Overloading in subclass

[Joseph Turian]

Let's say I have the following:

class Foo {
public:
void func();
};

class Bar : public Foo {
public:
void func() {
// do some work...

this->Foo::func();
}
}

Now, if I have an object b of type Bar, and I call b.func(), it will
call b.Bar::func(), which in turn will call b.Foo::func(), correct?
Joseph

[Rob Williscroft]

Joseph Turian wrote in news:1118876638.997229.221340
@z14g2000cwz.googlegroups.com in comp.lang.c++:

Let's say I have the following:

class Foo {
public:
void func();
};

class Bar : public Foo {
public:
void func() {
// do some work...

this->Foo::func();
}
}

Now, if I have an object b of type Bar, and I call b.func(), it will
call b.Bar::func(), which in turn will call b.Foo::func(), correct?

Correct.
Rob.
--
http://www.victim-prime.dsl.pipex.com/

[Rolf Magnus]

Joseph Turian wrote:

Let's say I have the following:

class Foo {
public:
void func();
};

class Bar : public Foo {
public:
void func() {
// do some work...

this->Foo::func();
}
}

Now, if I have an object b of type Bar, and I call b.func(), it will
call b.Bar::func(), which in turn will call b.Foo::func(), correct?

Right.

[David White]

"Joseph Turian" <tu****@gmail.com> wrote in message
news:11**********************@z14g2000cwz.googlegr oups.com...

Let's say I have the following:

class Foo {
public:
void func();
};

class Bar : public Foo {
public:
void func() {
// do some work...

this->Foo::func();
You can just write Foo::func(); here.
}
}

Now, if I have an object b of type Bar, and I call b.func(), it will
call b.Bar::func(), which in turn will call b.Foo::func(), correct?

Yes. You aren't really "overloading" here, since the compiler is not
choosing which of several functions to call based on their argument lists.

DW