[英]Call C++ member function pointer without knowing which class
I am trying to call a member function, possibly given the object pointer, without knowing what class the member function is from. 我试图调用一个成员函数,可能给出了对象指针,而不知道成员函数来自哪个类。 Is this possible? 这可能吗?
Basically I want something like the following to work. 基本上我想要像以下一样工作。
class Foo{
public:
Foo(void* object): obj(object) {}
void callFunc(void (*func)()){
obj->*func();
}
private:
void* obj;
};
class Bar{
public:
Bar(): foo(this) {}
void callSomeFunc(){
callFunc(someFunc);
}
void someFunc(){
cout << "hi\n";
}
private:
Foo foo;
};
int main(){
Bar bar;
bar.callSomeFunc();
return 0;
}
It looks a lot like an XY-problem. 它看起来很像XY问题。 Anyway, let's try to reply to your question as it is. 无论如何,让我们试着回答你的问题。
A function member is bound to the type of the class to which it belongs, unless it's a static one (the latter is treated just like a plain function pointer and you don't even have to pass a pointer to an instance to call it). 函数成员绑定到它所属的类的类型,除非它是静态函数(后者被视为普通函数指针,你甚至不必将指针传递给实例来调用它) 。
Therefore you can make callFunc
a function template and let it deduce the type for you: 因此,您可以将callFunc
作为函数模板,并让它为您推导出类型:
template<typename T>
void callFunc(void (T::*func)()){
(static_cast<T*>(obj)->*func)();
}
See it up and running on wandbox . 在wandbox上查看并运行它 。
Note that you can incur in errors when you static_cast
your obj
if its original type (the one you erased to put it in a void *
) isn't T
. 注意,你可以在错误招致当你static_cast
你的obj
如果原始类型(你擦除把它放在一个void *
)不T
。
Here is the full code you can see at the link above: 以下是您可以在上面链接中看到的完整代码:
#include<iostream>
class Foo{
public:
Foo(void* object): obj(object) {}
template<typename T>
void callFunc(void (T::*func)()){
(static_cast<T*>(obj)->*func)();
}
private:
void* obj;
};
class Bar{
public:
Bar(): foo(this) {}
void callSomeFunc(){
foo.callFunc(&Bar::someFunc);
}
void someFunc(){
std::cout << "hi\n";
}
private:
Foo foo;
};
int main(){
Bar bar;
bar.callSomeFunc();
return 0;
}
It's an XY problem. 这是一个XY问题。 Use a std::function
and/or a lambda. 使用std::function
和/或lambda。
#include <functional>
#include <iostream>
class Foo{
public:
template<class F>
void callFunc(F&& f){
f();
}
};
class Bar : public Foo{
public:
Bar(): foo() {}
void callSomeFunc(){
this->callFunc([this]{ someFunc(); });
}
void someFunc(){
std::cout << "hi\n";
}
private:
Foo foo;
};
int main(){
Bar bar;
bar.callSomeFunc();
return 0;
}
Although I find the solution provided by @skypjack more elegant, here a solution that templates the Foo
-class (not "only" the function) as a whole. 虽然我发现@skypjack提供的解决方案更加优雅,但这里有一个解决方案,它模拟了Foo
类(不仅仅是“函数”)的整体。 Thereby the type of obj
is known throughout the Foo
-class, which might be an advantage (or may be not). 从而类型obj
在整个已知Foo
-class,这可能是一个优点(或者可以是不)。
Further, see also a solution that stores the member together with the associated object. 此外,还可以参见将成员与关联对象一起存储的解决方案。 Maybe it's helpful in some way: 也许它在某些方面很有用:
#include <functional>
#include <iostream>
template<class T>
class Foo {
public:
Foo(T& obj) : _obj(obj) {}
void callFuncOnObj(void (T::*func)(void)) {
auto fn = mem_fn(func);
fn(_obj);
}
private:
T &_obj;
};
class Bar{
public:
Bar() : d(*this) {}
void callSomeFunc(){
d.callFuncOnObj(&Bar::someFunc);
}
void someFunc(){
cout << "hi Bar1\n";
}
private:
Foo<Bar> d;
};
class Foo2 {
public:
Foo2(std::function<void(void)> f) : _f(f) {}
void callFunction() {
_f();
}
private:
std::function<void(void)> _f;
};
class Bar2{
public:
Bar2() : d(std::bind(&Bar2::someFunc,this)) {}
void callSomeFunc(){
d.callFunction();
}
void someFunc(){
cout << "hi Bar2\n";
}
private:
Foo2 d;
};
int main(){
Bar bar;
bar.callSomeFunc();
Bar2 bar2;
bar2.callSomeFunc();
return 0;
}
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