Recitation 4 Outline Buffer overflow –Practical skills for Lab 3 Code optimization –Strength reduction –Common sub-expression –Loop unrolling Reminders.

Recitation 4

Outline• Buffer overflow

– Practical skills for Lab 3• Code optimization

– Strength reduction – Common sub-expression– Loop unrolling

Reminders• Lab 3: due Thursday• Exam1: next Tuesday

Minglong ShaoE-mail:[email protected]

Office hours:Thursdays 5-6PMWean Hall 1315

Buffer overflow: example1

void example1(){ volatile int n; char buf[4]; volatile int x;

n = 0x12345678; x = 0xdeadbeef;

strcpy(buf, “abcdefg"); // ‘a’ = 0x61, ‘b’ = 0x62

buf[4] = 0xab; buf[-4] = 0xcd;

}

1. n = ? x = ?

2. n = ? x = ?

3. n = ? x = ?

void example1(){ volatile int n; char buf[4]; volatile int x;

n = 0x12345678; x = 0xdeadbeef;

strcpy(buf, “abcdefg"); // ‘a’ = 0x61, ‘b’ = 0x62

buf[4] = 0xab; buf[-4] = 0xcd;

}

Buffer overflow: example1example1:push %ebpmov %esp,%ebpsub $0x18,%espmovl $0x12345678,0xfffffffc(%ebp)movl $0xdeadbeef,0xfffffff4(%ebp)sub $0x8,%esppush $0x8048424lea 0xfffffff8(%ebp),%eaxpush %eaxcall 0x8048268 <strcpy>add $0x10,%espmovb $0xab,0xfffffffc(%ebp)movb $0xcd,0xfffffff4(%ebp)leaveret

• n is stored at %ebp-4• x is stored at %ebp-12• buf is stored at %ebp-8

Old %ebpReturn addr

n 0xfc0xf80xf4

%ebp

bufx

Breakpoint 1: before calling strcpy

Old %ebp%ebp

Return addr…

%esp

56 7812 34

be efde ad

Stack frame of example1

Address high low

1. n = 0x12345678 x = 0xdeadbeef

n

bufx

void example1(){ volatile int n; char buf[4]; volatile int x;

n = 0x12345678; x = 0xdeadbeef;

strcpy(buf, “abcdefg"); // ‘a’ = 0x61, ‘b’ = 0x62

buf[4] = 0xab; buf[-4] = 0xcd;

}

0xfc0xf80xf4

0xe8

Breakpoint 2: after calling strcpy

Old %ebp%ebp

Return addr…

%esp

0xfc0xf80xf40xf0

66 6500 67

be efde ad

Stack frame of example1

n

bufx

2. n = 0x00676665 x = 0xdeadbeef

62 6164 63

void example1(){ volatile int n; char buf[4]; volatile int x;

n = 0x12345678; x = 0xdeadbeef;

strcpy(buf, “abcdefg"); // ‘a’ = 0x61, ‘b’ = 0x62

buf[4] = 0xab; buf[-4] = 0xcd;

}

Address high low

Breakpoint 3: before return

Old %ebp%ebp

Return addr…

%esp

0xfc0xf80xf40xf0

66 ab00 67

be cdde ad

Stack frame of example1

n

bufx

3. n = 0x006766ab x = 0xdeadbecd

62 6164 63

void example1(){ volatile int n; char buf[4]; volatile int x;

n = 0x12345678; x = 0xdeadbeef;

strcpy(buf, “abcdefg"); // ‘a’ = 0x61, ‘b’ = 0x62

buf[4] = 0xab; buf[-4] = 0xcd;

}

Address high low

70 6972 71Old ebp74 7300 75Return addr

70 6900 71Old ebp

Write more characters …

• What if we instead strcpy(buf, "abcdefghijk");

11+1 chars

• What if we instead strcpy(buf, "abcdefghijklmno");

15+1 chars

• Old ebp is overwritten

• Return addr is overwritten

Return addr…

66 6568 67

be efde ad62 6164 63

…

nbufx

66 6568 67

be efde ad62 6164 63

old ebp

ret addr

nbufx

old ebp

ret addr

Put code onto stack: example2

push %ebpmov %esp,%ebpsub $0x24,%esplea 0xffffffe8(%ebp),%eaxpush %eaxcall 0x8048254 <gets>mov $0x0,%eaxleaveret

int example2 ()

{

char buf[16];

gets (buf);

return 0;

}

Old %ebpReturn addr

0xfc0xf80xf4

%ebp

buf

0xf00xec0xe8

Old %ebpReturn addr

… 0xfc0xf80xf4

………

My codeMy code

0xf00xec0xe8

%ebp

Steps

1. Write assembly code

2. Get binary representation of the code

3. Generate ASCII for the binary code

4. Generate string according to ASCII code

5. Run the program with the input

Write assembly code

• Use your favorite text editor

• For example, my exploit code is

movl $0, -8(%ebp)

addl $0x12345678, %eax

• Save as *.s, e.g. input.s

Generate input string# generate binary code: input.ounix> gcc –c input.s

# generate ASCII representation for the codeunix> objdump –d input.o00000000 <.text>: 0: c7 45 f8 00 00 00 00 movl $0x0,0xfffffff8(%ebp) 7: 05 78 56 34 12 add $0x12345678,%eax

# put the ASCII code in a text fileunix> cat > input.txtc7 45 f8 00 00 00 00 05 78 56 34 12 <ctrl-D>

# generate characters according to the ASCII fileunix> sendstring -f input.txt > input.raw

# check whether it’s correct with “od” commandunix> od -t x1 input.raw0000000 c7 45 f8 00 00 00 00 05 78 56 34 12 0a

Run the program with the inputunix> gdb example2

(gdb) break example2

(gdb) run < input.raw

(gdb) x/16b $ebp-240xbffff860: 0xb8 0x87 0x16 0x40 0xc0 0x81 0x16 0x40

0xbffff868: 0x78 0xf8 0xff 0xbf 0x41 0x82 0x04 0x08

(gdb) nexti 3 # go to the inst. after “call gets”

(gdb) x/16b $ebp-240xbffff860: 0xc7 0x45 0xf8 0x00 0x00 0x00 0x00 0x05

0xbffff868: 0x78 0x56 0x34 0x12 0x00 0x82 0x04 0x08

(gdb) disas 0xbffff860 0xbffff86cDump of assembler code from 0xbffff860 to 0xbffff86c:

0xbffff860 movl $0x0,0xfffffff8(%ebp)

0xbffff867 add $0x12345678,%eax

f8 60bf ff

Buffer overflow: execute your code

Execute your code: how?• Overwrite return address w/

starting address of your code– Pad more chars to input string– Set last 4 bytes to the addr.Strings:

c7 45 f8 00 00 00 00 05

78 56 34 12 xx xx xx xx

xx xx xx xx xx xx xx xx

xx xx xx xx 60 f8 ff bf

– Need more code for a successful attack

Old %ebpReturn addr

0xfc0xf80xf40xf00xec0xe8

%ebpab 3442 34

1e 00fd 79df e689 1c

56 7812 34

45 c700 f800 0005 00

……%esp 0xdc

0xbffff860

movl $0, -8(%ebp)addl $0x12345678, %eaxmovl $o_ebp, %ebppush $ret_addrret

Assembly code example

if (g_val >= 0) return g_val << 2;else return –g_val << 2;

movl 0xg_val_addr,%eax testl %eax,%eax jge .L1 negl %eax.L1: sall $2,%eax ret

Security lessons learned

• Never trust the input you receive: use bounds-checking on all buffers– In particular, never ever use gets.

• Even the man page says so!• gcc also warns you

Code optimization: strength reduction

• Turn complex operations into cheap ones.– Expensive operations:

• multiplication, division, modulus

– Cheap operations:• addition, bit operations, shifting

Strength Reduction

int sum = 0;for (int i=0; i<size; i++) { sum += array[i]*2;}

int sum = 0;for (int i=0; i<size; i++) { sum += array[i] << 1;}

Before:

...imull $2, %eax...

...sarl $1, %eax...

• 9.2 cycles per iteration • 8.7 cycles per iteration

After:

Common sub-expression

#define COLS (4)

void transpose(int **m, int a, int b)

{

int i, j, t;

for(i = 0; i < COLS; ++i)

{

for(j = 0; j < i; ++j)

{

t = m[i][j];

m[i][j] = m[j][i]*a*b;

m[j][i] = t*a*b;

}

}

}

#define COLS (4)

void transpose_opt(int **m, int a, int b)

{

int i, j, t, c = a*b;

for(i = 0; i < COLS; ++i)

{

for(j = 0; j < i; ++j)

{

t = m[i][j];

m[i][j] = m[j][i]*c;

m[j][i] = t*c;

}

}

}

Before: After:

Loop Unrolling

• Reduces the loop overhead of computing the loop index and testing the loop condition

• Perform more data operations in each iteration• Make sure to change the loop condition to not

run over array bounds• Take care of the final few elements one at a time

Loop Unrolling: Bubble Sortint a[7] = {5, 7, 1, 3, 8, 2, 9};int tmp;

for(i=0; i < n-1; i++) { for(j=0; j < n-1-i; j++) { if(a[j+1] < a[j]) { tmp = a[j]; a[j] = a[j+1]; a[j+1] = tmp; } }}

Loop Unrolling: Bubble Sort

int a[7] = {5, 7, 1, 3, 8, 2, 9};int tmp;

for(i=0; i < n-1; i++) { for(j=0; j < n-3-i; j+=3) {

// Unroll three times if( a[j+1] < a[j] ) { tmp = a[j]; a[j] = a[j+1]; a[j+1] = tmp; }; if( a[j+2] < a[j+1]) { tmp = a[j+1]; a[j+1] = a[j+2]; a[j+2] = tmp; };

... if( a[j+3] < a[j+2]) { tmp = a[j+2]; a[j+2] = a[j+3]; a[j+3] = tmp; }; }

// Finish up the remaining elements for(; j< n-1-i; j++){ if( a[j+1] < a[j] ){ tmp = a[j]; a[j] = a[j+1]; a[j+1] = tmp; }; }}

Running time comparison

Pentium III, Linux, process time

0

5

10

15

20

1.E+04 2.E+04 3.E+04 4.E+04

array length

runn

ing

time

[s]

bubble unrolling