android - QAbstractButton 和 QPushButton* 之间的比较缺少强制转换

标签 android c++ qt gcc

我的代码出现在下面。当我尝试编译它时,我得到:

error: 29:38 comparison between distinct pointer types QAbstractButton and QPushButton* lacks a cast -> 'if (stdmetBox.clickedButton() == stdButton)'

系统:Amazon Kindle Fire 7"运行 Cyanogenmod 11.0 (Android 4.4.2 Kitkat) 编译器:G++/GCC(C4droid 插件版本 4.9.1 的 GCC) IDE:带有 SDL、GCC 和 Ministro 插件的 C4droid

如有任何帮助,我们将不胜感激。我已经搜索过,但我能找到的唯一实例是特定于 iOS 的。

#include <fstream>
#include <QApplication>
#include <QLabel>
#include <QMessageBox>
#include <QString>
#include <QAbstractButton>
#include <QInputDialog>
#include <QDebug>

using namespace std;

int setup() 
{
    string unitchar;
    string unitcharo;
    bool setupSuccess;
    int returncode;
    QMessageBox msgBox;
    QMessageBox stdmetBox;
    QMessageBox ynBox;

    msgBox.setText("Welcome to Tyler's Fitness App! This app will help you with your fitness goals, whatever they may be. Let's get you set up!");
    msgBox.exec();

    QPushButton *stdButton = stdmetBox.addButton(QT_TR_NOOP("Standard"), QMessageBox::ActionRole);
    QPushButton *metButton = stdmetBox.addButton(QT_TR_NOOP("Metric"), QMessageBox::ActionRole);                                                
    stdmetBox.exec();

    if (stdmetBox.clickedButton() == stdButton)
    {
        unitchar = "standard";
        unitcharo = "metric";
    }
    else if (stdmetBox.clickedButton() == metButton)
    {
        unitchar = "metric";
        unitcharo = "standard";
    }
    else
    {
        setupSuccess = false;
        returncode = 0;
        return returncode; 
    }

    if (unitchar == "standard")
    {
        ynBox.setInformativeText("Standard units include feet, miles, mph, pounds, etc.");
    }
    else 
    {
        ynBox.setInformativeText("Metric units include meters, kilometers, kph, kilograms, etc.");
    } 

    QPushButton *yesButton = ynBox.addButton(QT_TR_NOOP("Yes"), QMessageBox::ActionRole);
    QPushButton *noButton = ynBox.addButton(QT_TR_NOOP("No"), QMessageBox::ActionRole);
    ynBox.setText("Is that okay?")
    while (ynBox.clickedButton() == noButton)
    {
        if (unitchar == "standard")
        {
            unitchar = "metric";
            unitcharo = "standard";
        }
        else 
        {
            unitchar = "standard";
            unitcharo = "metric";
        }
        if (unitchar == "standard")
        {
            ynBox.setInformativeText("Standard units include feet, miles, mph, pounds, etc.");
        }
        else 
        {
            ynBox.setInformativeText("Metric units include meters, kilometers, kph, kilograms, etc.");
        }
        result = ynBox.exec();
    }

    if (unitchar == "standard")
    {
        msgBox.setText("Great! You've selected standard units.");
        msgBox.exec();
        setupSuccess = true;

    }
    else 
    {
        msgBox.setText("Great! You've selected metric units.");
        msgBox.exec();
        setupSuccess = true;
    }

    if (setupSuccess != true)
    {
        returncode = 0;
    }
    else if (unitchar == "standard")
    {
        returncode = 1;
    }
    else
    {
        returncode = 2;
    }
    return returncode;                      
}                       

int main(int argc, char* argv[])
{ 
    QApplication app(argc, argv);  
    unitmaster = setup();
    if ((unitmaster == 1) || (unitmaster == 2))
    {
        QMessageBox successBox;
        successBox.setText("The program has completed successfully!");
        successBox.exec();
    }
    else 
    {
        QMessageBox errorBox;
        errorBox.setText("The program has completed successfully!");
        errorBox.exec();
    }            
    return app.exec();
}

最佳答案

您似乎忘记包含 <QPushButton>即使你使用 QPushButton以一种需要它的定义的方式。其中一个 header 必须声明它,因为它被识别为一种类型,但如果它不完整,那么编译器就不会知道 QPushButton*可转换为 QAbstractButton* .

作为一般规则,请始终包含定义您使用的类型的 header ,除非您确定声明就足够了。基指针的隐式转换是需要定义的东西。

关于android - QAbstractButton 和 QPushButton* 之间的比较缺少强制转换,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/29649546/

相关文章:

c++ - 枚举的元编程问题

c++ - { Qt5.0.2/QML/QtQuick2.0/C++ } 运行没有错误的示例项目?

java - Android - forName() 方法 ClassNotFoundException

android - 使用 AlertDialog.Builder 时防止 StatusBar 显示

java - 命名空间 "classloader-namespace"无法访问库

c++ - 用 const 重载运算符 <,但不要作为 const 插入 map

使用 Dagger 2的Android生命周期库ViewModel

python - dll在使用memset时导致Python崩溃

javascript - Qt 2D Canvas 和 HTML5 canvas 的行为并不相同

c++ - 在按钮点击时控制 qt 中的循环