C++ 从超类调用子类方法
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C++ Call subclass method from superclass
提问by ChrisGeo
I have code of the following style:
我有以下样式的代码：
class SubClass;
class SuperClass;
class SuperClass {
private:
void bar() {
SubClass().foo();
}
};
class SubClass : SuperClass {
public:
void foo() {};
};
So basically I have a SuperClass from where I want to call a method foo() of the subclass. VS 2012 gives me the following errors:
所以基本上我有一个 SuperClass，我想从中调用子类的方法 foo()。VS 2012 给了我以下错误：
Error 1 error C2514: 'SubClass' : class has no constructors.
错误 1 错误 C2514：“子类”：类没有构造函数。
Error 2 error C2228: left of '.foo' must have class/struct/union.
错误 2 错误 C2228：'.foo' 的左边必须有类/结构/联合。
What is the correct structure for what I want to do?
我想做的事情的正确结构是什么？
回答by Kiril Kirov
You can't do this. You must (at least) declare the method in the base class. For example:
你不能这样做。您必须（至少）在基类中声明该方法。例如：
#include <iostream>
class SuperClass
{
public:
void bar()
{
foo();
}
private:
virtual void foo() // could be pure virtual, if you like
{
std::cout << "SuperClass::foo()" << std::endl;
}
};
class SubClass : public SuperClass // do not forget to inherit public
{
public:
virtual void foo() { std::cout << "SubClass::foo()" << std::endl; }
};
int main()
{
SuperClass* pTest = new SubClass;
pTest -> bar();
delete pTest;
}
will print SubClass::foo().
将打印SubClass::foo()。
回答by riv
First, you have to realise that calling SubClass().foo()has nothing to do with the current object - it will create a new SubClassobject and call its foomember. If you want the barfunction to call fooon the current object, you need to declare fooas a virtual function in the base class, as Kiril suggested.
首先，您必须意识到调用SubClass().foo()与当前对象无关 - 它会创建一个新SubClass对象并调用其foo成员。如果您希望bar函数foo在当前对象上调用，则需要foo在基类中声明为虚函数，如 Kiril 建议的那样。
However, if you do want to call fooon a new object (which doesn't make much sense, but whatever), you are doing it correctly. Your example didn't compile because SubClassis forward declared, but not defined before barfunction - you might want to move the implementation of barbelow the definition of SubClass, like so:
class SubClass;
class SuperClass;
但是，如果您确实想调用foo一个新对象（这没有多大意义，但无论如何），您就做对了。您的示例未编译，因为SubClass是前向声明的，但未在bar函数之前定义 - 您可能希望将 的实现移动bar到 的定义下方，如下所示SubClass：班级超班；
class SuperClass {
private:
void bar();
};
class SubClass : SuperClass {
public:
void foo() {};
};
void SuperClass::bar() {
SubClass().foo();
}
回答by asdru
you can try with some CRTP, ie:
您可以尝试使用一些 CRTP，即：
#include <iostream>
class SubClass;
template<typename T>
class SuperClass {
public:
void bar() {
((T *) this)->foo();
}
};
class SubClass : public SuperClass<SubClass> {
public:
void foo() {
std::cout << "hello crtp!" << std::endl;
};
};
int main(void){
SubClass a = SubClass();
a.bar();
return 0;
}
