[ Foro de C ]

Dividir en basic block

Cuantos bloques hay en este código de ensamblador??

0x0000054d <+0>: lea ecx,[esp+0x4]
0x00000551 <+4>: and esp,0xfffffff0
0x00000554 <+7>: push DWORD PTR [ecx-0x4]
0x00000557 <+10>: push ebp
0x00000558 <+11>: mov ebp,esp
0x0000055a <+13>: push ebx
0x0000055b <+14>: push ecx
0x0000055c <+15>: sub esp,0x10
0x0000055f <+18>: call 0x450 <__x86.get_pc_thunk.bx>
0x00000564 <+23>: add ebx,0x1a9c
0x0000056a <+29>: mov DWORD PTR [ebp-0x10],0x0
0x00000571 <+36>: lea eax,[ebx-0x19a0] ; “3jd9cjfk98hnd”
0x00000577 <+42>: mov DWORD PTR [ebp-0x14],eax
0x0000057a <+45>: sub esp,0xc
0x0000057d <+48>: push DWORD PTR [ebp-0x14]
0x00000580 <+51>: call 0x3e0 <strlen@plt>
0x00000585 <+56>: add esp,0x10
0x00000588 <+59>: mov DWORD PTR [ebp-0x18],eax
0x0000058b <+62>: mov DWORD PTR [ebp-0xc],0x0
0x00000592 <+69>: jmp 0x5ad <main+96>
0x00000594 <+71>: mov edx,DWORD PTR [ebp-0xc]
0x00000597 <+74>: mov eax,DWORD PTR [ebp-0x14]
0x0000059a <+77>: add eax,edx
0x0000059c <+79>: movzx eax,BYTE PTR [eax]
0x0000059f <+82>: movsx eax,al
0x000005a2 <+85>: imul eax,DWORD PTR [ebp-0x18]
0x000005a6 <+89>: add DWORD PTR [ebp-0x10],eax
0x000005a9 <+92>: add DWORD PTR [ebp-0xc],0x1
0x000005ad <+96>: mov eax,DWORD PTR [ebp-0xc]
0x000005b0 <+99>: cmp eax,DWORD PTR [ebp-0x18]
0x000005b3 <+102>: jl 0x594 <main+71>
0x000005b5 <+104>: sub esp,0x8
0x000005b8 <+107>: push DWORD PTR [ebp-0x10]
0x000005bb <+110>: lea eax,[ebx-0x1992] ; “[+] Codigo generado: %i\n”
0x000005c1 <+116>: push eax
0x000005c2 <+117>: call 0x3d0 <printf@plt>
0x000005c7 <+122>: add esp,0x10
0x000005ca <+125>: mov eax,0x0
0x000005cf <+130>: lea esp,[ebp-0x8]
0x000005d2 <+133>: pop ecx
0x000005d3 <+134>: pop ebx
0x000005d4 <+135>: pop ebp
0x000005d5 <+136>: lea esp,[ecx-0x4]
0x000005d8 <+139>: ret