I'm trying to overload the -> operator to eventually execute something along the lines:
MyInterface *myInstance = (MyInterface *)(new A());
myInstance->Toggle(); //this works wonderfully
std::shared_ptr<Wrapper<MyInterface>> sharedPtrWrapper = std::make_shared<Wrapper<MyInterface>>(pIOnAcess);
//the next line doesn't compile, I would like to achieve something like this, but even
//sharedPtrWrapper.get()->Toggle();
//would be nice to achieve, is this possible?
sharedPtrWrapper->Toggle();
//this works:
sharedPtrWrapper->operator->()->Toggle();
Note: I have no control over MyInterface, cannot implement the pure virtual destructor.
Here is what I tried (the below code runs):
#import <memory>
#import <iostream>
struct MyInterface {
virtual bool Toggle() = 0;
};
class A : public MyInterface {
public:
bool Toggle() {
stateEnabled = !stateEnabled;
std::cout<<"current state " << stateEnabled << std::endl;
return true;
}
private:
bool stateEnabled = false;
};
template <typename T>
class Wrapper {
private:
T *unsafePointer = nullptr;
public:
Wrapper<T>()
{ }
T *operator->() const {
return unsafePointer;
}
Wrapper<T>(T *stuff) {
unsafePointer = stuff;
}
~Wrapper<T>() {}
};
int main(int argc, const char * argv[]) {
MyInterface *myInstance = (MyInterface *)(new A());
myInstance->Toggle();
Wrapper<MyInterface> wrapperS(myInstance);
wrapperS->Toggle();
std::shared_ptr<Wrapper<MyInterface>> sharedPtrWrapper = std::make_shared<Wrapper<MyInterface>>(myInstance);
sharedPtrWrapper->operator->()->Toggle();
sharedPtrWrapper.operator->()->operator->()->Toggle();
sharedPtrWrapper.get()->operator->()->Toggle();
(*sharedPtrWrapper).operator->()->Toggle();
return 0;
}
Output:
current state 1
current state 0
current state 1
current state 0
current state 1
current state 0
Program ended with exit code: 0
Note: stackoverflow doesn't let me post this question, because my post has more code than text.
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