Machine-Level Programming IV: Structured Data

Transcription

1 Machine-Level Programming IV: Structured Data Topics Arrays Structs

2 Basic Data Types Integral 2 Stored & operated on in general registers Signed vs. unsigned depends on instructions used Intel GAS Bytes C byte b 1 [unsigned] char word w 2 [unsigned] short double word l 4 [unsigned] int Floating Point Stored & operated on in floating point registers Intel GAS Bytes C Single s 4 float Double l 8 double Extended t 10/12 long double

3 Array Allocation Basic Principle T A[L]; Array of data type T and length L Contiguously allocated region of L * sizeof(t) bytes char string[12]; int val[5]; double a[4]; x x + 12 x x + 4 x + 8 x + 12 x + 16 x + 20 x x + 8 x + 16 x + 24 x char *p[3]; x x + 4 x + 8

4 Array Access Basic Principle T A[L]; Array of data type T and length L Identifier A can be used as a pointer to array element 0 int val[5]; Reference Type val[4] int 3 val int * x Value val+1 int * x + 4 &val[2] int * x + 8 val[5] int?? *(val+1) int 5 x x + 4 x + 8 x + 12 x + 16 x + 20 val + i int * x + 4 i

5 Array Example typedef int zip_dig[5]; zip_dig cmu = { 1, 5, 2, 1, 3 }; zip_dig mit = { 0, 2, 1, 3, 9 }; zip_dig ucb = { 9, 4, 7, 2, 0 }; 5 Notes zip_dig cmu; zip_dig mit; zip_dig ucb; Declaration zip_dig cmu equivalent to int cmu[5] Example arrays were allocated in successive 20 byte blocks Not guaranteed to happen in general

6 Array Accessing Example Computation Register %edx contains starting address of array Register %eax contains array index Desired digit at 4*%eax + %edx Use memory reference (%edx,%eax,4) int get_digit (zip_dig z, int dig) { return z[dig]; } Memory Reference Code # %edx = z # %eax = dig movl (%edx,%eax,4),%eax # z[dig] 6

7 7 Referencing Examples zip_dig cmu; zip_dig mit; zip_dig ucb; Code Does Not Do Any Bounds Checking! Reference Address mit[3] * 3 = 48 3 mit[5] * 5 = 56 9 mit[-1] *-1 = 32 3 cmu[15] *15 = 76?? Value Out of range behavior implementation-dependent No guaranteed relative allocation of different arrays Guaranteed? Yes No No No

8 Array Loop Example Original Source Transformed Version 8 As generated by GCC Eliminate loop variable i Convert array code to pointer code Express in do-while form No need to test at entrance int zd2int(zip_dig z) { int i; int zi = 0; for (i = 0; i < 5; i++) { zi = 10 * zi + z[i]; } return zi; } int zd2int(zip_dig z) { int zi = 0; int *zend = z + 4; do { zi = 10 * zi + *z; z++; } while(z <= zend); return zi; }

9 Array Loop Implementation Registers 9 %ecx %eax %ebx z zi zend Computations 10*zi + *z implemented as *z + 2*(zi+4*zi) z++ increments by 4 int zd2int(zip_dig z) { int zi = 0; int *zend = z + 4; do { zi = 10 * zi + *z; z++; } while(z <= zend); return zi; } # %ecx = z xorl %eax,%eax # zi = 0 leal 16(%ecx),%ebx # zend = z+4.l59: leal (%eax,%eax,4),%edx # 5*zi movl (%ecx),%eax # *z addl $4,%ecx # z++ leal (%eax,%edx,2),%eax # zi = *z + 2*(5*zi) cmpl %ebx,%ecx # z : zend jle.l59 # if <= goto loop

10 Structures Concept Contiguously-allocated region of memory Refer to members within structure by names Members may be of different types struct rec { int i; int a[3]; int *p; }; Memory Layout i a p Accessing Structure Member void set_i(struct rec *r, int val) { r->i = val; } Assembly # %eax = val # %edx = r movl %eax,(%edx) # Mem[r] = val 10

11 Generating Pointer to Struct. Member struct rec { int i; int a[3]; int *p; }; r i a p r *idx Generating Pointer to Array Element Offset of each structure member determined at compile time int * find_a (struct rec *r, int idx) { return &r->a[idx]; } # %ecx = idx # %edx = r leal 0(,%ecx,4),%eax # 4*idx leal 4(%eax,%edx),%eax # r+4*idx+4 11

12 Structure Referencing (Cont.) C Code struct rec { int i; int a[3]; int *p; }; void set_p(struct rec *r) { r->p = &r->a[r->i]; } i a p i a Element i # %edx = r movl (%edx),%ecx # r->i leal 0(,%ecx,4),%eax # 4*(r->i) leal 4(%edx,%eax),%eax # r+4+4*(r->i) movl %eax,16(%edx) # Update r->p 12

13 Alignment Aligned Data 13 Primitive data type requires K bytes Address must be multiple of K Required on some machines; advised on IA32 treated differently by Linux and Windows! Motivation for Aligning Data Memory accessed by (aligned) double or quad-words Compiler Inefficient to load or store datum that spans quad word boundaries Virtual memory very tricky when datum spans 2 pages Inserts gaps in structure to ensure correct alignment of fields

14 Specific Cases of Alignment Size of Primitive Data Type: 14 1 byte (e.g., char) no restrictions on address 2 bytes (e.g., short) lowest 1 bit of address must be bytes (e.g., int, float, char *, etc.) lowest 2 bits of address must be bytes (e.g., double) Windows (and most other OS s & instruction sets):» lowest 3 bits of address must be Linux:» lowest 2 bits of address must be 00 2» i.e., treated the same as a 4-byte primitive data type 12 bytes (long double) Linux:» lowest 2 bits of address must be 00 2» i.e., treated the same as a 4-byte primitive data type

15 Satisfying Alignment with Structures Offsets Within Structure Must satisfy element s alignment requirement Overall Structure Placement Each structure has alignment requirement K Largest alignment of any element Initial address & structure length must be multiples of K struct S1 { char c; int i[2]; double v; } *p; Example (under Windows): K = 8, due to double element c i[0] i[1] v p+0 p+4 p+8 p+16 p Multiple of 4 Multiple of 8 Multiple of 8 Multiple of 8

16 Linux vs. Windows Windows (including Cygwin): K = 8, due to double element struct S1 { char c; int i[2]; double v; } *p; Linux: c i[0] i[1] v p+0 p+4 p+8 p+16 p+24 Multiple of 4 Multiple of 8 Multiple of 8 Multiple of 8 K = 4; double treated like a 4-byte data type c i[0] i[1] p+0 p+4 p+8 v p+12 p Multiple of 4 Multiple of 4 Multiple of 4 Multiple of 4

17 Overall Alignment Requirement struct S2 { double x; int i[2]; char c; } *p; p must be multiple of: 8 for Windows 4 for Linux x i[0] i[1] c p+0 p+8 p+12 p+16 Windows: p+24 Linux: p+20 struct S3 { float x[2]; int i[2]; char c; p must be multiple of 4 (in either OS) } *p; x[0] x[1] i[0] i[1] c p+0 p+4 p+8 p+12 p+16 p+20 17

18 Ordering Elements Within Structure struct S4 { char c1; double v; char c2; int i; } *p; 10 bytes wasted space in Windows c1 v p+0 p+8 p+16 p+20 p+24 c2 i struct S5 { double v; char c1; char c2; int i; } *p; 2 bytes wasted space v c1 c2 p+0 p+8 p+12 p+16 i 18

19 Arrays of Structures Principle Allocated by repeating allocation for array type In general, may nest arrays & structures to arbitrary depth struct S6 { short i; float v; short j; } a[10]; a[1].i a[1].v a[1].j a+12 a+16 a+20 a+24 a[0] a[1] a[2] a+0 a+12 a+24 a+36 19

20 Satisfying Alignment within Structure Achieving Alignment Starting address of structure array must be multiple of worst-case alignment for any element a must be multiple of 4 Offset of element within structure must be multiple of element s alignment requirement v s offset of 4 is a multiple of 4 Overall size of structure must be multiple of worst-case alignment for any element Structure padded with unused space to be 12 bytes a+0 a[0] struct S6 { short i; float v; short j; } a[10]; a[i] a+12i Multiple of 4 a[1].i a+12i a+12i+4 a[1].v a[1].j 20 Multiple of 4

21 Summary Arrays in C Contiguous allocation of memory Pointer to first element No bounds checking Compiler Optimizations Compiler often turns array code into pointer code (zd2int) Uses addressing modes to scale array indices Structures Allocate bytes in order declared Pad in middle and at end to satisfy alignment 21

22 Machine-Level Programming V: Miscellaneous Topics Topics Linux Memory Layout Understanding Pointers Buffer Overflow Floating Point Code 22

23 Upper 2 hex digits of address 23 FF C0 BF 80 7F Red Hat v. 6.2 ~1920MB memory limit 40 3F Stack Heap DLLs Heap Data Text Stack Heap DLLs Data Text Linux Memory Layout Runtime stack (8MB limit) Dynamically allocated storage When call malloc, calloc, new Dynamically Linked Libraries Library routines (e.g., printf, malloc) Linked into object code when first executed Statically allocated data E.g., arrays & strings declared in code Executable machine instructions Read-only

24 BF Linux Memory Allocation Initially Stack BF Linked Stack BF Some Heap Stack BF More Heap Stack 80 7F 80 7F 80 7F 80 7F Heap Heap 40 3F 40 3F DLLs 40 3F DLLs 40 3F DLLs Data Text Data Text Data Text Heap Data Text

25 C operators Operators Associativity () [] ->. left to right! ~ * & (type) sizeof right to left * / % left to right + - left to right << >> left to right < <= > >= left to right ==!= left to right & left to right ^ left to right left to right && left to right left to right?: right to left = += -= *= /= %= &= ^=!= <<= >>= right to left, left to right Note: Unary +, -, and * have higher precedence than binary forms 25

26 C pointer declarations int *p int *p[13] int *(p[13]) int **p int (*p)[13] int *f() int (*f)() int (*(*f())[13])() int (*(*x[3])())[5] 26 p is a pointer to int p is an array[13] of pointer to int p is an array[13] of pointer to int p is a pointer to a pointer to an int p is a pointer to an array[13] of int f is a function returning a pointer to int f is a pointer to a function returning int f is a function returning ptr to an array[13] of pointers to functions returning int x is an array[3] of pointers to functions returning pointers to array[5] of ints

27 Internet Worm and IM War November, 1988 Internet Worm attacks thousands of Internet hosts. How did it happen? July, 1999 Microsoft launches MSN Messenger (instant messaging system). Messenger clients can access popular AOL Instant Messaging Service (AIM) servers AIM client MSN server MSN client AIM server 27 AIM client

28 Internet Worm and IM War (cont.) August 1999 Mysteriously, Messenger clients can no longer access AIM servers. Microsoft and AOL begin the IM war: AOL changes server to disallow Messenger clients Microsoft makes changes to clients to defeat AOL changes. At least 13 such skirmishes. How did it happen? The Internet Worm and AOL/Microsoft War were both based on stack buffer overflow exploits! many Unix functions do not check argument sizes. allows target buffers to overflow. 28

29 String Library Code 29 Implementation of Unix function gets No way to specify limit on number of characters to read Similar problems with other Unix functions /* Get string from stdin */ char *gets(char *dest) { int c = getc(); char *p = dest; while (c!= EOF && c!= '\n') { *p++ = c; c = getc(); } *p = '\0'; return dest; } strcpy: Copies string of arbitrary length scanf, fscanf, sscanf, when given %s conversion specification

30 Vulnerable Buffer Code /* Echo Line */ void echo() { char buf[4]; /* Way too small! */ gets(buf); puts(buf); } int main() { printf("type a string:"); echo(); return 0; } 30

31 Buffer Overflow Executions unix>./bufdemo Type a string: unix>./bufdemo Type a string:12345 Segmentation Fault unix>./bufdemo Type a string: Segmentation Fault 31

32 Buffer Overflow Stack Stack Frame for main Return Address Saved %ebp [3][2][1][0] buf %ebp /* Echo Line */ void echo() { char buf[4]; /* Way too small! */ gets(buf); puts(buf); } 32 Stack Frame for echo echo: pushl %ebp movl %esp,%ebp subl $20,%esp pushl %ebx addl $-12,%esp leal -4(%ebp),%ebx pushl %ebx call gets... # Save %ebp on stack # Allocate space on stack # Save %ebx # Allocate space on stack # Compute buf as %ebp-4 # Push buf on stack # Call gets

33 Buffer Overflow Stack Example unix> gdb bufdemo (gdb) break echo Breakpoint 1 at 0x (gdb) run Breakpoint 1, 0x in echo () (gdb) print /x *(unsigned *)$ebp $1 = 0xbffff8f8 (gdb) print /x *((unsigned *)$ebp + 1) $3 = 0x804864d Stack Frame for main Stack Frame for main Before call to gets Return Address Saved %ebp [3][2][1][0] buf %ebp Return Address 86 4d bfsaved ff %ebpf8 0xbffff8d8 [3][2][1][0] xx xx xx xx buf Stack Frame for echo Stack Frame for echo : call c <echo> d: mov 0xffffffe8(%ebp),%ebx # Return Point

34 Buffer Overflow Example #1 Before Call to gets Input = 123 Stack Frame for main Stack Frame for main Return Address Saved %ebp [3][2][1][0] buf %ebp Return Address 86 4d bfsaved ff %ebpf8 0xbffff8d8 [3][2][1][0] buf Stack Frame for echo Stack Frame for echo No Problem 34

35 Buffer Overflow Stack Example #2 Stack Frame for main Stack Frame for main Input = Return Address Saved %ebp [3][2][1][0] buf Stack Frame for echo echo code: %ebp Return Address 86 4d bfsaved ff 00 %ebp35 0xbffff8d8 [3][2][1][0] buf Stack Frame for echo Saved value of %ebp set to 0xbfff0035 Bad news when later attempt to restore %ebp : push %ebx : call 80483e4 <_init+0x50> # gets : mov 0xffffffe8(%ebp),%ebx b: mov %ebp,%esp d: pop %ebp # %ebp gets set to invalid value e: ret 35

36 Buffer Overflow Stack Example #3 Stack Frame for main Stack Frame for main Input = Return Address Saved %ebp [3][2][1][0] buf %ebp Return Address Saved %ebp35 0xbffff8d8 [3][2][1][0] buf Stack Frame for echo Stack Frame for echo Invalid address %ebp and return address corrupted No longer pointing to desired return point : call c <echo> d: mov 0xffffffe8(%ebp),%ebx # Return Point 36

37 Malicious Use of Buffer Overflow Stack after call to gets() return address A void foo(){ bar();... } void bar() { char buf[64]; gets(buf);... } data written by gets() B B pad exploit code foo stack frame bar stack frame 37 Input string contains byte representation of executable code Overwrite return address with address of buffer When bar() executes ret, will jump to exploit code

38 Exploits Based on Buffer Overflows Buffer overflow bugs allow remote machines to execute arbitrary code on victim machines. Internet worm Early versions of the finger server (fingerd) used gets() to read the argument sent by the client: finger Worm attacked fingerd server by sending phony argument: finger exploit-code padding new-return-address exploit code: executed a root shell on the victim machine with a direct TCP connection to the attacker. 38

39 Exploits Based on Buffer Overflows Buffer overflow bugs allow remote machines to execute arbitrary code on victim machines. IM War AOL exploited existing buffer overflow bug in AIM clients exploit code: returned 4-byte signature (the bytes at some location in the AIM client) to server. When Microsoft changed code to match signature, AOL changed signature location. 39

40 Date: Wed, 11 Aug :30: (PDT) From: Phil Bucking Subject: AOL exploiting buffer overrun bug in their own software! To: Mr. Smith, I am writing you because I have discovered something that I think k you might find interesting because you are an Internet security expert with experience in this area. I have also tried to contact AOL but received no response. I am a developer who has been working on a revolutionary new instant messaging client that should be released later this year.... It appears that the AIM client has a buffer overrun bug. By itself this might not be the end of the world, as MS surely has had its share. But AOL is now *exploiting their own buffer overrun bug* to help in its efforts to block MS Instant Messenger.... Since you have significant credibility with the press I hope that t you can use this information to help inform people that behind AOL's friendly exterior they are nefariously compromising peoples' security. Sincerely, Phil Bucking Founder, Bucking Consulting philbucking@yahoo.com It was later determined that this originated from within Microsoft! 40

41 Code Red Worm History June 18, Microsoft announces buffer overflow vulnerability in IIS Internet server July 19, over 250,000 machines infected by new virus in 9 hours White house must change its IP address. Pentagon shut down public WWW servers for day When We Set Up CS:APP Web Site Received strings of form GET /default.ida?nnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnnn...nnnn NNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNNN%u9090%u6858%ucbd3%u780 1%u9090%u6858%ucbd3%u7801%u9090%u6858%ucbd3%u7801%u9090%u909 0%u8190%u00c3%u0003%u8b00%u531b%u53ff%u0078%u0000%u00=a HTTP/1.0" "-" "-" 41

42 Code Red Exploit Code Starts 100 threads running Spread self Generate random IP addresses & send attack string Between 1st & 19th of month Attack Send 98,304 packets; sleep for 4-1/2 hours; repeat» Denial of service attack Between 21st & 27th of month Deface server s home page After waiting 2 hours 42

43 Code Red Effects Later Version Even More Malicious Code Red II As of April, 2002, over 18,000 machines infected Still spreading Paved Way for NIMDA Variety of propagation methods One was to exploit vulnerabilities left behind by Code Red II 43

44 Avoiding Overflow Vulnerability /* Echo Line */ void echo() { char buf[4]; /* Way too small! */ fgets(buf, 4, stdin); puts(buf); } Use Library Routines that Limit String Lengths fgets instead of gets strncpy instead of strcpy Don t use scanf with %s conversion specification Use fgets to read the string 44

45 IA32 Floating Point History 8086: first computer to implement IEEE FP separate 8087 FPU (floating point unit) 486: merged FPU and Integer Unit onto one chip Summary Hardware to add, multiply, and divide Floating point data registers Various control & status registers Integer Unit Instruction decoder and sequencer FPU Floating Point Formats 45 single precision (C float): 32 bits double precision (C double): 64 bits extended precision (C long double): 80 bits Memory

46 FPU Data Register Stack FPU register format (extended precision) s exp frac 0 FPU registers 8 registers Logically forms shallow stack Top called %st(0) When push too many, bottom values disappear Top %st(3) %st(2) %st(1) %st(0) stack grows down 46

47 FPU instructions Large number of floating point instructions and formats ~50 basic instruction types load, store, add, multiply sin, cos, tan, arctan, and log! Sample instructions: Instruction Effect Description fldz push 0.0 Load zero flds Addr push M[Addr] Load single precision real fmuls Addr %st(0) <- %st(0)*m[addr] Multiply faddp %st(1) <- %st(0)+%st(1); pop Add and pop 47

48 Floating Point Code Example Compute Inner Product of Two Vectors 48 Single precision arithmetic Common computation float ipf (float x[], float y[], int n) { int i; float result = 0.0; } for (i = 0; i < n; i++) { result += x[i] * y[i]; } return result; pushl %ebp movl %esp,%ebp pushl %ebx # setup movl 8(%ebp),%ebx # %ebx=&x movl 12(%ebp),%ecx # %ecx=&y movl 16(%ebp),%edx # %edx=n fldz # push +0.0 xorl %eax,%eax # i=0 cmpl %edx,%eax # if i>=n done jge.l3.l5: flds (%ebx,%eax,4) # push x[i] fmuls (%ecx,%eax,4) # st(0)*=y[i] faddp # st(1)+=st(0); pop incl %eax # i++ cmpl %edx,%eax # if i<n repeat jl.l5.l3: movl -4(%ebp),%ebx # finish movl %ebp, %esp popl %ebp ret # st(0) = result

49 Inner Product Stack Trace Initialization 1. fldz 0.0 %st(0) Iteration 0 Iteration 1 2. flds (%ebx,%eax,4) 0.0 %st(1) x[0] %st(0) 3. fmuls (%ecx,%eax,4) 0.0 %st(1) x[0]*y[0] %st(0) 5. flds (%ebx,%eax,4) x[0]*y[0] x[1] 6. fmuls (%ecx,%eax,4) x[0]*y[0] x[1]*y[1] %st(1) %st(0) %st(1) %st(0) faddp 0.0+x[0]*y[0] %st(0) 7. faddp x[0]*y[0]+x[1]*y[1] %st(0)

50 Final Observations Memory Layout OS/machine dependent (including kernel version) Basic partitioning: stack/data/text/heap/dll found in most machines Type Declarations in C Notation obscure, but very systematic Working with Strange Code Important to analyze nonstandard cases E.g., what happens when stack corrupted due to buffer overflow Helps to step through with GDB IA32 Floating Point Strange shallow stack architecture 50