c++ - 使用Qt编译测试程序时出错

Question

我是 C++/Qt 的新手

我正在阅读 Jasmin Blanchette 和 Mark Summerfield 合着的“C++ GUI Programming with Qt 4”一书。

我在编写示例程序时遇到了一些我无法解决的编译错误。下面的代码和错误。感谢您的帮助。

谢谢。

查找对话框.h

#ifndef FINDDIALOG_H
#define FINDDIALOG_H

#include <QDialog>

class QCheckBox;
class QLabel;
class QLineEdit;
class QPushButton;
class QWidget;

class FindDialog : QDialog
{
    Q_OBJECT
public:
    FindDialog(QWidget *parent = 0);
signals:
    void findNext(const QString &str, Qt::CaseSensitivity cs);
    void findPrevious(const QString &str, Qt::CaseSensitivity cs);
private slots:
    void findClicked();
    void enableFindButton(const QString &text);
private:
    QLabel *label;
    QLineEdit *lineEdit;
    QCheckBox *caseCheckBox;
    QCheckBox *backwardCheckBox;
    QPushButton *findButton;
    QPushButton *closeButton;
};

#endif // FINDDIALOG_H

查找对话框.cpp

#include <QtGui>

#include "finddialog.h"

FindDialog::FindDialog(QWidget *parent) : QDialog(parent)
{
    label = new QLabel(tr("Find &what:"));
    lineEdit = new QLineEdit;
    label->setBuddy(lineEdit);

    caseCheckBox = new QCheckBox(tr("Match &case"));
    backwardCheckBox = new QCheckBox(tr("Search &backward"));

    findButton = new QPushButton(tr("&Find"));
    findButton->setDefault(true);
    findButton->setEnabled(false);

    closeButton = new QPushButton(tr("Close"));

    connect(lineEdit, SIGNAL(textChanged(QString)), this, SLOT(enableFindButton(QString)));
    connect(findButton, SIGNAL(clicked()), this, SLOT(findClicked()));
    connect(closeButton, SIGNAL(clicked()), this, SLOT(close()));

    QHBoxLayout *topLeftLayout = new QHBoxLayout;
    topLeftLayout->addWidget(label);
    topLeftLayout->addWidget(lineEdit);

    QVBoxLayout *leftLayout = new QVBoxLayout;
    leftLayout->addLayout(topLeftLayout);
    leftLayout->addWidget(caseCheckBox);
    leftLayout->addWidget(backwardCheckBox);

    QVBoxLayout *rightLayout = new QVBoxLayout;
    rightLayout->addWidget(findButton);
    rightLayout->addWidget(closeButton);
    rightLayout->addStretch();

    QHBoxLayout *mainLayout = new QHBoxLayout;
    mainLayout->addLayout(leftLayout);
    mainLayout->addLayout(rightLayout);
    setLayout(mainLayout);

    setWindowTitle(tr("Find"));
    setFixedHeight(sizeHint().height());
}

void FindDialog::findClicked()
{
    QString text = lineEdit->text();
    Qt::CaseSensitivity cs =
            caseCheckBox->isChecked() ? Qt::CaseSensitive
                : Qt::CaseInsensitive;
    if (backwardCheckBox->isChecked()) {
        emit findPrevious(text, cs);
    } else {
        emit findNext(text, cs);
    }
}

void FindDialog::enableFindButton(const QString &text)
{
    findButton->setEnabled(!text.isEmpty());
}

主要.cpp

#include <QApplication>
#include "finddialog.h"
int main(int argc, char *argv[])
{
    QApplication app(argc, argv);
    FindDialog *dialog = new FindDialog;
    dialog->show();
    return app.exec();
}

下面的错误..

c:/Qt/2010.02.1/qt/include/QtGui/../../src/gui/kernel/qwidget.h: In function 'int qMain(int, char**)': c:/Qt/2010.02.1/qt/include/QtGui/../../src/gui/kernel/qwidget.h:485: error: 'void QWidget::show()' is inaccessible main.cpp:7: error: within this context main.cpp:7: error: 'QWidget' is not an accessible base of 'FindDialog'

Answer 1

您应该从QWidget 或QDialog 公开继承:

 class FindDialog : public QDialog {
      // ...

show() 实际上是由 FindDialog、QWidget 的基础实现的 - 但您没有从它公开继承，因此不能'访问它。

类的继承默认是私有(private)的，即

class A : B {}; 

和

class A : private B {};

是等价的。