当类名存储在另一个变量中时，如何创建类的实例 [英] How to create instance of a class when the class name is stored in another variable

问题描述

可能重复：

C ++中的最小反射

基本上我有类名存储在 char * ，如何创建该类的实例。
以下是我在想什么。当我通过 Test 作为模板参数这是很好，但如果我尝试这样 Test * t = CreateType< ptr> 它不会工作。

Basically I have class name stored in char * , how Do i create instance of that class. Following is what I was thinking. It works when I pass Test as template parameter which is fine, but if I try something like this Test *t = CreateType<ptr> it will not work.

有任何方法使这项工作。

Is there any way to make this work.

class Test{
public:
    Test() {
    }
    ~Test() {
    }
};

template <typename T> 
T* CreateType() {
    return new T;
}

int main ( int argc, char **argv) {
    char *ptr = "Test";
    Test *T = CreateType<Test>();
    return 0;
}

推荐答案

直接可能在C ++中，因为语言不是内省的。

It is not directly possible in C++, because the language is not introspective.

您可以尝试做类似的 。例如，您可以使用地图< string，baseClassPointer> 其中键将是类的名称，并且该值将是指向该类对象的指针。然后，您可以通过从地图复制对象（您可以通过类型名称获取）创建其他对象。我不建议做这样的事情，虽然。

You can try to do something similar though. For example, you can have a map< string, baseClassPointer > where the key will be the class' name, and the value will be a pointer to that class' object. Then you can create other objects by copying an object from the map (which you can get by the type name). I wouldn't suggest doing things like that though.

工厂模式可以满足您的需求。

Factory pattern may suit your needs though.

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