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我打了一些弯路,功能挂钩,我有下面的代码一个奇怪的问题发送():功能仅挂接钩子的recv()和不使用迂回
基本上是,无论发生了什么DetourTransactionCommit()是成功的,但实际上只有recv()函数被挂钩,而发送不是,因为OutputDebugStringA(“Sent packet!”);
从未触发
#include "stdafx.h"
#include "stdio.h"
#include "WinInet.h"
#include "tchar.h"
#include "windows.h"
#include "detours.h"
#include <Winsock2.h>
#include <WS2tcpip.h>
#include <crtdbg.h>
#pragma comment(lib, "detours.lib")
#pragma comment(lib, "WinInet.lib")
#pragma comment(lib, "ws2_32.lib")
int (WINAPI *pSend)(SOCKET s, const char* buf, int len, int flags) = send;
int WINAPI MySend(SOCKET s, const char* buf, int len, int flags);
int (WINAPI *pRecv)(SOCKET s, char* buf, int len, int flags) = recv;
int WINAPI MyRecv(SOCKET s, char* buf, int len, int flags);
BOOL APIENTRY DllMain(HMODULE hModule,
DWORD ul_reason_for_call,
LPVOID lpReserved
)
{
LONG errore;
switch (ul_reason_for_call)
{
case DLL_PROCESS_ATTACH:
DetourTransactionBegin();
DetourUpdateThread(GetCurrentThread());
DetourAttach(&(PVOID&)pSend, MySend);
if (DetourTransactionCommit() == NO_ERROR) {
OutputDebugStringA("Send function hooked successfully");
}
else{
OutputDebugStringA("Failed to hook Send function");
}
DetourTransactionBegin();
DetourUpdateThread(GetCurrentThread());
DetourAttach(&(PVOID&)pRecv, MyRecv);
if (DetourTransactionCommit() == NO_ERROR) {
OutputDebugStringA("Recv function hooked successfully");
}
else{
OutputDebugStringA("Failed to hook Recv function");
}
case DLL_THREAD_ATTACH:
case DLL_THREAD_DETACH:
case DLL_PROCESS_DETACH:
break;
}
return TRUE;
}
int WINAPI MySend(SOCKET s, const char* buf, int len, int flags) {
OutputDebugStringA("Sent packet!");
return pSend(s, buf, len, flags);
}
int WINAPI MyRecv(SOCKET s, char* buf, int len, int flags) {
OutputDebugStringA("Received packet!");
return pRecv(s, buf, len, flags);
}
UPDATE: Appearently与功能问题是关系到过程中,我试图注入DLL到。 它看起来像试图挂钩发送()在Internet Explorer 11 x86失败的原因,我仍然要弄清楚。 我尝试使用winsock2(putty)将完全相同的DLL注入到另一个程序中,并且该函数正确连接。
也许有人知道发生这种情况的原因?
也许钩从未运行,因为'发送()'不会被调用?还有'WSASend','WSASendMsg' ......多种写入套接字的方式。 –
不管'WSASend *'函数是否在引擎盖下调用'send'? –
很高兴看到你的绕行功能。 –