Question

我正在尝试使用QDataStream对象加载和存储自定义QMetaType。这是一个例子：

# necessary includes...
from mpl_toolkits.mplot3d import axes3d
import matplotlib.pyplot as plt
import numpy as np

raw_data = np.loadtxt('data.txt', delimiter=';', dtype=np.string_)

angle    = raw_data[0 , 1:].astype(float)
distance = raw_data[1:, 0 ].astype(float)
data     = raw_data[1:, 1:].astype(float)


# make the example data a bit more interesting...
data = (50 + angle[np.newaxis,:]) / (10 + np.sqrt(distance[:,np.newaxis])) 

# setting up the plot
fig = plt.figure()
ax = fig.add_subplot(111, projection='3d')

# the trickey part creating the data that plot_wireframe wants
Z = data
X, Y = np.meshgrid(angle, distance)

ax.plot_wireframe(X, Y, Z)

# fancing it up a bit
ax.set_xticks(angle)
ax.set_yticks(distance[::2])
ax.set_xlabel('angle')
ax.set_ylabel('distance')

# and showing the plot ...
plt.show()

幻数回来了，但有一条消息int main(int argc, char *argv[]) { QApplication a(argc, argv); QString test1("/test1/"); const QString path1(QDir::homePath()+test1); qDebug() << "path1 variable is " << path1; QDir apphome(path1); qDebug() << "apphome path is " << apphome.absolutePath(); if (!apphome.mkdir(path1)) { qDebug() << "mkdir returned false. directory already exists?"; } if(!apphome.setCurrent(path1)) { qDebug() << "did not set current directory"; } qDebug() << "apphome path is " << apphome.absolutePath(); Basic basic1; Basic basic2; basic1.value = 14; QFile file1("file1name"); if (!file1.open(QIODevice::WriteOnly)) { qDebug() << "file1 not open."; } QDataStream dataStream1(&file1); QVariant qvar1; qvar1.setValue(basic1); dataStream1 << (quint32)12345; dataStream1 << qvar1; file1.close(); file1.open(QIODevice::ReadOnly); QDataStream dataStream2(&file1); quint32 magic; QVariant qvar2; dataStream1 >> magic; qDebug() << "magic number is " << magic; dataStream2 >> qvar2; file1.close(); basic2 = qvar2.value<Basic>(); qDebug() << "14 = " << basic1.value << " = " << basic2.value << "."; //MainWindow w; //w.show(); return a.exec(); }，然后是QVariant::save: unable to save type 'Basic' (type id: 1026).然后是QVariant::load: unable to load type 1026.。 Basic类只来自QMetaType文档：

14 = 14 = 0

我的想法已经用完了，有谁知道造成这个问题的原因是什么？ Qt的版本是5.10.1。

Answer 1

因为编译器无法读懂您的想法，您需要描述序列化的可能性，例如：

struct Basic
{
    Basic();
    Basic(const Basic &basic);
    ~Basic();
    int value;

    friend QDataStream & operator << (QDataStream &arch, const Basic& object)
    {
        arch << object.value;
        return arch;
    }

    friend QDataStream & operator >> (QDataStream &arch, Basic& object)
    {
        arch >> object.value;
        return arch;
    }
};
Q_DECLARE_METATYPE(Basic);

在main（）

中

qRegisterMetaType<Basic>("Basic");
qRegisterMetaTypeStreamOperators<Basic>("Basic");

保存\加载操作之前

会发生。 Q_DECLARE_METATYPE需要用QVariant存储类型，这两者需要注册对象的“三巨头”来管理它作为资源及其序列化方法。输出：

path1 variable is  "C:/Users/Yaroslav/test1/"
apphome path is  "C:/Users/Yaroslav/test1"
mkdir returned false. directory already exists?
apphome path is  "C:/Users/Yaroslav/test1"
magic number is  12345
14 =  14  =  14 .

P.S。请注意，如果您在没有窗口的情况下离开return a.exec();，则会让程序永远停留在内存中，直到您停止它为止。

在QDataStream中使用自定义QMetaType

1 个答案: