c ++函数指针，用于将QVariant转换为{自定义类型 [英] c++ function pointer used to convert a QVariant to {a custom type

问题描述

QVariant包含要转换为自定义类型MyClass或MyClass2的QMap对象。

A QVariant is holding a QMap object that is to be converted into a custom type, MyClass or MyClass2.

示例：

class MyClass{
   int item1;
   int item2;
   QString string1;
   AnotherClass subclass;
};

class MyClass2{
   int item1;
   QString string1;
   AnotherClass subclass;
};

函数已被写入以将QVariant转换为关联类

functions have been written to convert the QVariant to the associated classes

MyClass QVariantToMyClass1(QVariant);
MyClass2 QVariantToMyClass1(QVariant);

我的问题是，在模板函数中，传递函数指针的正确方法是什么？下面显示的代码返回一个错误'const class QVariant没有成员名为convFunct'

My question is, in a template function what is the proper way to pass in the function pointer? The code shown below returns an error 'const class QVariant has no member named convFunct'

template<class T>
QList<T> QVariantToQList(QVariant & qv,T (* convFunct)() )
{
    // Create the list that will hold the return values
    QList<T> qListOfMembers;
    if(qv.typeName() == "QVariantMap"){
        foreach(QVariant const& mapMember,qv.toMap())
        {
            qListOfMembers.append(mapMember.convFunct());
        }
    }
    else if (qv.typeName() == "QVariantList"){
        foreach(QVariant const& listMember,qv.toList())
        {
            qListOfMembers.append(listMember.convFunct());
        }
    }
    else
    {
        qDebug()<< "QVariantToQList currently is implemented only for QMap and QList types";
        throw ;
    }
    return qListOfMembers;
}

这是一个后续问题到以前的问题这个问题和这一个的区别是T是'MyClass'或'MyClass2'而不是

This is a follow-up question to a previous question The difference between that question and this one is the T is 'MyClass' or 'MyClass2' instead of a type that is normally held by a QVariant.

推荐答案

如果我正确理解您的问题， convFunct 应该是一个获得 QVariant 并返回 MyClass 或 MyClass2 ，是否正确？如果你的答案是肯定的，那么这个函数应该得到一个类型 QVariant 的参数，你的函数没有参数，所以结果是：

If I understand your question correctly, convFunct supposed to be a function that get a QVariant and return an instance of MyClass or MyClass2, is it correct? if your answer is yes, then this function should get a parameter of type QVariant and your function get no parameter, so the result is:

template<class T>
QList<T> QVariantToQList(QVariant & qv,T (*convFunct)(QVariant const&) )
{
    // Create the list that will hold the return values
    QList<T> qListOfMembers;
    if(qv.typeName() == "QVariantMap"){
        foreach(QVariant const& mapMember,qv.toMap())
        {
            qListOfMembers.append(convFunct(mapMember));
        }
    }
    else if (qv.typeName() == "QVariantList"){
        foreach(QVariant const& listMember,qv.toList())
        {
            qListOfMembers.append(convFunct(listMember));
        }
    }
    else
    {
        qDebug()<< "QVariantToQList currently is implemented only for QMap and QList types";
        throw ;
    }
    return qListOfMembers;
}

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