1. College of Computing
Georgia Institute of Technology
Supported in part by:
NSF awards CCF-0541080 and CCR-0205422 to Georgia Tech,
DHS and US Air Force Contract No. FA8750-05-2-0214
Effective Memory Protection
Using Dynamic Tainting
Ioannis Doudalis
Milos Prvulovic
(hardware)
James Clause
Alessandro Orso
(software)
and

2. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory

3. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory

4. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory

5. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:

6. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:

7. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:

8. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:

9. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n: 3

10. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n: 3

11. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n: 3

12. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n: 3

13. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:
9
3
0

14. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:
9
8
2
3
2

15. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
i <= n ➜ i < n
buf:
np:
i:
n:
9
8
2
3
2

16. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:
9
8
2
3
2

17. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:
9
8
2
3
2

18. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:
9
8
2
3
3
7

19. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:
9
8
2
3
3
7

20. void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
Illegal memory accesses (IMA)
Memory
buf:
np:
i:
n:
9
8
2
3
3
7
Illegal memory accesses
• Caused by common programming mistakes
• Cause non-deterministic failures
• Cause security vulnerabilities

21. Previous work
Static techniques
•Language based
e.g., Jim et al. 02, Necula et al. 05
•Analysis based
e.g., Dor et al. 03, Hallem et al. 02, Heine and Lam
03, Xie et al. 03
Dynamic techniques
•Analysis based
e.g., Dhurjati and Adve 06, Ruwase and Lam 04, Xu
et al. 04, Hastings and Joyce 92, Seward and
Nethercote 05
•Hardware based
e.g., Qin et al. 05,Venkataramani et al. 07, Crandall
and Chong 04, Dalton et al. 07,Vachharajani et al. 04

22. Previous work
Static techniques
•Language based
e.g., Jim et al. 02, Necula et al. 05
•Analysis based
e.g., Dor et al. 03, Hallem et al. 02, Heine and Lam
03, Xie et al. 03
Dynamic techniques
•Analysis based
e.g., Dhurjati and Adve 06, Ruwase and Lam 04, Xu
et al. 04, Hastings and Joyce 92, Seward and
Nethercote 05
•Hardware based
e.g., Qin et al. 05,Venkataramani et al. 07, Crandall
and Chong 04, Dalton et al. 07,Vachharajani et al. 04
}Require source code

23. Previous work
Static techniques
•Language based
e.g., Jim et al. 02, Necula et al. 05
•Analysis based
e.g., Dor et al. 03, Hallem et al. 02, Heine and Lam
03, Xie et al. 03
Dynamic techniques
•Analysis based
e.g., Dhurjati and Adve 06, Ruwase and Lam 04, Xu
et al. 04, Hastings and Joyce 92, Seward and
Nethercote 05
•Hardware based
e.g., Qin et al. 05,Venkataramani et al. 07, Crandall
and Chong 04, Dalton et al. 07,Vachharajani et al. 04
}
}
Require source code
Unacceptable overhead

24. Previous work
Static techniques
•Language based
e.g., Jim et al. 02, Necula et al. 05
•Analysis based
e.g., Dor et al. 03, Hallem et al. 02, Heine and Lam
03, Xie et al. 03
Dynamic techniques
•Analysis based
e.g., Dhurjati and Adve 06, Ruwase and Lam 04, Xu
et al. 04, Hastings and Joyce 92, Seward and
Nethercote 05
•Hardware based
e.g., Qin et al. 05,Venkataramani et al. 07, Crandall
and Chong 04, Dalton et al. 07,Vachharajani et al. 04
}
}
Require source code
Unacceptable overhead
} Extensive modification

25. Previous work
Static techniques
•Language based
e.g., Jim et al. 02, Necula et al. 05
•Analysis based
e.g., Dor et al. 03, Hallem et al. 02, Heine and Lam
03, Xie et al. 03
Dynamic techniques
•Analysis based
e.g., Dhurjati and Adve 06, Ruwase and Lam 04, Xu
et al. 04, Hastings and Joyce 92, Seward and
Nethercote 05
•Hardware based
e.g., Qin et al. 05,Venkataramani et al. 07, Crandall
and Chong 04, Dalton et al. 07,Vachharajani et al. 04
}
}
Require source code
Unacceptable overhead
We define our approach to overcome these
limitations
• Operate at the binary level
• Use hardware to reduce overhead
• Minimal, practical modifications
} Extensive modification

26. Approach overview

27. 1 Assign
taint marks
Approach overview

28. 1 Assign
taint marks
Approach overview
P1
A

29. P1
1
A
1
1 Assign
taint marks
Approach overview

30. 1 Assign
taint marks
Approach overview
P1
P2
1
2
AB
12

31. 1 Assign
taint marks
Approach overview
P3
P1
P2
3
1
2
AC B
123

32. 1 Assign
taint marks
Approach overview
P3
P1
P2
3
1
2
P5
AC B
123
P4

33. 1 Assign
taint marks
Approach overview
2 Propagate
taint marks
P3
P1
P2
3
1
2
P5
AC B
123
P4

34. 1 Assign
taint marks
Approach overview
2 Propagate
taint marks
P3
P1
P2
3
1
2
P5
AC B
123
P4

35. 1 Assign
taint marks
Approach overview
2 Propagate
taint marks
P3
P1
P2
3
1
2
P5
3
AC B
123
P4
1

36. 1 Assign
taint marks
Approach overview
3 Check
taint marks
2 Propagate
taint marks
P3
P1
P2
3
1
2
P5
3
AC B
123
P4
1

37. 1 Assign
taint marks
Approach overview
3 Check
taint marks
2 Propagate
taint marks
P3
P1
P2
3
1
2
P5
3
AC B
123
P4
1

38. 1 Assign
taint marks
Approach overview
3 Check
taint marks
2 Propagate
taint marks
P3
P1
P2
3
1
2
P5
3
AC B
123
P4
1

39. 1 Assign
taint marks
Approach overview
3 Check
taint marks
2 Propagate
taint marks
P3
P1
P2
3
1
2
P5
3
AC B
123
P4
1
✔

40. 1 Assign
taint marks
Approach overview
3 Check
taint marks
2 Propagate
taint marks
P3
P1
P2
3
1
2
AC B
123
P4
P5
3
1

41. 1 Assign
taint marks
Approach overview
3 Check
taint marks
2 Propagate
taint marks
P3
P1
P2
3
1
2
AC B
123
P4
P5
3
1

42. 1 Assign
taint marks
Approach overview
3 Check
taint marks
2 Propagate
taint marks
P3
P1
P2
3
1
2
AC B
123
P4
✘
P5
3
1

43. Outline
• Our approach
1. Assigning taint marks
2. Propagating taint marks
3. Checking taint marks
• Empirical evaluation
• Conclusions

44. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory

45. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range

46. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:

47. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
{
[&np, &np + sizeof(int *))

48. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:

49. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321

50. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory

51. Memory Pointers
1 Assigning taint marksStaticDynamic
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to

52. Memory Pointers
1 Assigning taint marksStaticDynamic
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
address-of operator (&)

53. Memory Pointers
1 Assigning taint marksStaticDynamic
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to

54. Memory Pointers
1 Assigning taint marksStaticDynamic
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4321

55. Memory Pointers
1 Assigning taint marksStaticDynamic
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to

56. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory

57. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations

58. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
buf:
np:
i:
n:
4
2
321
3

59. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
buf:
np:
i:
n:
4
2
321
3
[ret, ret + arg0)
{

60. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
buf:
np:
i:
n:
4
2
321
3

61. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
buf:
np:
i:
n:
4
2
321555
3

62. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory

63. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory

64. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
return value of malloc

65. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory

66. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
buf:
np:
i:
n:
4
2
321555
3

67. Memory Pointers
1 Assigning taint marksStaticDynamic
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory

68. Memory Pointers
1 Assigning taint marksStaticDynamic
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Static memory allocations
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4321
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to statically allocated memory
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
1 Identify the ranges
of allocated memory
2 Assign a unique taint
mark to each range
buf:
np:
i:
n:
4
2
321
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Dynamic memory allocations
1 Identify pointer
creation sites
2 Assign the pointer the same taint
mark as the memory it points to
buf:
np:
i:
n:
4
2
321555
3
void main() {
int *np, n, i, *buf;
np = &n;
printf(“Enter size: “);
scanf(“%d”, np);
buf = malloc(n * sizeof(int));
for(i = 0; i <= n; i++)
*(buf + i) = rand()%10;
...
}
Pointers to dynamically allocated memory

69. AND
Addition, Subtraction
Overview
Propagating taint marks2
Multiplication, Division,
OR, XOR
Overview
P1
1
P2

70. AND
Addition, Subtraction
Overview
Propagating taint marks2
Multiplication, Division,
OR, XOR
Overview +, −, ×, ÷,
and, or, xor,
...
P1
1
P2

71. AND
Addition, Subtraction
Overview
Propagating taint marks2
Multiplication, Division,
OR, XOR
Overview +, −, ×, ÷,
and, or, xor,
...
P1
1
P2
Should the result be tainted?
If so, how?

72. AND
Addition, Subtraction
Overview
Propagating taint marks2
Multiplication, Division,
OR, XOR
Overview
• Propagation must take into account both
operation semantics and programmer intent
+, −, ×, ÷,
and, or, xor,
...
P1
1
P2
Should the result be tainted?
If so, how?

73. AND
Addition, Subtraction
Overview
Propagating taint marks2
Multiplication, Division,
OR, XOR
Overview
• Propagation must take into account both
operation semantics and programmer intent
• Our policy is based on knowledge of
C/C++/assembly and patterns observed in
real software
+, −, ×, ÷,
and, or, xor,
...
P1
1
P2
Should the result be tainted?
If so, how?

74. AND
Addition, Subtraction
Overview
Propagating taint marks2
Multiplication, Division,
OR, XOR
Addition, Subtraction
A +/− B = C
A B C
Most common use of addition and
subtraction is to add or subtract a
pointer and an offset
1 1
1 /1 no
taint
...

75. AND
Addition, Subtraction
Overview
Propagating taint marks2
Multiplication, Division,
OR, XOR
AND The result of anding a pointer and a
mask should be treated differently
depending on the value of the mask
c = a & 0xffffff00 - base address
c = a & 0x000000ff - offset
A & B = C
A B C
1 or1
no
taint
...

76. AND
Addition, Subtraction
Overview
Propagating taint marks2
Multiplication, Division,
OR, XOR
Multiplication, Division,
OR, XOR
We found zero cases where the
result of any of these operations
was a pointer

77. Checking taint marks
When memory is accessed through a pointer:
compare the memory taint mark and the pointer taint mark
3
Pointer Memory IMA?
no
yes
yes
yes
yes
1
2 2
2
3
3

78. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}

79. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}

80. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}

81. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
i:
n:
4321

82. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
i:
n:
4321

83. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
i:
n:
4 2321

84. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
i:
n:
4 2321

85. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
i:
n:
4 2321

86. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
i:
n:
4 2321
✔

87. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
i:
n:
4 2321

88. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
i:
n:
4 2321
3

89. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
i:
n:
4 2321
3

90. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
5
i:
n:
4 2321555
3

91. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
5
i:
n:
4 2321555
3

92. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
5
i:
n:
4 2321555
3
0

93. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
+ =5 5
buf:
np:
5
i:
n:
4 2321555
3
0

94. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
+ =5 5
buf:
np:
5
i:
n:
4 2321555
3
0
✔

95. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
5
i:
n:
4 2321555
3
0

96. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
5
i:
n:
4 2321555
9
3
0

97. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
5
i:
n:
4 2321555
9
8
2
3
2

98. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
5
i:
n:
4 2321555
9
8
2
3
2

99. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
5
i:
n:
4 2321555
9
8
2
3
3
+ =5 5

100. Preventing IMAs
void main() {
1. int *np, n, i, *buf;
2. np = &n;
3. printf(“Enter size: “);
4. scanf(“%d”, np);
5. buf = malloc(n * sizeof(int));
6. for(i = 0; i <= n; i++)
7. *(buf + i) = rand()%10;
...
}
buf:
np:
5
i:
n:
4 2321555
9
8
2
3
3
+ =5 5
✘

101. Limiting the number of
taint marks
An unlimited number of taint marks makes a
hardware implementation infeasible
• increases the overhead (time and space)
• complicates the design

102. Limiting the number of
taint marks
An unlimited number of taint marks makes a
hardware implementation infeasible
• increases the overhead (time and space)
• complicates the design
➡ Assign taint marks from a limited, reusable
pool

103. Effects on the approach
With an random assignment of n taint marks the
detection probability is:
IMAs are detected probabilistically
p = 1
1
n

104. Effects on the approach
With an random assignment of n taint marks the
detection probability is:
IMAs are detected probabilistically
2 marks = 50%, 4 marks = 75%, 16 marks = 93.75%, 256 marks = 99.6%
p = 1
1
n

105. Effects on the approach
With an random assignment of n taint marks the
detection probability is:
1. The technique can be tuned by increasing or decreasing
the number of taint marks
IMAs are detected probabilistically
2 marks = 50%, 4 marks = 75%, 16 marks = 93.75%, 256 marks = 99.6%
p = 1
1
n

106. Effects on the approach
With an random assignment of n taint marks the
detection probability is:
1. The technique can be tuned by increasing or decreasing
the number of taint marks
2. In practice the approach is successful with only a small
number (2) of taint marks
IMAs are detected probabilistically
2 marks = 50%, 4 marks = 75%, 16 marks = 93.75%, 256 marks = 99.6%
p = 1
1
n

107. Empirical evaluation
RQ1: Is the efficiency of our approach sufficient for it
to be applied to deployed software?
RQ2:What is the effectiveness of our technique when
using limited number of taint marks?

108. RQ1: experimental method
• Hardware implementation
• Cycle accurate simuator (SESC)
• Treat taint marks as first class citizens
• Subjects
• SPEC CPU2000 benchmark (12 applications)
• Calculate the overhead imposed by our
approach for each subject application

109. RQ1: experimental method
• Hardware implementation
• Cycle accurate simuator (SESC)
• Treat taint marks as first class citizens
• Subjects
• SPEC CPU2000 benchmark (12 applications)
• Calculate the overhead imposed by our
approach for each subject application
Current implementation assigns taint marks only to
dynamically allocated memory, but propagation and
checking are fully implemented

110. RQ1: results
0
5
10
15
20
25
30
bzip2
crafty
eon
gap
gcc
gzip
m
cf
parser
perlbm
k
twolf
vortex
vpr
average
%overhead(time)
2 marks
8 marks
16 marks
256 marks

111. RQ1: results
0
5
10
15
20
25
30
bzip2
crafty
eon
gap
gcc
gzip
m
cf
parser
perlbm
k
twolf
vortex
vpr
average
%overhead(time)
2 marks
8 marks
16 marks
256 marks
• Even with 256 marks, the average overhead is
in the single digits

112. RQ1: results
0
5
10
15
20
25
30
bzip2
crafty
eon
gap
gcc
gzip
m
cf
parser
perlbm
k
twolf
vortex
vpr
average
%overhead(time)
2 marks
8 marks
16 marks
256 marks
• All attacks were detected with two taint marks
• Even with 256 marks, the average overhead is
in the single digits

113. RQ1: results
0
5
10
15
20
25
30
bzip2
crafty
eon
gap
gcc
gzip
m
cf
parser
perlbm
k
twolf
vortex
vpr
average
%overhead(time)
2 marks
8 marks
16 marks
256 marks
• All attacks were detected with two taint marks
• Software-only implementations impose two
orders of magnitude more overhead
• Even with 256 marks, the average overhead is
in the single digits

114. RQ2: experimental method
• Software implementation
• Binary instrumenter (Pin)
• Use instrumentation to assign, propagate, and check
taint marks
• Subjects
• SPEC CPU2000 benchmark (12 applications)
• 5 applications with 7 known IMAs
• Run both each applications protected by our
software implementation and check that only the
known illegal memory accesses are detected (5
times)

115. RQ2: results
Application IMA location Type Detected
bc-1.06 more_arrays: 177 buffer overflow ✔ (5/5)
bc-1.06 lookup: 577 buffer overflow ✔ (5/5)
gnupg-1.4.4 parse_comment: 2095 integer overflow ✔ (5/5)
mutt-1.4.2.li utf8_to_utf7: 199 buffer overflow ✔ (5/5)
php-5.2.0 php_char_to_str_ex: 3152 integer overflow ✔ (5/5)
pine-4.44 rfc882_cat: 260 buffer overflow ✔ (5/5)
squid-2.3 ftpBuildTitleUrl: 1024 buffer overflow ✔ (5/5)
Applications with known IMAs

116. RQ2: results
Application IMA location Type Detected
bc-1.06 more_arrays: 177 buffer overflow ✔ (5/5)
bc-1.06 lookup: 577 buffer overflow ✔ (5/5)
gnupg-1.4.4 parse_comment: 2095 integer overflow ✔ (5/5)
mutt-1.4.2.li utf8_to_utf7: 199 buffer overflow ✔ (5/5)
php-5.2.0 php_char_to_str_ex: 3152 integer overflow ✔ (5/5)
pine-4.44 rfc882_cat: 260 buffer overflow ✔ (5/5)
squid-2.3 ftpBuildTitleUrl: 1024 buffer overflow ✔ (5/5)
Applications with known IMAs
All attacks were detected with two taint marks

117. RQ2: results
Application IMA location Type Detected
bc-1.06 more_arrays: 177 buffer overflow ✔ (5/5)
bc-1.06 lookup: 577 buffer overflow ✔ (5/5)
gnupg-1.4.4 parse_comment: 2095 integer overflow ✔ (5/5)
mutt-1.4.2.li utf8_to_utf7: 199 buffer overflow ✔ (5/5)
php-5.2.0 php_char_to_str_ex: 3152 integer overflow ✔ (5/5)
pine-4.44 rfc882_cat: 260 buffer overflow ✔ (5/5)
squid-2.3 ftpBuildTitleUrl: 1024 buffer overflow ✔ (5/5)
Application IMA location Type Detected
vortex SendMsg: 279
null-pointer
dereference
✔ (5/5)
Applications with known IMAs
SPEC Benchmarks (“IMA free”)
All attacks were detected with two taint marks

118. Future work
• Complete implementation that handles
static memory
• Additional experiments with a wider range
of IMAs
• Further optimization of the hardware
implementation

119. Conclusions
• Definition of an approach for preventing illegal
memory accesses in deployed software
• uses dynamic taint analysis to protect memory
• uses probabilistic detection to achieve acceptable
overhead
• Empirical evaluation showing that the approach
• is effective at detecting IMA in real applications
• can be implemented efficiently in hardware