Re: calling class method
shaun roe wrote:
In article <2008010311314116807-kirakun@earthlinknet>,
Kira Yamato <kirakun@earthlink.net> wrote:
On 2008-01-03 11:02:00 -0500, mosfet <john.doe@anonymous.org> said:
Hi,
Let's say I have two classes A and B with B inheriting from A
with one non virtual method :
class A
{
public:
void MethodA()
{
cout << "MethodeA in class A" << endl
}
};
class B : public class A
{
public:
void MethodA()
{
cout << "MethodeA in class B" << endl
}
};
int main()
{
B testB;
testB.MethodA(); // Should display "MethodeA in class B"
}
IF I try this example the MethodA from class B will be called.
And what if I want from a B object to call the A implementation ?
Is it possible ?
testB.A::MethodA();
like so?:
#include <iostream>
using namespace std;
class A{
public:
void methodA(){
cout << "Method A in class A" << endl;
}
};
class B : public A{
public:
void methodA(){
cout << "Method A in class B" << endl;
}
void methodAfromA(){
A::methodA();
}
};
int main (int argc, char * const argv[]) {
// insert code here...
std::cout << "Hello, World!\n";
B testB;
testB.methodA(); // Should display "Method A in class B"
testB.methodAfromA();
return 0;
}
Why bother with all this? Just do as Kira has suggested, in the
'main' function:
int main() {
B testB;
testB.A::MethodA();
}
Or do you have a problem with that?
V
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