x32dbg/x64dbg逆向之反汇编do-while分析1反汇编do-while分析do-while语句在开发中同样十分常见它和while区别主要先执行循环体再判断条件是否继续循环的场景。相比whiledo-while至少会先执行一次循环体因此在一些特定逻辑处理中非常实用。编译后映射到汇编代码里具体是什么样子咱们也通过x32dbg/x64dbg一步步调试也看下哦2C语言案例代码单个do-while咱们在学习逆向分析的同时同步梳理C和C基础知识点一正一反正向开发逆向分析为后续深耕底层技术打好基础哦。#define _CRT_SECURE_NO_WARNINGS#include stdio.hint main(void){int n;int i 1;int sum 0;printf(Please input an number: );scanf(%d, n);do{sum sum i;i i 1;} while (i n);printf(Sum from 1 to %d %d\n, n, sum);return 0;}4C语言案例代码多个do-while咱们在学习逆向分析的同时同步梳理C和C基础知识点一正一反正向开发逆向分析为后续深耕底层技术打好基础哦。#include stdio.hint main(){int x 1;// Single do-whiledo{printf(%d , x);x;} while (x 5);printf(\n);printf(\n);x 0;// First do-while layerdo{int y 1;printf(| );// Second do-while layerdo{int z 1;// Third do-while layerdo{printf(* );z;} while (z 3);printf(| );y;} while (y 2);printf(\n);x;} while (x 2);return 0;}5反向分析指针单个do-while之前咱们把单个while和多个while分析外以后再来看会发现底层几乎一样但是一旦嵌套多了咱们会发现不同的地方哦所以咱们以后尽量从最底层入手这样理解任何东西就相当easy哈哈哈x64Debug版本00007FF60D851930 | 40:55 | push rbp | FileName.cpp:500007FF60D851932 | 57 | push rdi |00007FF60D851933 | 48:81EC 48010000 | sub rsp,148 |00007FF60D85193A | 48:8D6C24 20 | lea rbp,qword ptr ss:[rsp20] |00007FF60D85193F | 48:8D7C24 20 | lea rdi,qword ptr ss:[rsp20] |00007FF60D851944 | B9 1A000000 | mov ecx,1A |00007FF60D851949 | B8 CCCCCCCC | mov eax,CCCCCCCC |00007FF60D85194E | F3:AB | rep stosd |00007FF60D851950 | 48:8B05 E9B60000 | mov rax,qword ptr ds:[__security_cooki | rax:__enc$textbss$end5A00007FF60D851957 | 48:33C5 | xor rax,rbp | rax:__enc$textbss$end5A00007FF60D85195A | 48:8985 18010000 | mov qword ptr ss:[rbp118],rax | rax:__enc$textbss$end5A00007FF60D851961 | 48:8D0D A0060100 | lea rcx,qword ptr ds:[__14F49BB1_FileN | FileName.cpp:1573248000007FF60D851968 | E8 3FFAFFFF | call project1.7FF60D8513AC |00007FF60D85196D | 90 | nop | 上面全是debug特有的就不多少了00007FF60D85196E | C745 24 01000000 | mov dword ptr ss:[rbp24],1 | FileName.cpp:7开始赋值变量int i 1;00007FF60D851975 | C745 44 00000000 | mov dword ptr ss:[rbp44],0 | FileName.cpp:8int sum 0;00007FF60D85197C | 48:8D0D A5930000 | lea rcx,qword ptr ds:[Please input an | FileName.cpp:10, 00007FF60D85AD28:Please input an number: print参数00007FF60D851983 | E8 D0F8FFFF | call project1.7FF60D851258 | printf00007FF60D851988 | 90 | nop |00007FF60D851989 | 48:8D55 04 | lea rdx,qword ptr ss:[rbp4] | FileName.cpp:11, rdx:__enc$textbss$end5Aint n作为scanf存储数字的变量地址第二个参数rdx00007FF60D85198D | 48:8D0D B4930000 | lea rcx,qword ptr ds:[%d...] | 00007FF60D85AD48:%d第一个参数00007FF60D851994 | E8 D3F8FFFF | call project1.7FF60D85126C | scanf00007FF60D851999 | 90 | nop |00007FF60D85199A | 8B45 24 | mov eax,dword ptr ss:[rbp24] | FileName.cpp:15获取局部变量i00007FF60D85199D | 8B4D 44 | mov ecx,dword ptr ss:[rbp44] | 获取局部变量sum00007FF60D8519A0 | 03C8 | add ecx,eax | sum sum i;00007FF60D8519A2 | 8BC1 | mov eax,ecx |00007FF60D8519A4 | 8945 44 | mov dword ptr ss:[rbp44],eax | 把结果赋值给sum00007FF60D8519A7 | 8B45 24 | mov eax,dword ptr ss:[rbp24] | FileName.cpp:16同理获取局部变量i00007FF60D8519AA | FFC0 | inc eax | i i 1;00007FF60D8519AC | 8945 24 | mov dword ptr ss:[rbp24],eax | 把结果赋值给i00007FF60D8519AF | 8B45 04 | mov eax,dword ptr ss:[rbp4] | FileName.cpp:17获取局部变量i进行比较00007FF60D8519B2 | 3945 24 | cmp dword ptr ss:[rbp24],eax | while (i n);00007FF60D8519B5 | 7E E3 | jle project1.7FF60D85199A | 如果小于等于继续执行循环语句内容00007FF60D8519B7 | 44:8B45 44 | mov r8d,dword ptr ss:[rbp44] | FileName.cpp:19获取sum00007FF60D8519BB | 8B55 04 | mov edx,dword ptr ss:[rbp4] | 获取n00007FF60D8519BE | 48:8D0D 8B930000 | lea rcx,qword ptr ds:[Sum from 1 to % | 00007FF60D85AD50:Sum from 1 to %d %d\n00007FF60D8519C5 | E8 8EF8FFFF | call project1.7FF60D851258 | printf00007FF60D8519CA | 90 | nop |00007FF60D8519CB | 33C0 | xor eax,eax | FileName.cpp:2000007FF60D8519CD | 8BF8 | mov edi,eax | FileName.cpp:2100007FF60D8519CF | 48:8D4D E0 | lea rcx,qword ptr ss:[rbp-20] |00007FF60D8519D3 | 48:8D15 26930000 | lea rdx,qword ptr ds:[7FF60D85AD00] | rdx:__enc$textbss$end5A00007FF60D8519DA | E8 8AF6FFFF | call project1.7FF60D851069 |00007FF60D8519DF | 8BC7 | mov eax,edi |00007FF60D8519E1 | 48:8B8D 18010000 | mov rcx,qword ptr ss:[rbp118] |
