qt4 connect const int 出错

Qt4 中的 `connect` 函数用于创建信号槽连接，通常用于处理事件驱动的应用程序。当你尝试用 `connect(const int signalId, ...)` 的形式连接信号到函数时，可能会遇到错误，因为 `const int` 并不是一个合法的信号或槽。

在Qt中，信号通常是对象的成员函数，格式如 `QObject::signalName()` 或 `self->signalName()`, 而不是硬编码的整数。槽则是处理这些信号的成员函数，它们应该有明确的参数列表，以便与信号匹配。如果你试图将一个整数值作为信号来连接，这在语法上是无效的，并且会引发编译错误。

正确的做法是确保你有一个实际存在的信号，并将其名称传递给 `connect`，比如：

QObject *object = ...;
QObject::connect(object, &MyClass::mySignal, this, &MyClass::slotFunction);

其中 `mySignal` 是 MyClass 类的一个公开信号，而 `slotFunction` 是处理这个信号的槽函数。

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) ]

这个错误是由于尝试在QObject::connect()函数中将 QFuture<void> 类型的对象传递给 QObject 类型的参数引起的。QObject::connect()函数期望其第三个参数是 QObject 类型的指针，但是在这里传递的是 QFuture<void> 类型的对象，导致类型不匹配的编译错误。

要解决这个错误，您需要在QObject::connect()函数中传递一个QObject类型的指针作为第三个参数。如果您需要将 QFuture<void> 类型的对象连接到QObject类的信号或槽上，您可以使用 QFutureWatcher 类来监视 QFuture 对象并在其完成时发出信号。在这种情况下，您可以使用 QFutureWatcher::finished() 信号并将其连接到 QObject 类的槽。

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.