Question

我在迭代QList时遇到段错误。我不明白我做错了什么。

我有一个QList of Conversation。在一个对话中我有一个Msg的QList。以下是课程描述：

消息类：

class Msg {
public:
    Msg();
    Msg(const Msg& other);
    Msg& operator=(const Msg& other);
    virtual ~Msg();
    bool operator==(const Msg& other);

    QString         id()                   const { return _id; }
    MsgContact      author()               const { return _author; }
    MsgContact      dest()                 const { return _dest; }
    QDateTime       date()                 const { return _receivedDate; }
    MsgDirection    direction()            const { return _direction; }
    QString         text()                 const { return _text; }
    bool            transmitted()          const { return _transmitted; }

    void setId(const QString& id)                       { _id = id; }
    void setAuthor(const MsgContact& author)            { _author        = author; }
    void setDest(const MsgContact& dest)                { _dest          = dest; }
    void setDate(const QDateTime& receivedDate)         { _receivedDate  = receivedDate; }
    void setDirection(const MsgDirection& direction)    { _direction     = direction; }
    void setText(const QString& text)                   { _text          = text; }
    void setTransmitted(const bool& transmitted)        { _transmitted   = transmitted; }

private:
    QString _id;
    MsgContact _author;
    MsgContact _dest;
    QDateTime _receivedDate;
    MsgDirection _direction;
    QString _text;
    bool _transmitted; //indique que le message a été transmis
    bool _read; //indique la lecture
};

会话课程：

class Conversation
{
public:
    Conversation();
    Conversation(const Conversation& other);
    virtual ~Conversation();
    Conversation& operator=(const Conversation& other);
    bool operator==(const Conversation& other);

    bool                  isNull()        const { return (NULL == _title || NULL == _destId); }
    const QString         title()         const       { return _title; }
    const QString         destId()        const       { return _destId; }
    QList<Msg>            messages()      const       { return _messages; }

    void setDestId(const QString& destId)    { _destId      = destId; }
    void setTitle(const QString& title)      { _title       = title; }

    void addMsg(const Msg& msg);

    static Conversation INVALID_CONVERSATION;

private:
    QList<Msg>   _messages;
    QString      _title;
    QString      _destId;
};

void Conversation::addMsg(const Msg& msg)
{
    _messages.append(msg);
}

生成段错误的代码。我创建了一条消息，我遍历对话列表以在相关对话中添加消息。然后，我想迭代消息列表，我得到一个段错误。我使用不同的方式来访问工作正常的消息。

    Msg *m = new Msg();
    m->setId(xmppMsg.id());
    m->setDest(findContactById(conversationId));
    m->setDirection(MsgOutgoing);
    m->setAuthor(_myContact);
    m->setText(message);
    m->setDate(xmppMsg.stamp());

    QList<Conversation>::iterator it;
    for(it = _conversations.begin(); _conversations.end() != it; it++)
    {
        if((*it).destId() == conversationId)
        {
            (*it).addMsg(*m);
            Q_EMIT(conversationChanged((*it)));
            break;
        }
    }


    qDebug() << "NB : " <<(*it).messages().size(); // ok, the number of message is incremented.

    //test several way of accessing a message, these works fine.
    qDebug() << "doSend " << it->messages().at(0).id();  
    qDebug() << "doSend " << it->messages().begin()->id();
    qDebug() << "doSend " << (*(it->messages().begin())).id();


    //try to iterate
    QList<Msg>::iterator msgIt = it->messages().begin();
    if(msgIt != it->messages().end())
    {
        qDebug() << "TEST - "<<  msgIt->id();  //segfault. 
    }

感谢您的帮助

Answer 1

（首先编辑“回答”，这是对答案的实际尝试）

我的猜测：

require(data.table)
subject <- data.table(interval = paste("int", 1:4, sep=""), 
                      start = c(2,10,12,25), 
                      end = c(7,14,18,28))
query <- data.table(start = c(0,5,10,15,20,25), 
                    end = c(4,9,14,19,24,29))

setkey(subject, start, end)
ans = foverlaps(query, subject, type="any")
ans[, .(count = sum(!is.na(start)), 
        which = paste(interval, collapse=", ")), 
     by = .(i.start, i.end)]

#    i.start i.end count      which
# 1:       0     4     1       int1
# 2:       5     9     1       int1
# 3:      10    14     2 int2, int3
# 4:      15    19     1       int3
# 5:      20    24     0         NA
# 6:      25    29     1       int4

它返回了QList _messages的副本，而不是对它的引用。我不确定它会给你看到的结果，但它看起来不对我。也许尝试这样的事情？

QList<Msg>            messages()      const       { return _messages; }

segfault通过迭代器访问qlist元素

1 个答案: