我正在尝试通过套接字连接发送图像,但我遇到以下代码的问题:
//stream to char array
STATSTG myStreamStats;
ULONG bytesSaved;
myStream->Stat(&myStreamStats, 0);
char* streamData = new char[myStreamStats.cbSize.QuadPart];
if(myStream->Read(streamData, myStreamStats.cbSize.QuadPart, &bytesSaved) == S_OK)
cout<<"OK!"<<endl;
else
cout<<"Not OK!"<<endl;
//char array to stream
if(myStreamR->Write(streamData, myStreamStats.cbSize.QuadPart, &bytesSaved) == S_OK)
cout<<"OK!"<<endl;
else
cout<<"Not OK!"<<endl;
//saving the image to a file
myImage = Image::FromStream(myStreamR);
myImage->Save(lpszFilename, &imageCLSID, NULL);
程序编译并运行,但我没有得到我的图像。如果我使用原始的“myStream”而不是“myStreamR”,我会得到它,它是从原始流中读取的char数组构造的。
输出是两个“OK!”,这意味着所有字节都被复制到数组中,并且所有字节都被粘贴到新流中。但是,我检查了savedBytes,我发现在read()之后它的值是0(不好),而在write()之后,它等于我给出的流大小。那么为什么在地球上读取()如果没有读取任何内容会给我一个“S_OK”标志?
答案 0 :(得分:6)
在向数据写入数据后,您并不是在寻找MyStreamR
。 Image::FromStream()
开始读取流的当前位置,所以如果你不回头,那么就没有数据可供阅读。
试试这个:
STATSTG myStreamStats = {0};
if (FAILED(myStream->Stat(&myStreamStats, 0)))
cout << "Stat failed!" << endl;
else
{
char* streamData = new char[myStreamStats.cbSize.QuadPart];
ULONG bytesSaved = 0;
if (FAILED(myStream->Read(streamData, myStreamStats.cbSize.QuadPart, &bytesSaved)))
cout << "Read failed!" << endl;
else
{
//char array to stream
if (FAILED(myStreamR->Write(streamData, bytesSaved, &bytesSaved)))
cout << "Write failed!" << endl;
else
{
LARGE_INTEGER li;
li.QuadPart = 0;
if (FAILED(myStreamR->Seek(li, STREAM_SEEK_SET, NULL)))
cout << "Seek failed!" << endl;
else
{
//saving the image to a file
myImage = Image::FromStream(myStreamR);
if (myImage1->GetLastStatus() != Ok)
cout << "FromStream failed!" << endl;
else
{
if (myImage->Save(lpszFilename, &imageCLSID, NULL) != Ok)
cout << "Save failed!" << endl;
else
cout << "OK!" << endl;
}
}
}
}
}