qt5.12.8实例

您好！以下是一个简单的Qt 5.12.8实例，演示了一个基本的窗口应用程序：

#include <QApplication>
#include <QWidget>
#include <QPushButton>

int main(int argc, char *argv[])
{
    QApplication app(argc, argv);

    QWidget window;
    window.setWindowTitle("Qt 5.12.8 实例");
    window.resize(250, 150);

    QPushButton button("点击我", &window);
    button.setGeometry(10, 10, 100, 30);

    QObject::connect(&button, &QPushButton::clicked, [&]() {
        button.setText("已点击");
    });

    window.show();

    return app.exec();
}

这个例子创建了一个简单的窗口应用程序，窗口标题为 "Qt 5.12.8 实例"，大小为250x150。在窗口中创建了一个按钮，按钮上显示 "点击我"，并且按钮被点击后会改变文本为 "已点击"。

请注意，编写Qt应用程序通常需要在.pro文件中正确配置项目依赖关系，并使用qmake或CMake生成构建系统文件，然后进行构建和运行。

如果您有任何问题，请随时提问！

相关问题

这是什么错误，D:\WorkSoftware\Qt\Qt5.12.12\5.12.12\msvc2015_64\include\QtCore\qobject.h:254: error: C2664: “QMetaObject::Connection QObject::connectImpl(const QObject *,void **,const QObject *,void **,QtPrivate::QSlotObjectBase *,Qt::ConnectionType,const int *,const QMetaObject *)”: 无法将参数 3 从“const QFuture<void> ”转换为“const QObject ” D:\WorkSoftware\Qt\Qt5.12.12\5.12.12\msvc2015_64\include\QtCore/qobject.h(254): note: 与指向的类型无关；转换要求 reinterpret_cast、C 样式转换或函数样式转换 mainwindow.cpp(14): note: 参见对正在编译的函数 模板 实例化“QMetaObject::Connection QObject::connect<void(__cdecl MainWindow:: )(void),void(__cdecl QFuture<void>:: )(void)>(const MainWindow ,Func1,const QFuture<void> ,Func2,Qt::ConnectionType)”的引用 with [ Func1=void (__cdecl MainWindow:: )(void), Func2=void (__cdecl QFuture<void>:: )(void) ]

这个错误是编译错误，可能是因为在使用`QObject::connect`连接两个对象时，其中一个参数的类型不正确。

根据错误提示，它在尝试将类型为`const QFuture<void>`的第三个参数转换为类型为`const QObject*`的参数。这意味着该参数应该是一个`QObject`对象，但是却传递了一个`QFuture<void>`对象，因此编译器无法将其转换为正确的类型。

要解决这个错误，你需要检查你的代码，并确保你传递正确的参数类型。在这种情况下，你需要将类型为`QFuture<void>`的参数替换为类型为`QObject*`的对象，或者使用`QFutureWatcher`类来监视`QFuture`对象并在其完成时发出信号。

D:\WorkSoftware\Qt\Qt5.12.12\5.12.12\msvc2015_64\include\QtCore\qobject.h:254: error: C2664: “QMetaObject::Connection QObject::connectImpl(const QObject *,void **,const QObject *,void **,QtPrivate::QSlotObjectBase *,Qt::ConnectionType,const int *,const QMetaObject *)”: 无法将参数 3 从“const QFuture<void> *”转换为“const QObject *” D:\WorkSoftware\Qt\Qt5.12.12\5.12.12\msvc2015_64\include\QtCore/qobject.h(254): note: 与指向的类型无关；转换要求 reinterpret_cast、C 样式转换或函数样式转换 mainwindow.cpp(14): note: 参见对正在编译的函数 模板 实例化“QMetaObject::Connection QObject::connect<void(__cdecl MainWindow::* )(void),void(__cdecl QFuture<void>::* )(void)>(const MainWindow *,Func1,const QFuture<void> *,Func2,Qt::ConnectionType)”的引用 with [ Func1=void (__cdecl MainWindow::* )(void), Func2=void (__cdecl QFuture<void>::* )(void) ]

It looks like you're encountering a compilation error in your C++ code. The error message indicates that there is a problem with a call to the `QObject::connect()` function.

The error message suggests that you're trying to connect a signal from a `QFuture<void>` object to a slot in a `MainWindow` object. However, the third argument to `connect()` should be a pointer to a `QObject`, not a `QFuture<void>` object.

To resolve this error, you'll need to change the third argument to `connect()` to be a pointer to a `QObject` object that emits the signal you're trying to connect. If the signal is emitted by the `QFuture<void>` object, you may need to create a separate `QObject` subclass to emit the signal on behalf of the `QFuture<void>` object.

Alternatively, you may be able to use a lambda expression to connect the signal and slot without needing to create a separate `QObject` subclass. Here's an example of how to use a lambda expression to connect a signal from a `QFuture<void>` object to a slot in a `MainWindow` object:

QFuture<void> future;
MainWindow mainWindow;
QObject::connect(&future, &QFuture<void>::finished, &mainWindow, [](){
    // Slot code here
});

This connects the `finished` signal of the `QFuture<void>` object to a lambda function that contains the code for the slot. The lambda function is then connected to the `MainWindow` object using the `connect()` function.