Quick introduction to reverse engineering for beginners

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is turned off. Now let’s compile all this in MSVC 2010, with optimization turned on (/Ox): _str$ = 8 ; size = 4 _strlen PROC mov ecx, DWORD PTR _str$[esp-4] ; ECX -> pointer to the string mov eax, ecx ; move to EAX $LL2@strlen_: mov dl, BYTE PTR [eax] ; DL = *EAX inc eax ; EAX++ test dl, dl ; DL==0? jne SHORT $LL2@strlen_ ; no, continue loop sub eax, ecx ; calculate pointers difference dec eax ; decrement EAX ret 0 _strlen_ ENDP Now it’s all simpler. But it is needless to say that compiler could use registers such efficiently only in small functions with small number of local variables. INC/DEC — are increment/decrement instruction, in other words: add 1 to variable or subtract. Let’s check GCC 4.4.1 with optimization turned on (-O3 key): public strlen strlen proc near arg_0 = dword ptr 8 push ebp mov ebp, esp mov ecx, [ebp+arg_0] mov eax, ecx loc_8048418: movzx edx, byte ptr [eax] add eax, 1 test dl, dl jnz short loc_8048418 not ecx add eax, ecx pop retn ebp strlen endp Here GCC is almost the same as MSVC, except of MOVZX presence. However, MOVZX could be replaced here to mov dl, byte ptr [eax]. Probably, it is simpler for GCC compiler’s code generator to remember that whole register is allocated for char variable and it can be sure that highest bits will not contain noise at any point. After, we also see new instruction NOT. This instruction inverts all bits in operand. It can be said, it is synonym to XOR ECX, 0ffffffffh instruction. NOT and following ADD calculating pointer difference and subtracting 1. At the beginning ECX, where pointer to str is stored, inverted and 1 is subtracted from it. See also: “Signed number representations” 2.4. In other words, at the end of function, just after loop body, these operations are executed: ecx=str; eax=eos; ecx=(-ecx)-1; 33
eax=eax+ecx return eax ... and this is equivalent to: ecx=str; eax=eos; eax=eax-ecx; eax=eax-1; return eax Why GCC decided it would be better? I cannot be sure. But I’m assure that both variants are equivalent in efficiency sense. 34
Page 1 and 2: Quick introduction to reverse engin
Page 3 and 4: 1.16 C++ classes . . . . . . . . .
Page 5 and 6: Preface Here (will be) some of my n
Page 7 and 8: 1.1 Hello, world! Let’s start wit
Page 9 and 10: 1.2 Stack Stack — is one of the m
Page 11 and 12: (_snprintf() function works just li
Page 13 and 14: 1.3 printf() with several arguments
Page 15 and 16: 1.4 scanf() Now let’s use scanf()
Page 17 and 18: GCC replaced first printf() call to
Page 19 and 20: }; return 0; printf ("What you ente
Page 21 and 22: 1.5 Passing arguments via stack Now
Page 23 and 24: 1.6 One more word about results ret
Page 25 and 26: push OFFSET $SG741 ; ’a
Page 27 and 28: 1.8 switch()/case/default 1.8.1 Few
Page 29 and 30: 1.8.2 A lot of cases If switch() st
Page 31 and 32: loc_804840C: ; DATA XREF: .rodata:0
Page 33 and 34: et 0 _main ENDP Nothing very specia
Page 35 and 36: add esp, 1Ch xor eax, eax ; return
Page 37: C/C++ standard defines char type as
Page 41 and 42: mov eax, ecx imul edx sar edx, 1 mo
Page 43 and 44: ; current stack state: ST(0) = resu
Page 45 and 46: ; in local stack here 8 bytes still
Page 47 and 48: fld QWORD PTR _a$[esp-4] ; current
Page 49 and 50: In other words, fnstsw ax / sahf in
Page 51 and 52: 1.13 Arrays Array is just a set of
Page 53 and 54: mov [esp+70h+var_70], eax call _pri
Page 55 and 56: jge SHORT $LN1@main mov ecx, DWORD
Page 57 and 58: ebp 0x15 0x15 esi 0x0 0 edi 0x0 0 e
Page 59 and 60: 1.14 Bit fields A lot of functions
Page 61 and 62: So, let’s download Linux Kernel 2
Page 63: There are some redundant code prese
Page 66 and 67: y 10 just by dropping last digit (3
Page 68 and 69: _key$ = 8 ; size = 4 _len$ = 12 ; s
Page 70 and 71: 1.15 Structures It can be defined t
Page 72 and 73: call DWORD PTR __imp__GetSystemTime
Page 74 and 75: 1.15.4 Fields packing in structure
Page 76 and 77: }; int b; struct outer_struct { cha
Page 78 and 79: }; printf ("stepping=%d\n", tmp->st
Page 80 and 81: call ___printf_chk mov eax, esi shr
Page 82 and 83: mov edx, DWORD PTR _t$[ebp] or edx,
Page 84 and 85: 1.16 C++ classes I placed a C++ cla
Page 86 and 87: push ecx mov DWORD PTR _this$[ebp],
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call _ZN1c4dumpEv lea eax, [esp+20h
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1.17 Unions 1.17.1 Pseudo-random nu
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xor eax, eax pop esi mov esp, ebp p
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Let’s compile it in MSVC 2010 (I
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comp() function: public comp comp p
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Some compilers can do vectorization
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mov ecx, [esp+10h+var_10] mov edx,
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GCC In all other cases, non-SSE2 co
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add ebx, edx lea esi, [esi+0] loc_8
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movdqa xmm1, XMMWORD PTR [ecx+16] a
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1.20 x86-64 It’s a 64-bit extensi
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} x40 = a3 | x31; x41 = x24 & ~x37;
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or ebx, DWORD PTR _x6$[esp+36] mov
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and r11, r8 and rsi, r8 and r10, r1
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pop rdi pop rsi ret 0 s1 ENDP Nothi
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2.1 LEA instruction LEA (Load Effec
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2.3 npad It’s an assembler macro
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2.4 Signed number representations T
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call function function: .. do somet
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3.2 String Debugging messages are o
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.text:3011E91B DD 1E fstp qword ptr
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loc_80483F2: pop ebp retn _f endp 4
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loc_8048444: loc_8048448: loc_80484
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public f2 f2 proc near push ebp mov
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loc_804840A: loc_804842E: loc_80484
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mov esi, DWORD PTR _k$[esp+16] push
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or eax, edx retn f endp 4.1.8 Task
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4.2 Middle level 4.2.1 Task 2.1 Wel
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mov [edx+eax], bl mov ebx, edi mov
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loc_80485AB: ; CODE XREF: f+1E0 ; f
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mov [ecx+eax], dl inc eax cmp ebx,
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Chapter 5 Tools ∙ IDA as disassem
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Chapter 7 More examples 147
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.text:00541050 arg_0 = dword ptr 4
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.text:005410F3 .text:005410F3 loc_5
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.text:005411A9 pop ebx .text:005411
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.text:00541249 .text:00541249 next_
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.text:005412FC mov esi, offset cube
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.text:005413DC call _fwrite ; write
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.text:005414A7 push ecx ; Filename
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.text:005413A3 mov ecx, [esp+44h+pa
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.text:00541408 push offset aRb ; "r
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}; }; return; fseek (f, 0, SEEK_END
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.text:005411B5 mov ebp, eax Check f
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.text:0054124C inc ebp .text:005412
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This instruction clears bit, in oth
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.text:005410CD add esp, 40h .text:0
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}; for (i=0; i
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}; tmp[y][z]=get_bit (row, y, z); f
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}; fread (buf, flen, 1, f); fclose
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7.2 SAP client network traffic comp
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.text:64413F68 .text:64413F68 loc_6
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.text:64405171 call dbg .text:64405
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.text:64404FF7 push offset aLogging
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.text:64405113 loc_64405113: .text:
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Now let’s dig deeper and find con
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.text:64411E0B .text:64411E0B loc_6
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Chapter 8 Other things 8.1 Compiler
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Chapter 9 Tasks solutions 9.1 Easy
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9.1.5 Task 1.5 Hint #1: Keep in min
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} 9.1.8 Task 1.8 Solution: two 100*
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#define PUSH(low, high) ((void) ((t
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} } ignore one or both. Otherwise,
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Quick introduction to reverse engineering for beginners

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