Question

我有一个大问题。我创建了一个虚拟项目来隔离我的错误。因为是一个包含6个文件的项目，并且在这里添加所有代码是不可行的，所以我创建了一个github project并在此处添加了所有代码，同时描述了这个想法。你可以看看那里的代码。

main.cpp 的源代码：

#include <QApplication>
#include <QObject>

#include "mainwindow.h"
#include "testclass.h"

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

    MainWindow w;
    TestClass c;

    QObject::connect(&w, &MainWindow::mySignal,
                     &c, &TestClass::mySlot);

    w.show();

    return a.exec();
}

类MainWindow有一个Q_SIGNAL，当我按下按钮上的按钮时会发出mySlot Q_SLOT只是TestClass中的QObject（继承qDebug）QtCreator消息。

如果我从$ qmake QMAKE_TEST.pro $ make g++ -c -m64 -pipe -O2 -Wall -W -D_REENTRANT -DQT_NO_DEBUG -DQT_GUI_LIB -DQT_CORE_LIB -DQT_SHARED -I/usr/share/qt4/mkspecs/linux-g++-64 -I. -I/usr/include/qt4/QtCore -I/usr/include/qt4/QtGui -I/usr/include/qt4 -I. -I. -o main.o main.cpp In file included from main.cpp:4:0: mainwindow.h: In function ‘int main(int, char**)’: mainwindow.h:19:7: error: ‘void MainWindow::mySignal()’ is protected void mySignal(); ^ main.cpp:14:36: error: within this context QObject::connect(&w, &MainWindow::mySignal, ^ main.cpp:15:41: error: no matching function for call to ‘QObject::connect(MainWindow*, void (MainWindow::*)(), TestClass*, void (TestClass::*)())’ &c, &TestClass::mySlot); ^ main.cpp:15:41: note: candidates are: In file included from /usr/include/qt4/QtCore/qcoreapplication.h:45:0, from /usr/include/qt4/QtGui/qapplication.h:45, from /usr/include/qt4/QtGui/QApplication:1, from main.cpp:1: /usr/include/qt4/QtCore/qobject.h:204:17: note: static bool QObject::connect(const QObject*, const char*, const QObject*, const char*, Qt::ConnectionType) static bool connect(const QObject *sender, const char *signal, ^ /usr/include/qt4/QtCore/qobject.h:204:17: note: no known conversion for argument 2 from ‘void (MainWindow::*)()’ to ‘const char*’ /usr/include/qt4/QtCore/qobject.h:217:17: note: static bool QObject::connect(const QObject*, const QMetaMethod&, const QObject*, const QMetaMethod&, Qt::ConnectionType) static bool connect(const QObject *sender, const QMetaMethod &signal, ^ /usr/include/qt4/QtCore/qobject.h:217:17: note: no known conversion for argument 2 from ‘void (MainWindow::*)()’ to ‘const QMetaMethod&’ /usr/include/qt4/QtCore/qobject.h:337:13: note: bool QObject::connect(const QObject*, const char*, const char*, Qt::ConnectionType) const inline bool QObject::connect(const QObject *asender, const char *asignal, ^ /usr/include/qt4/QtCore/qobject.h:337:13: note: no known conversion for argument 2 from ‘void (MainWindow::*)()’ to ‘const char*’ make: *** [main.o] Error 1 $ $构建它，我的项目构建正常但是当我从命令行构建它时，我得到一个奇怪的错误。

我从命令行做什么：

mySignal

首先为什么说protected si Q_SIGNALS，因为AFAIK所有Qt Creator都是公开的？其次，如果我从function afterAjaxResponse(responseText) { cache[ajaxIdentifier]=responseText; }运行它并且从命令行运行它时会出现错误，那么为什么这样可以正常工作？

有人可以帮帮我吗？

Answer 1

发现问题...我正在使用Qt5.5版本4构建qmake项目......

仅在从命令行构建Qt项目时出错

1 个答案: