我已经从帖子here中修改了这段代码的一部分,它将截图jpg图像保存到缓冲区。我使用sento()通过UDP发送此缓冲图像。我无法转换此char数组数据发送回图片。感谢您提供的任何帮助
void gdiscreen()
{
char buffer[61400];
using namespace Gdiplus;
wchar_t filename[200];
memset(filename,0,sizeof(filename));
cnt++;
GdiplusStartupInput gdiplusStartupInput;
ULONG_PTR gdiplusToken;
GdiplusStartup(&gdiplusToken, &gdiplusStartupInput, NULL);
EncoderParameters encoderParameters;
ULONG quality;
{
HDC scrdc, memdc;
HBITMAP membit;
scrdc = ::GetDC(0);
int Height = GetSystemMetrics(SM_CYSCREEN);
int Width = GetSystemMetrics(SM_CXSCREEN);
memdc = CreateCompatibleDC(scrdc);
membit = CreateCompatibleBitmap(scrdc, Width, Height);
HBITMAP hOldBitmap =(HBITMAP) SelectObject(memdc, membit);
BitBlt(memdc, 0, 0, Width, Height, scrdc, 0, 0, SRCCOPY);
Gdiplus::Bitmap bitmap(membit, NULL);
CLSID clsid;
GetEncoderClsid(L"image/jpeg", &clsid);
encoderParameters.Count = 1;
encoderParameters.Parameter[0].Guid = EncoderQuality;
encoderParameters.Parameter[0].Type = EncoderParameterValueTypeLong;
encoderParameters.Parameter[0].NumberOfValues = 1;
quality = 20;
encoderParameters.Parameter[0].Value = &quality;
IStream *pStream = NULL;
LARGE_INTEGER liZero = {};
ULARGE_INTEGER pos = {};
STATSTG stg = {};
ULONG bytesRead=0;
HRESULT hrRet=S_OK;
// this is your buffer that will hold the jpeg bytes
DWORD dwBufferSize = 0; // this is the size of that buffer;
hrRet = CreateStreamOnHGlobal(NULL, TRUE, &pStream);
bitmap.Save(pStream, &clsid, &encoderParameters);
hrRet = pStream->Seek(liZero, STREAM_SEEK_SET, &pos);
hrRet = pStream->Stat(&stg, STATFLAG_NONAME);
dwBufferSize = stg.cbSize.LowPart;
// copy the stream into memory
hrRet = pStream->Read(buffer, stg.cbSize.LowPart, &bytesRead);
}
}
sendto()函数是:
sendto(s,buffer,sizeof(buffer) , 0 , (struct sockaddr *) &si_other, slen)
答案 0 :(得分:1)
终于解决了......就像保存文件一样简单
std::ofstream("D:\\abc.jpg", std::ios::binary).write(buf,recv_len);
这里第一个参数是用于保存文件的位置,第二个是操作模式,因为这里我们正在处理图像数据二进制被选择。写入方法是自解释的