Quick introduction to reverse engineering for beginners

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This instruction clears bit, in other words, it saves all bits in CL which are also set in DL except those in DL which are cleared. This means that if DL is, for example, 11101111 in binary form, all bits will be saved except 5th (counting from lowest bit). .text:00541036 and cl, dl Store it back: .text:00541038 mov cube64[eax+esi*8], cl .text:0054103F pop esi .text:00541040 retn .text:00541040 set_bit endp It is almost the same as get_bit(), except, if arg_C is zero, the function clears specific bit in array, or sets it otherwise. We also know that array size is 64. First two arguments both in set_bit() and get_bit() could be seen as 2D cooridnates. Then array will be 8*8 matrix. Here is C representation of what we already know: #define IS_SET(flag, bit) ((flag) & (bit)) #define SET_BIT(var, bit) ((var) |= (bit)) #define REMOVE_BIT(var, bit) ((var) &= ~(bit)) char cube[8][8]; void set_bit (int x, int y, int shift, int bit) { if (bit) SET_BIT (cube[x][y], 1
EBX is a pointer to internal array: .text:0054107D lea ebx, [esp+50h+internal_array_64] .text:00541081 Two nested loops are here: .text:00541081 first_loop1_begin: .text:00541081 xor esi, esi ; ESI is loop 2 counter .text:00541083 .text:00541083 first_loop2_begin: .text:00541083 push ebp ; arg_0 .text:00541084 push esi ; loop 1 counter .text:00541085 push edi ; loop 2 counter .text:00541086 call get_bit .text:0054108B add esp, 0Ch .text:0054108E mov [ebx+esi], al ; store to internal array .text:00541091 inc esi ; increment loop 1 counter .text:00541092 cmp esi, 8 .text:00541095 jl short first_loop2_begin .text:00541097 inc edi ; increment loop 2 counter .text:00541098 add ebx, 8 ; increment internal array pointer by 8 at each loop 1 iteration .text:0054109B cmp edi, 8 .text:0054109E jl short first_loop1_begin ... we see that both loop counters are in range 0..7. Also, they are used as first and second arguments of get_bit(). Third argument of get_bit() is the only argument of rotate1(). What get_bit() returns, is being placed into internal array. Prepare pointer to internal array again: .text:005410A0 lea ebx, [esp+50h+internal_array_64] .text:005410A4 mov edi, 7 ; EDI is loop 1 counter, initial state is 7 .text:005410A9 .text:005410A9 second_loop1_begin: .text:005410A9 xor esi, esi ; ESI is loop 2 counter .text:005410AB .text:005410AB second_loop2_begin: .text:005410AB mov al, [ebx+esi] ; value from internal array .text:005410AE push eax .text:005410AF push ebp ; arg_0 .text:005410B0 push edi ; loop 1 counter .text:005410B1 push esi ; loop 2 counter .text:005410B2 call set_bit .text:005410B7 add esp, 10h .text:005410BA inc esi ; increment loop 2 counter .text:005410BB cmp esi, 8 .text:005410BE jl short second_loop2_begin .text:005410C0 dec edi ; decrement loop 2 counter .text:005410C1 add ebx, 8 ; increment pointer in internal array .text:005410C4 cmp edi, 0FFFFFFFFh .text:005410C7 jg short second_loop1_begin .text:005410C9 pop edi .text:005410CA pop esi .text:005410CB pop ebp .text:005410CC pop ebx 174
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Quick introduction to reverse engin
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1.16 C++ classes . . . . . . . . .
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Preface Here (will be) some of my n
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1.1 Hello, world! Let’s start wit
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1.2 Stack Stack — is one of the m
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(_snprintf() function works just li
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1.3 printf() with several arguments
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1.4 scanf() Now let’s use scanf()
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GCC replaced first printf() call to
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}; return 0; printf ("What you ente
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1.5 Passing arguments via stack Now
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1.6 One more word about results ret
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push OFFSET $SG741 ; ’a
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1.8 switch()/case/default 1.8.1 Few
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1.8.2 A lot of cases If switch() st
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loc_804840C: ; DATA XREF: .rodata:0
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et 0 _main ENDP Nothing very specia
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add esp, 1Ch xor eax, eax ; return
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C/C++ standard defines char type as
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eax=eax+ecx return eax ... and this
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mov eax, ecx imul edx sar edx, 1 mo
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; current stack state: ST(0) = resu
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; in local stack here 8 bytes still
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fld QWORD PTR _a$[esp-4] ; current
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In other words, fnstsw ax / sahf in
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1.13 Arrays Array is just a set of
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mov [esp+70h+var_70], eax call _pri
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jge SHORT $LN1@main mov ecx, DWORD
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ebp 0x15 0x15 esi 0x0 0 edi 0x0 0 e
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1.14 Bit fields A lot of functions
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So, let’s download Linux Kernel 2
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There are some redundant code prese
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y 10 just by dropping last digit (3
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_key$ = 8 ; size = 4 _len$ = 12 ; s
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1.15 Structures It can be defined t
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call DWORD PTR __imp__GetSystemTime
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1.15.4 Fields packing in structure
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}; int b; struct outer_struct { cha
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}; printf ("stepping=%d\n", tmp->st
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call ___printf_chk mov eax, esi shr
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mov edx, DWORD PTR _t$[ebp] or edx,
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1.16 C++ classes I placed a C++ cla
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push ecx mov DWORD PTR _this$[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$[esp+36] mov
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and r11, r8 and rsi, r8 and r10, r1
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pop rdi pop rsi ret 0 s1 ENDP Nothi
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2.1 LEA instruction LEA (Load Effec
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2.3 npad It’s an assembler macro
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2.4 Signed number representations T
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call function function: .. do somet
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3.2 String Debugging messages are o
Page 128 and 129: .text:3011E91B DD 1E fstp qword ptr
Page 130 and 131: loc_80483F2: pop ebp retn _f endp 4
Page 132 and 133: loc_8048444: loc_8048448: loc_80484
Page 134 and 135: public f2 f2 proc near push ebp mov
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Page 138 and 139: mov esi, DWORD PTR _k$[esp+16] push
Page 140 and 141: or eax, edx retn f endp 4.1.8 Task
Page 142 and 143: 4.2 Middle level 4.2.1 Task 2.1 Wel
Page 144 and 145: mov [edx+eax], bl mov ebx, edi mov
Page 146 and 147: loc_80485AB: ; CODE XREF: f+1E0 ; f
Page 148 and 149: mov [ecx+eax], dl inc eax cmp ebx,
Page 150 and 151: Chapter 5 Tools ∙ IDA as disassem
Page 152 and 153: Chapter 7 More examples 147
Page 154 and 155: .text:00541050 arg_0 = dword ptr 4
Page 156 and 157: .text:005410F3 .text:005410F3 loc_5
Page 158 and 159: .text:005411A9 pop ebx .text:005411
Page 160 and 161: .text:00541249 .text:00541249 next_
Page 162 and 163: .text:005412FC mov esi, offset cube
Page 164 and 165: .text:005413DC call _fwrite ; write
Page 166 and 167: .text:005414A7 push ecx ; Filename
Page 168 and 169: .text:005413A3 mov ecx, [esp+44h+pa
Page 170 and 171: .text:00541408 push offset aRb ; "r
Page 172 and 173: }; }; return; fseek (f, 0, SEEK_END
Page 174 and 175: .text:005411B5 mov ebp, eax Check f
Page 176 and 177: .text:0054124C inc ebp .text:005412
Page 180 and 181: .text:005410CD add esp, 40h .text:0
Page 182 and 183: }; for (i=0; i
Page 184 and 185: }; tmp[y][z]=get_bit (row, y, z); f
Page 186 and 187: }; fread (buf, flen, 1, f); fclose
Page 188 and 189: 7.2 SAP client network traffic comp
Page 190 and 191: .text:64413F68 .text:64413F68 loc_6
Page 192 and 193: .text:64405171 call dbg .text:64405
Page 194 and 195: .text:64404FF7 push offset aLogging
Page 196 and 197: .text:64405113 loc_64405113: .text:
Page 198 and 199: Now let’s dig deeper and find con
Page 200 and 201: .text:64411E0B .text:64411E0B loc_6
Page 202 and 203: Chapter 8 Other things 8.1 Compiler
Page 204 and 205: Chapter 9 Tasks solutions 9.1 Easy
Page 206 and 207: 9.1.5 Task 1.5 Hint #1: Keep in min
Page 208 and 209: } 9.1.8 Task 1.8 Solution: two 100*
Page 210 and 211: #define PUSH(low, high) ((void) ((t
Page 212 and 213: } } ignore one or both. Otherwise,
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