I was wondering if it's possible to override just one function in a class without creating an entirely new class.
I would like bObj1.foo();
to output "foo!" and bObj2.foo()
to output "foo?", but currently they both output "foo!".
#include <iostream>
using namespace std;
class B {
public:
virtual void foo() { cout << 开发者_C百科"foo!" << endl; }
};
class A {
public:
B f();
};
B A::f() {
B bObj;
return bObj;
}
class C : public A {
};
int main()
{
A aObj;
B bObj1 = aObj.f();
bObj1.foo(); // "foo!"
C cObj;
B bObj2 = cObj.f();
bObj2.foo(); // "foo?"
}
You can get the behavior that you want with a simple change, which consists in moving the "virtual" behavior to the A and C classes.
Here I modified your application to return the expected result:
#include <iostream>
using namespace std;
class A;
class B {
public:
B(A& a) : aref(a) {}
void foo();
private:
A& aref;
};
class A {
public:
B f();
virtual void foo() { cout << "foo!" << endl; }
};
B A::f() {
B bObj(*this);
return bObj;
}
class C : public A {
public:
virtual void foo() { cout << "foo?" << endl; }
};
void B::foo() { aref.foo(); }
int main()
{
A aObj;
B bObj1 = aObj.f();
bObj1.foo(); // "foo!"
C cObj;
B bObj2 = cObj.f();
bObj2.foo(); // "foo?"
}
In order to change the virtual function, you have to create a new type - there's no way around that in C++. However, an alternate mechanism - function objects - may do what you want here.
#include <functional>
#include <iostream>
using namespace std;
class B {
public:
B(function<void ()> foo_impl) : foo_impl(foo_impl) {}
void foo() {foo_impl();}
private:
function<void()> foo_impl;
};
class A {
public:
virtual B f();
};
B A::f() {
B bObj([](){cout << "foo!" << endl;});
return bObj;
}
class C : public A {
public:
virtual B f() override;
};
B C::f() {
B bObj([](){cout << "foo?" << endl;});
return bObj;
}
int main()
{
A aObj;
B bObj1 = aObj.f();
bObj1.foo(); // "foo!"
C cObj;
B bObj2 = cObj.f();
bObj2.foo(); // "foo?"
}
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