Visual Studio 2022 C＆＃x2B;＆＃x2B;中的错误QT项目

发布于 2025-02-03 06:21:29 字数 804 浏览 6 评论 0原文

我正在使用QT和OPENCV的Visual Studio 2022上的C ++项目。我正在使用videocapture使用Open（）函数打开RTSP流。

但是，当我尝试打开RTSP流时，我会得到以下例外：

Exception thrown at 0x00007FFB37BB4F69 in faulty.exe: Microsoft C++ exception: MONZA::DdiThreadingContext<MONZA::AdapterTraits_Gen12LP>::msg_end at memory location 0x00000029218FF9F0.
The thread 0x45d8 has exited with code 0 (0x0).

相关代码如下：

string un = port_ipDialog->devInfo.szUserName;
string stri = "rtsp://" + un + ":" + PWD + "@" + IP + "/gen/fakescreen?channel=5&subtype=2";
memcpy(rtsp, stri.c_str(), 256);
if (!capture.open(rtsp)) {
    QMessageBox::warning(nullptr, "notice", "fail!!", QMessageBox::Yes | QMessageBox::Yes);
    return;
}

奇怪的是，同一件事在另一台计算机上正常工作。

有人可以深入了解这个例外吗？为什么会发生这种情况？我会很感激。

原文

I am working on a C++ project on Visual Studio 2022 with Qt and OpenCV.
I am using VideoCapture to open an rtsp stream using the open() function.

However, I get an exception as follows when I try to open the rtsp stream:

Exception thrown at 0x00007FFB37BB4F69 in faulty.exe: Microsoft C++ exception: MONZA::DdiThreadingContext<MONZA::AdapterTraits_Gen12LP>::msg_end at memory location 0x00000029218FF9F0.
The thread 0x45d8 has exited with code 0 (0x0).

The relevant code is as follows:

string un = port_ipDialog->devInfo.szUserName;
string stri = "rtsp://" + un + ":" + PWD + "@" + IP + "/gen/fakescreen?channel=5&subtype=2";
memcpy(rtsp, stri.c_str(), 256);
if (!capture.open(rtsp)) {
    QMessageBox::warning(nullptr, "notice", "fail!!", QMessageBox::Yes | QMessageBox::Yes);
    return;
}

The strange thing is that the same thing is working perfectly on another computer.

Can someone give some insight into this exception and why it might be occurring?
I would be grateful.

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