錯誤在編譯時使用Qt

Question

一個測試程序，我是很新的C++/QT

我下面的書茉莉布蘭切特和Mark夏季科幻場「C++ GUI編程的Qt 4」。

我正在研究一個示例程序，並遇到了一些我無法解決的編譯錯誤。下面的代碼和錯誤。任何幫助表示讚賞。

謝謝。

finddialog.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 

finddialog.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()); 
} 

的main.cpp下面

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

誤差..

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

您應該QWidget或QDialog公開繼承：

class FindDialog : public QDialog { 
     // ... 

show()實際上是由FindDialog，QWidget基本實現 - 但你是不是繼承公開的，並因此不能訪問它。

繼承的類是通過默認的專用的，即

class A : B {}; 

和

class A : private B {}; 

是等價的。