在std :: map中继承基类作为值 [英] Inheritance in std::map with base class as value

问题描述

这是一个代码：

class Base {
    public:
        long index;
};
class Derived : public Base {
    public:
        bool value;
};

void call(map<char *, Base *> *base) {
    map<char *, Base *>::iterator it = base->begin();
    cout << it->second->index << endl;
}
void test(void) {
    map<char *, Derived *> *d = new map<char *, Derived *>;

    call(d);
}

编译器提醒一个错误：

error C2664: 'call' : cannot convert parameter 1 from 'std::map<_Kty,_Ty> *' to 'std::map<_Kty,_Ty> *'
1>        with
1>        [
1>            _Kty=char *,
1>            _Ty=Derived *
1>        ]
1>        and
1>        [
1>            _Kty=char *,
1>            _Ty=Base *
1>        ]
1>        Types pointed to are unrelated; conversion requires reinterpret_cast, C-style cast or function-style cast

我明白为什么会发生这种错误。
我不明白如何使它工作。什么样的演员和如何使用？

I understand why this error is happened. I do not understand how to make it work. What kind of cast and how to use it?

UPD

对不精确的抱歉，让我解释一下更多的细节。
我有两组由ClassA和ClassB表示的数据。
这两个类都有一个共同的成员 - 例如索引。
两套都包装成一个地图（特别感谢Rob对char *进行了重大修正）：

I'm sorry for imprecision, let me explain more details. I have two sets of data represented by ClassA and ClassB. Both of these classes have one common member - an "index", for example. Both sets are wrapped into a maps (special thanks to Rob for a significant correction with char*):

std::map<char, ClassA>
std::map<char, ClassB>

有时候，我需要迭代两个地图才能获得一个普通成员index的值。
我试图避免代码重复，并且只需要一个函数来迭代这两个地图。

Sometime I need to iterate over both maps to get a value of a common member "index". I'm trying to avoid code duplication and make just one function to iterate over both maps.

我以为我可以用一个普通成员提取一个超类，使用如下参数的函数：

I thought I may extract a superclass with a common member and make a function with parameter like this one:

std::map<char, SuperClassAB>

或

std::map<char, SuperClassAB>::iterator

但是看起来像是

UPD2

一个聪明的人给了我解决方案： / p>

One smart guy gave me the solution:

template <class T>
void call(map<char, T> *base) {
    map<char, T>::iterator it = base->begin();
    cout << it->second->index << endl;
}
void test(void) {
    map<char, Derived *> d;
    call(&d);
}

推荐答案

没有人似乎建议但是您也可以使调用一个函数模板。

No one seemed to suggest this yet, but you could also make call a function template.

    template <class Type>
    void call(const std::map<char*,Type*> & base)
    {
        static_assert(std::is_base_of<Base, Type>::value,
                      "Function is only callable with maps "
                      "whose mapped_type is derived from Base");
        /* More stuff */
    }

使用 Base ，派生以及从 Base 派生的任何其他内容。

This way the function is callable with Base, Derived and anything else that derives from Base.

这篇关于在std :: map中继承基类作为值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！