Reputation: 3474
I understand that the following code doesn't work -- can't convert base to foo.
Is there something I can do, or some pattern to employ which would get me close the behavior I'm trying to achieve in the code below? IOW, if I have a base class pointer to a derived type, how can I invoke a specific method that matches the derived type (not the base type)?
What patterns might I use? I looked into Curiously Recursive (or recurring) Template Pattern, however this imposed other limitations itself.
class base {};
class foo : public base {};
void method(const foo& f){}
int main(){
base* b = new foo();
method(*b);
}
Upvotes: 1
Views: 631
Reputation: 15334
The easiest way is probably to just make method()
a virtual
member function on foo
:
class base {
public:
virtual void method() const = 0;
};
class foo : public base {
public:
void method() const override { }
};
int main(){
foo f;
base* b = &f;
b->method();
}
But if for some reason that is not possible (perhaps you don't have access to method()
or perhaps you want to keep the logic in method()
separate from foo
) you could use a simplified version of the Visitor Pattern.
The visitor pattern relies on all the classes in your hierarchy having a virtual function to dispatch based on class type.
In your case you don't need double-dispatch so you don't actually need the visitor object and can just call your method
function directly from a virtual dispatch
function:
class base {
public:
virtual void dispatch() const = 0;
};
class foo : public base {
public:
void dispatch() const override;
};
void method(const foo& f){}
void foo::dispatch() const {
method(*this);
}
int main(){
foo f;
base* b = &f;
b->dispatch();
}
You have to remember that in most contexts the compiler doesn't know that your base
pointer is actually of type foo
.
Upvotes: 2
Reputation: 42858
Use virtual
functions to solve these kinds of problems:
class base {
public:
virtual void func() const { /* A */ }
};
class foo : public base {
public:
void func() const override { /* B */ }
};
void method(const base& f) {
f.func();
}
int main(){
base* b = new foo();
method(*b);
}
Now, depending on the actual type of f
, either A
or B
code will be executed in method
.
Upvotes: 1