intrestring puzzle

[dada]

The following source file does not compile.
#include <cstdio>
class X {
public:
X() { i = 100; }
private:
int i;
};
int main()
{
X x;
printf("%d\n", x.i);
}

Add one line (having one semicolon) to the definition of X that in
turn makes the file compilable. itsValue shud b private.

[Alf P. Steinbach]

* us********@yahoo.com (dada) schriebt:

The following source file does not compile.
Add one line (having one semicolon) to the definition of X that in
turn makes the file compilable. itsValue shud b private.

Yes, it's friday, but don't drink and post at the same time.

[Leor Zolman]

On 27 Feb 2004 10:19:59 -0800, us********@yahoo.com (dada) wrote:

The following source file does not compile.
#include <cstdio>
class X {
public:
X() { i = 100; }
private:
int i;
};
int main()
{
X x;
printf("%d\n", x.i);
}

Add one line (having one semicolon) to the definition of X that in
turn makes the file compilable. itsValue shud b private.

class X {
public:
X() { i = 100; }
private:
public: int x; // added this
int i;
};

Problem is, I have no idea what you mean by
itsValue shud b private.
Is that the name of a variable? If so, where did it come from, and if not,
values aren't public or private, so what are you talking about? And could
you use English, please? ;-)
-leor

Leor Zolman
BD Software
le**@bdsoft.com
www.bdsoft.com -- On-Site Training in C/C++, Java, Perl & Unix
C++ users: Download BD Software's free STL Error Message
Decryptor at www.bdsoft.com/tools/stlfilt.html

[Julián Albo]

dada escribió:

Add one line (having one semicolon) to the definition of X that in
turn makes the file compilable. itsValue shud b private.

friend int main ();

Regards.

[Jonathan Turkanis]

"Julián Albo" <JU********@terra.es> wrote in message news:40***************@terra.es..."dada escribió: "> Add one line (having one semicolon) to the definition of X that in
"> turn makes the file compilable. itsValue shud b private. "friend int main ();

also std::printf.

Jonathan

[Claudio Puviani]

"Jonathan Turkanis" <te******@kangaroologic.com> wrote

"Julián Albo" <JU********@terra.es> wrote in message

news:40***************@terra.es...

"dada escribió:

"> Add one line (having one semicolon) to the definition of X that in
"> turn makes the file compilable. itsValue shud b private.

"friend int main ();

also std::printf.

Making printf() a friend wouldn't help because it's not the one trying to access
x.i. main() is accessing x.i and passing its value to printf(). :-)

Claudio Puviani

[Jonathan Turkanis]

"Claudio Puviani" <pu*****@hotmail.com> wrote in message
news:Sh********************@news4.srv.hcvlny.cv.ne t...

"Jonathan Turkanis" <te******@kangaroologic.com> wrote

"Julián Albo" <JU********@terra.es> wrote in message news:40***************@terra.es...

"dada escribió:

"> Add one line (having one semicolon) to the definition of X that in"> turn makes the file compilable. itsValue shud b private.

"friend int main ();

also std::printf.

Making printf() a friend wouldn't help because it's not the one

trying to access x.i. main() is accessing x.i and passing its value to printf(). :-)

I wasn't suggesting making it a friend. I was just pointing out that
namespace qualification was missing.

Jonathan

[Thomas Matthews]

dada wrote:

The following source file does not compile.
#include <cstdio>
class X {
public:
X() { i = 100; }
private:
int i;
};
int main()
{
X x;
printf("%d\n", x.i);
}

Add one line (having one semicolon) to the definition of X that in
turn makes the file compilable. itsValue shud b private.

Firdays [sp] are bad days to post issues that can
be resolved by consulting the FAQ.
http://www.parashift.com/c++-faq-lit...d-objects.html
http://www.parashift.com/c++-faq-lite/friends.html
http://www.parashift.com/c++-faq-lite/newbie.html

Here is the reason. You have declared the variable 'i'
as private (having private accessibility). "private"
means that only class member functions have access
to those items. Any function outside of the class
cannot see nor access a private item.

The compiler is complaining that the main() function
is trying to access a private data member.

If you want a private member to be displayed, you
either have to create a member function to do that:
class X
{
// as above
public:
void print(void)
{ printf("%d\n", i);}
void print_cpp_style(void)
{ cout << i << endl;}
void print_better_style(ostream& out)
{ out << i; }
};

int main(void)
{
X junk;
junk.print();
junk.print_cpp_style();
junk.print_better_style(cout);
cout << endl;
return 0;
}
or you will have to place the member variable
into a public section:
class X
{
public:
int i;
};

Also, due yourself a favor and look up "initialization
list" in your favorite text book or help file.

--
Thomas Matthews

C++ newsgroup welcome message:
http://www.slack.net/~shiva/welcome.txt
C++ Faq: http://www.parashift.com/c++-faq-lite
C Faq: http://www.eskimo.com/~scs/c-faq/top.html
alt.comp.lang.learn.c-c++ faq:
http://www.raos.demon.uk/acllc-c++/faq.html
Other sites:
http://www.josuttis.com -- C++ STL Library book

[dada]

yes u r rite.but its possible to access a private member directly by
using pointers.like for eg
X x
X *ptr = &x;
int *info;
info=(int*)ptr;
info will be able to print out i.but i want to
access the private data by including just 1 line in the program
to the definiton of X with only 1 semicolon. this is the requirement
of the puzzle.
u shud not change the main function
class X {

public:
X() { i = 100; }
private:
int i;
};
int main()
{
X x;
printf("%d\n", x.i);
}

Thomas Matthews <Th****************************@sbcglobal.net> wrote in message news:<hU*******************@newssvr16.news.prodigy .com>... dada wrote:

The following source file does not compile.
#include <cstdio>
class X {
public:
X() { i = 100; }
private:
int i;
};
int main()
{
X x;
printf("%d\n", x.i);
}

Add one line (having one semicolon) to the definition of X that in
turn makes the file compilable. itsValue shud b private.

Firdays [sp] are bad days to post issues that can
be resolved by consulting the FAQ.
http://www.parashift.com/c++-faq-lit...d-objects.html
http://www.parashift.com/c++-faq-lite/friends.html
http://www.parashift.com/c++-faq-lite/newbie.html

Here is the reason. You have declared the variable 'i'
as private (having private accessibility). "private"
means that only class member functions have access
to those items. Any function outside of the class
cannot see nor access a private item.

The compiler is complaining that the main() function
is trying to access a private data member.

If you want a private member to be displayed, you
either have to create a member function to do that:
class X
{
// as above
public:
void print(void)
{ printf("%d\n", i);}
void print_cpp_style(void)
{ cout << i << endl;}
void print_better_style(ostream& out)
{ out << i; }
};

int main(void)
{
X junk;
junk.print();
junk.print_cpp_style();
junk.print_better_style(cout);
cout << endl;
return 0;
}
or you will have to place the member variable
into a public section:
class X
{
public:
int i;
};

Also, due yourself a favor and look up "initialization
list" in your favorite text book or help file.

--
Thomas Matthews

C++ newsgroup welcome message:
http://www.slack.net/~shiva/welcome.txt
C++ Faq: http://www.parashift.com/c++-faq-lite
C Faq: http://www.eskimo.com/~scs/c-faq/top.html
alt.comp.lang.learn.c-c++ faq:
http://www.raos.demon.uk/acllc-c++/faq.html
Other sites:
http://www.josuttis.com -- C++ STL Library book

[Richard Herring]

In message <f7**************************@posting.google.com >, dada
<us********@yahoo.com> writes

yes u r rite.but its possible to access a private member directly by
using pointers.like for eg
X x
X *ptr = &x;
int *info;
info=(int*)ptr;
info will be able to print out i.

#include <iostream>
#include <ostream>
class X
{
public:
X() : i(100) {}
virtual ~X() {}
private:
int i;
};

int main()
{
X x;
X *ptr = &x;
int *info;
info=(int*)ptr;
std::cout << *info << std::endl;
return 0;
}

Strange. I tried that here but it printed 4202796. Any idea why that
might be?

--
Richard Herring

[Mike Wahler]

"Richard Herring" <ju**@[127.0.0.1]> wrote in message
news:Dw**************@baesystems.com...

In message <f7**************************@posting.google.com >, dada
<us********@yahoo.com> writes

yes u r rite.but its possible to access a private member directly by
using pointers.like for eg
X x
X *ptr = &x;
int *info;
info=(int*)ptr;
info will be able to print out i.
#include <iostream>
#include <ostream>
class X
{
public:
X() : i(100) {}
virtual ~X() {}
private:
int i;
};

int main()
{
X x;
X *ptr = &x;
int *info;
info=(int*)ptr;

What do you think this assignment is supposed to do?
All you've done with the cast, is lie to the compiler.
Saying it doesn't make it so.
std::cout << *info << std::endl;
return 0;
}

Strange. I tried that here but it printed 4202796. Any idea why that
might be?

Nobody can explain undefined behavior. It's undefined.
(or at least implementation-defined, in which case you
need to read your documentation to find out).

-Mike