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简单的使用Windows提供的Hook API SetWindowsHookExA，实现弹窗提示每次按下的按键
步骤：
1.编写一个Dll调用SetWindowsHookExA，实现弹窗
2.编写一个exe调用LoadLibrary加载生成的Dll
dll代码如下
#include<stdio.h>
#include<windows.h>
HHOOK shook;
LRESULT WINAPI keyboard(int nCode,WPARAM wParam,LPARAM lParam);
BOOL APIENTRY DllMain( HINSTANCE hModule,DWORD ul_reason_for_call,LPVOID lpReserved )
{switch(ul_reason_for_call){case DLL_PROCESS_ATTACH:{shook = SetWindowsHookExA(WH_KEYBOARD,keyboard,hModule,0);}case DLL_THREAD_ATTACH:case DLL_THREAD_DETACH:case DLL_PROCESS_DETACH:break;}return TRUE;
}
LRESULT __stdcall keyboard(int nCode,WPARAM wParam,LPARAM lParam)
{if(nCode >=0){if(!(lParam & 0x80000000)){char tcKey[MAX_PATH] = {0};GetKeyNameTextA(lParam,tcKey,MAX_PATH);MessageBoxA(NULL,tcKey,tcKey,MB_OK);return CallNextHookEx(shook,nCode,wParam,lParam);}}elsereturn CallNextHookEx(shook,nCode,wParam,lParam);
}
exe代码如下
#include<stdio.h>
#include<windows.h>
void main()
{LoadLibraryA("sethook.dll");getchar();
}