what will happen if we print the following...

[Shraddha]

int *s=(int *)2000;
int *p=(int *)1000;

printf("%d",p-s);

[=?iso-8859-1?q?Erik_Wikstr=F6m?=]

On 25 Maj, 14:47, Shraddha <shraddhajosh.. [e-mail] e:

int *s=(int *)2000;
int *p=(int *)1000;

printf("%d",p-s);

What do you think will happen?

By the way, this is probably better asked in comp.lang.c

--
Erik Wikström

[shadowman]

Shraddha wrote:

int *s=(int *)2000;
int *p=(int *)1000;

printf("%d",p-s);

This question doesn't make any sense. You'll take the time to post this
to a newsgroup, but you won't compile the code and see for yourself what
happens?

[Salt_Peter]

On May 25, 8:47 am, Shraddha <shraddhajosh.. [e-mail] e:

int *s=(int *)2000;
int *p=(int *)1000;

printf("%d",p-s);

What will happen is nowhere near as important as what the intention of
the code is. What are you trying to do?

[osmium]

"Salt_Peter " writes:

On May 25, 8:47 am, Shraddha <shraddhajosh.. [e-mail] e:

>int *s=(int *)2000;
int *p=(int *)1000;

printf("%d", p-s);

What will happen is nowhere near as important as what the intention of
the code is. What are you trying to do?

It looks to me like he is trying to learn the rules of some computer
language. And perhaps its interaction with its native habitat.

[Default User]

osmium wrote:

"Salt_Peter " writes:

On May 25, 8:47 am, Shraddha <shraddhajosh.. [e-mail] e:

int *s=(int *)2000;
int *p=(int *)1000;
>
printf("%d",p-s);

What will happen is nowhere near as important as what the intention
of the code is. What are you trying to do?

It looks to me like he is trying to learn the rules of some computer
language. And perhaps its interaction with its native habitat.

Learning a programming language, especially a large one like C++, by
random usenet posts is a fool's game. It wastes everyone's time.

He needs to get a decent book and work through it, asking questions
about things he can't figure out.


Brian

[James Kanze]

On May 25, 2:47 pm, Shraddha <shraddhajosh.. [e-mail] e:

int *s=(int *)2000;
int *p=(int *)1000;

printf("%d",p-s);

It depends very much on the implementation, and it could core
dump, or crash the system. On most systems, however, it will
probably output the value of 1000/sizeof(int).

--
James Kanze (Gabi Software) email: ja*********@gma il.com
Conseils en informatique orientée objet/
Beratung in objektorientier ter Datenverarbeitu ng
9 place Sémard, 78210 St.-Cyr-l'École, France, [phone]

[Salt_Peter]

On May 25, 3:07 pm, "osmium" <r124c4u...@com cast.netwrote:

"Salt_Peter " writes:

On May 25, 8:47 am, Shraddha <shraddhajosh.. [e-mail] e:

int *s=(int *)2000;
int *p=(int *)1000;

printf("%d",p-s);

What will happen is nowhere near as important as what the intention of
the code is. What are you trying to do?

It looks to me like he is trying to learn the rules of some computer
language. And perhaps its interaction with its native habitat.

Unfortunately, that appears to be the case (although i'ld replace
'rules' with 'effects'). I'ld suggest that setting pointers to
unallocated memory using magic numbers is begging for a whiplashing.
Calculating the offset between 2 memory locations and dividing a
hypothetical sizeof(int) is not a constructive discussion for the OP.
Thats not what C++ is about, IMHO.

He might as well do:
#include <iostream>

int main()
{
size_t t(1000);
std::cout << t/sizeof(int) << std::endl;
}

I'ld rather he come forth and explain the purpose or expectations of
the code before we start hitting him with semantics (prefer
static_cast<>() to old casts, magic pointers are evil, etc).

[osmium]

"Shraddha" wrote:

int *s=(int *)2000;
int *p=(int *)1000;

printf("%d",p-s);

You would do well to compile this, run it, see what happens and remeber it.
#include <stdio.h>

int main(void)
{
double x = 1.e35;
int n;
n = (int) x;
printf("%d\n", n);
return 0;
}

The message is clear. A cast says: "compiler, do this conversion. I am a
smart human and I know what I am doing." The compiler believes you and does
its best to comply. The results may be nonsense, which is what I get on my
compiler. It prints -250. The compilers I have used say "can not convert'
in the pre-cast error message. It really means: I *will not*, unless you
explicitly tell me to.

[osmium]

"osmium" writes:

#include <stdio.h>

int main(void)
{
double x = 1.e35;
int n;
n = (int) x;
printf("%d\n", n);
return 0;
}

Oops. That's C code, I read it as a C problem and answered accordingly.