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What would be the O/P? and Why?

#include <iostream>
using namespace std;

class A {
public:
void foo() { cerr << "this is done\n"; }
};

int
main(int argc, char* argv[])
{
void *p = NULL;

((A*)p)->foo();
((A*)NULL)->foo();

return 0;
}

i amn't completely convinced with the behaviour of this program. what
kinda o/p can i expect from this? does all compilers give us the same
kinda behaviour from this code?
with regards,
sreenivas t.
Jul 22 '05 #1
2 1027
Sreenivas wrote:

#include <iostream>
using namespace std;

class A {
public:
void foo() { cerr << "this is done\n"; }
};

int
main(int argc, char* argv[])
{
void *p = NULL;

((A*)p)->foo();
((A*)NULL)->foo();

return 0;
}

i amn't completely convinced with the behaviour of this program.


The behaviour is undefined.
Anything migth happen, from 'crashing your computer' to 'works
as expected.'

--
Karl Heinz Buchegger
kb******@gascad .at
Jul 22 '05 #2
> #include <iostream>
using namespace std;

class A {
public:
void foo() { cerr << "this is done\n"; }
};

int
main(int argc, char* argv[])
{
void *p = NULL;

((A*)p)->foo();
((A*)NULL)->foo();

return 0;
}

i amn't completely convinced with the behaviour of this program. what
kinda o/p can i expect from this? does all compilers give us the same
kinda behaviour from this code?
with regards,


This is undefined behavior. If it works, don't rely on it.

The thing is, compilers usually implement member functions as ordinary
functions with an added 'this' pointer. So foo() actually is

// in global namespace
void foo_mangled_nam e(A *this) {}

Calling that function on a null pointer actually works on your
implementation, because you don't use the this pointer. In your
example, 'this' would be 0, which is a bad thing.

Don't do that.
Jonathan
Jul 22 '05 #3

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