•
Bounded Model Checking:

–
Merge

both

branches into 1 formula


Conditional
(if) statements

Example

int

example1(
int

x) {


int

ret;



if (x > 0)


ret = x;


else


ret =
-
x;





assert(ret
>= 0);



return ret;

}

SSA



ret1
= x0


ret2
=
-
x0

ret3 = (
x0>0
? ret1 : ret2)

Q: Is !(ret3 >= 0)
satisfiable
?

Is this program correct?

Constructing Logic Formulas from Code

•
Convert statements into Static Single Assignment
(SSA) form

= Bit
-
vector Equations in quantifier
-
free 1
st

order logic


•
Encode SSA into target solver input format

–
Bit
-
vector arithmetic logic

–
“SMT” Solver

–
SMT
-
LIB 1.0 standard

Example SMT
-
LIB

:
extrafuns
(x0
BitVec
[32
])

:
extrafuns
(ret1
BitVec
[32
])

:
extrafuns
(ret2
BitVec
[32
])

:
extrafuns
(ret3
BitVec
[32
])

:
extrapreds
(branchcond1)

:assumption (=
ret1 x0
)

:
assumption
(=
ret2 (
bvneg

x0)

:assumption (
iff

branchcond1

(
bvsgt

x0 bv0[32
])

:
assumption
(=
ret3 (
ite

branchcond1

ret1
ret2)

(
not

(
bvsge

ret3 bv0[32
])

:formula true

SSA




ret1
=
x0

ret2
=
-
x0


ret3 = (
x0>0
? ret1 : ret2)


Is
!(ret3 >= 0)
satisfiable
?

Querying the Solver

$ ./z3 example1.smt
–
m


ret3
-
> bv2147483648[32]

ret1
-
> bv2147483648[32]

branchcond1
-
> false

ret2
-
> bv2147483648[32]

x0
-
> bv2147483648[32]

sat

2147483648


0x80000000


int

example1(
int

x
) {

…


•
32 bits Two’s Complement system

–
Positive range: [0 .. 2
N
-
1

–

1]

–
Or: [0x00 .. 0x7FFFFFFF]

–
0x80000000 is a
negative

signed 32
-
bit
value:
-
2147483648



Example

int

example1(
int

x) {


int

ret;



if (x > 0)


ret = x;


else


ret =
-
x;





assert(ret
>= 0);



return ret;

}

SSA



ret1
= x0


ret2
=
-
x0

ret3 = (
x0>0
? ret1 : ret2)

Q: Is !(ret3 >= 0)
satisfiable
?

Assertion violated if

x =
-
2147483648

Slightly Modified Example

int

example1(
char

x) {


int

ret;



if (x > 0)


ret = x;


else


ret =
-
x;





assert(ret
>= 0);



return ret;

}

SSA



ret1
= x0


ret2
=
-
x0

ret3 = (
x0>0
? ret1 : ret2)

Q: Is !(ret3 >= 0)
satisfiable
?

Example

:
extrafuns
(x0
BitVec
[32
])

:
extrafuns
(ret1
BitVec
[32
])

:
extrafuns
(ret2
BitVec
[32
])

:
extrafuns
(ret3
BitVec
[32
])

:
extrapreds
(branchcond1)

:assumption (=
ret1
(
sign_extend
[24]


x0))

:
assumption
(=
ret2 (
bvneg

(
sign_extend
[24]

x0))

:assumption (
iff

branchcond1 (
bvsgt

x0 bv0[32
])

:
assumption
(=
ret3 (
ite

branchcond1 ret1
ret2)

(
not

(
bvsge

ret3 bv0[32
])

:formula true

SSA




ret1
=
x0

ret2
=
-
x0


ret3 = (
x0>0
? ret1 : ret2)


Is
!(ret3 >= 0)
satisfiable
?

Querying the Solver

$ ./z3 example1.smt
–
m


unsat

int

example1(
char

x) {


int

ret;



if (x > 0)


ret = x;


else


ret =
-
x;




assert(ret >= 0);



return ret;

}

No satisfying assignment exists


==> Assertion holds for all
possible inputs!

SMT
-
LIB “Cheat” Sheet:
Bit
-
vectors

•
Declare

32
-
bit “variable” ‘a’: n
-
bits
Sign Extension
to ‘a’:

•
:
extrafuns
( a
BitVec
[32] )
sign_extend
[n] a


•
32
-
bit
constant

‘1234’

•
bv1234[32
]


•
Unary

functions:

•
~a



bvnot

(a
)

•
-
a



bvneg

(
a)


•
Binary

functions:
Binary

predicates:

•
bvand

bvor

bvxor

bvadd

bvshl

bvlshr

bvsgt

bvsge



bvfoo

(
a b)

•


& | ^ + << >> > >=


SMT
-
LIB “Cheat” Sheet:
Booleans

•
Declare

a predicate ‘C’:

•
:
extrapreds
( C )


•
Unary

connectives:

•
! C



not (C)


•
Binary

connectives:

•
Implies and or
xor

iff





foo (C D)

•


=> && ||


•
Ternary

connectives:

•
C ? a : b






ite

(C a b) where a, b can be bit
-
vectors

+

Exercise: C Operator Precedence

1.
SSA equations?

2.
SMT
-
LIB formula?

a = (b >> c) + d;

b =
-
(a ^ ~c);

Exercise: C Operator Precedence

i
nt

a,b
;

c
har d;

a = (b >> 3) + d;

b =
-
(a ^ ~d);

SSA

a1

= (b0 >> 3) + d0;

b1

=
-
(
a1

^ ~d0);

SMT
-
LIB

:
extrafuns
(a1
BitVec
[32])

:
extrafuns
(b0
BitVec
[32])

:
extrafuns
(b1
BitVec
[32])

:
extrafuns
(d0
BitVec
[8
])

:assumption(= a1 (
bvadd

(
bvlshr

b0 bv3[32]) (
sign_extend
[24] d0))

:assumption(= b1 (
bvneg

(
bvxor

(
bvnot

(
sign_extend
[24] d0) a1 )))

Additional References

•
An enjoyable read on verification history:

–
Vijay
D’Silva
,
Tales from Verification History


•
More about “constraint
s
olvers”:

–
Daniel
Kroening

&
Ofer

Strichman
,
Decision Procedures: An
Algorithmic Point of View