polymorphic method override

[stf]

Why does this code not compile?

#include <iostream>

using namespace std;

class A {
public:
virtual void myMethod(const string& s) const;
};

class B : public A {
public:
virtual void myMethod(int x) const;
};

void A::myMethod(const string& s) const {
cerr << "(A) myMethod: s is: " << s << endl;
}
void B::myMethod(int x) const {
cerr << "(B) myMethod: x = " << x << endl;
}

int main() {
std::string s("my string one");
B b;
b.myMethod(s);
return 0;
}

The error is:

$ g++ -Wall ./test-5.cpp -o test-5.out ./test-5.cpp: In function
‘int main()’:
../test-5.cpp:25: error: no matching function for call to
‘B::myMethod(std::string&)’
../test-5.cpp:18: note: candidates are: virtual void B::myMethod(int)
const

It looks that if I change:

b.myMethod(s);

to:

b.A::myMethod(s);

then it works OK. However this is bulls*t of course, because later
if someone implements B::myMethod(const string& s) const, the
purpose of virtual method will be defeated.

How to cope with this??!

Thanks!

STF

 

[Alf P. Steinbach]

Why does this code not compile?

#include<iostream>

using namespace std;

class A {
public:
virtual void myMethod(const string& s) const;
};

class B : public A {
public:
virtual void myMethod(int x) const;
};

void A::myMethod(const string& s) const {
cerr<< "(A) myMethod: s is: "<< s<< endl;
}
void B::myMethod(int x) const {
cerr<< "(B) myMethod: x = "<< x<< endl;
}

int main() {
std::string s("my string one");
B b;
b.myMethod(s);
return 0;
}

The error is:

$ g++ -Wall ./test-5.cpp -o test-5.out ./test-5.cpp: In function
‘int main()’:
./test-5.cpp:25: error: no matching function for call to
‘B::myMethod(std::string&)’
./test-5.cpp:18: note: candidates are: virtual void B::myMethod(int)
const

This is a FAQ.

It's often a good idea to check the FAQ (or a textbook) before posting.

See http://www.parashift.com/c++-faq-lite/strange-inheritance.html#faq-23.9


Cheers & hth.,

- Alf

 

[Salt_Peter]

Why does this code not compile?

#include <iostream>

using namespace std;

class A {
   public:
       virtual void myMethod(const string& s) const;

};

class B : public A {
   public:
       virtual void myMethod(int x) const;

};

void A::myMethod(const string& s) const {
   cerr << "(A) myMethod: s is: " << s << endl;}

void B::myMethod(int x) const {
   cerr << "(B) myMethod: x = " << x << endl;

}

int main() {
   std::string s("my string one");
   B b;
   b.myMethod(s);
   return 0;

}

The error is:

$ g++ -Wall ./test-5.cpp -o test-5.out ./test-5.cpp: In function
‘int main()’:
./test-5.cpp:25: error: no matching function for call to
‘B::myMethod(std::string&)’
./test-5.cpp:18: note: candidates are: virtual void B::myMethod(int)
const

It looks that if I change:

b.myMethod(s);

to:

b.A::myMethod(s);

then it works OK. However this is bulls*t of course, because later
if someone implements B::myMethod(const string& s) const, the
purpose of virtual method will be defeated.

How to cope with this??!

what you get is an overide, not an overload.
The result you get is expected.
Now try it with a pointer to base.