James Kanze said:
That's the compiler writers problem, not mine.
It certainly is a problem for the compiler, and perhaps the program too.
Especially if you didn't initialised the empty pointer.
Let me put it another way, where would you get the address for the
virtual function?
It simply can't be done as the concept of virtual functions only lives
in the world of objects.
Please show some basic code. I guarantee you cannot.
Paul said:
Pete Becker said:
Garrett Hartshaw said:-----BEGIN PGP SIGNED MESSAGE-----
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James Kanze said:[...]
I can understand the concept you express but
a) how do you get the address of a member function?
&ClassName::functionName
Concretely:
struct C { void f(); };
void (C::*pf)() = &C::f;
b) what happens if this member function is virtual?
It works correctly. That's why pointer to member functions are
often larger than any other pointer types (but there are other
ways of solving the problem).
What would your pointer point to ?
That's the compiler writers problem, not mine
It certainly is a problem for the compiler, and perhaps the program too.
Especially if you didn't initialised the empty pointer.
Let me put it another way, where would you get the address for the
virtual function?
It simply can't be done as the concept of virtual functions only lives
in the world of objects.
Please show some basic code. I guarantee you cannot.
#include <iostream>
class C1 {
public:
virtual void f() {
std::cout << "C1::f" << std::endl;
}
};
class C2 : public C1 {
public:
virtual void f() {
std::cout << "C2::f" << std::endl;
}
};
int main () {
void (C1::*p)() = &C1::f; //create a pointer to a member function
C1 a; //create a object of type C1
C1 * b; //create a object of type pointer to C1
Paul said:
Garrett said:
Garrett Hartshaw said:
Garrett Hartshaw said:-----BEGIN PGP SIGNED MESSAGE-----
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[...]
I can understand the concept you express but
a) how do you get the address of a member function?
&ClassName::functionName
Concretely:
struct C { void f(); };
void (C::*pf)() = &C::f;
b) what happens if this member function is virtual?
It works correctly. That's why pointer to member functions are
often larger than any other pointer types (but there are other
ways of solving the problem).
What would your pointer point to ?
That's the compiler writers problem, not mine
It certainly is a problem for the compiler, and perhaps the program too.
Especially if you didn't initialised the empty pointer.
Let me put it another way, where would you get the address for the
virtual function?
You cannot do this with virtual functions and you are wrong
to suggest it works correctly.
It does work, and I've done it. More than once.
It simply can't be done as the concept of virtual functions only lives
in the world of objects.
Please show some basic code. I guarantee you cannot.
#include <iostream>
class C1 {
public:
virtual void f() {
std::cout << "C1::f" << std::endl;
}
};
class C2 : public C1 {
public:
virtual void f() {
std::cout << "C2::f" << std::endl;
}
};
int main () {
void (C1::*p)() = &C1::f; //create a pointer to a member function
C1 a; //create a object of type C1
C1 * b; //create a object of type pointer to C1
The challenge put forward was to invoke a virtual function without
creating an object. Here you have created an object.
I applaud your abilities nonetheless, assuming the code is correct.
The very nature of virtual functions require objects and perhaps you
have demostrated this.
This was what i was referring when I initially stated 'it won't work
with virtual functions':
"You can take the address of a member function (and assign it to a
function pointer of the proper type), and it will *not* be tied to any
specific object."
Dog is a class
More fully:
Dog is a class,that is the definition of an object type.
If fido is an object, and Dog is a class
How can fido be a Dog , because an object is not a class?
You can juggle these terms around till the cows come home.
But you cannot state that "fido is *not* an object type" because fido is
an object type in the same sense it's a Dog type'
Consider someone asking the question , what type is fido?
You could reply ... fido is an object type.
You are changing the context :
Dog fido;
to
type-name identifier;
You can say that the identifier fido is not an object type, but then its
not an object either because you have defined it as an identifier.
You implicitly imply a new context.
Paul said:
Paul said:
You can't just make statements to define the C++ language as you interpretLeigh Johnston said:
Garrett Hartshaw said:-----BEGIN PGP SIGNED MESSAGE-----
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Garrett Hartshaw said:-----BEGIN PGP SIGNED MESSAGE-----
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On 01/07/2011 02:06 PM, Paul wrote:
[...]
I can understand the concept you express but
a) how do you get the address of a member function?
&ClassName::functionName
Concretely:
struct C { void f(); };
void (C::*pf)() = &C::f;
b) what happens if this member function is virtual?
It works correctly. That's why pointer to member functions are
often larger than any other pointer types (but there are other
ways of solving the problem).
What would your pointer point to ?
That's the compiler writers problem, not mine
It certainly is a problem for the compiler, and perhaps the program
too.
Especially if you didn't initialised the empty pointer.
Let me put it another way, where would you get the address for the
virtual function?
You cannot do this with virtual functions and you are wrong
to suggest it works correctly.
It does work, and I've done it. More than once.
It simply can't be done as the concept of virtual functions only lives
in the world of objects.
Please show some basic code. I guarantee you cannot.
#include <iostream>
class C1 {
public:
virtual void f() {
std::cout << "C1::f" << std::endl;
}
};
class C2 : public C1 {
public:
virtual void f() {
std::cout << "C2::f" << std::endl;
}
};
int main () {
void (C1::*p)() = &C1::f; //create a pointer to a member function
C1 a; //create a object of type C1
C1 * b; //create a object of type pointer to C1
The challenge put forward was to invoke a virtual function without
creating an object. Here you have created an object.
I applaud your abilities nonetheless, assuming the code is correct.
The very nature of virtual functions require objects and perhaps you
have demostrated this.
This was what i was referring when I initially stated 'it won't work
with virtual functions':
"You can take the address of a member function (and assign it to a
function pointer of the proper type), and it will *not* be tied to any
specific object."
The (virtual) member function was *not* tied to a specific object, as it
(the same function) was used by two different objects. If it was *a part
of* the object, you would not be able to have a pointer to it.
Garrett Hartshaw said:-----BEGIN PGP SIGNED MESSAGE-----
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-----BEGIN PGP SIGNED MESSAGE-----
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On 01/07/2011 02:06 PM, Paul wrote:
[...]
I can understand the concept you express but
a) how do you get the address of a member function?
&ClassName::functionName
Concretely:
struct C { void f(); };
void (C::*pf)() = &C::f;
b) what happens if this member function is virtual?
It works correctly. That's why pointer to member functions are
often larger than any other pointer types (but there are other
ways of solving the problem).
What would your pointer point to ?
That's the compiler writers problem, not mine
It certainly is a problem for the compiler, and perhaps the program
too.
Especially if you didn't initialised the empty pointer.
Let me put it another way, where would you get the address for the
virtual function?
You cannot do this with virtual functions and you are wrong
to suggest it works correctly.
It does work, and I've done it. More than once.
It simply can't be done as the concept of virtual functions only lives
in the world of objects.
Please show some basic code. I guarantee you cannot.
#include <iostream>
class C1 {
public:
virtual void f() {
std::cout << "C1::f" << std::endl;
}
};
class C2 : public C1 {
public:
virtual void f() {
std::cout << "C2::f" << std::endl;
}
};
int main () {
void (C1::*p)() = &C1::f; //create a pointer to a member function
C1 a; //create a object of type C1
C1 * b; //create a object of type pointer to C1
The challenge put forward was to invoke a virtual function without
creating an object. Here you have created an object.
I applaud your abilities nonetheless, assuming the code is correct.
b = new C2(); //allocate a object of type C2
(a.*p)(); //call the member function pointed to
//by p, with &a as this
(b->*p)(); //call the member function pointed to
//by p (virtually), with b as this
return 0;
}
This code prints the following.
C1::f
C2::f
The very nature of virtual functions require objects and perhaps you
have demostrated this.
This was what i was referring when I initially stated 'it won't work
with virtual functions':
"You can take the address of a member function (and assign it to a
function pointer of the proper type), and it will *not* be tied to any
specific object."
The (virtual) member function was *not* tied to a specific object, as it
(the same function) was used by two different objects. If it was *a part
of* the object, you would not be able to have a pointer to it.
With:
(a.*p)(); //call the member function pointed to by p, with &a as this
(b->*p)(); //call the member function pointed to by p (virtually),
with b as this
There are two functions here.
As I understand it you seem to think there is only one function . ????
Ulrich Eckhardt said:
We *are* staying in the C++ context , you are the only one who is suggesting
Are you suggesting the standards describe a built-in type as an object type?