Have any solution

[sojin]

#include<stdio.h>
void change()
{
/* Write something in this function so that the output of printf in
main
function should give 5 . Do not change the main function */
}

void main()
{
int i=5;
change();
i=10;
printf("%d",i);
}

[Karl Heinz Buchegger]

sojin wrote:

#include<stdio.h>
void change()
{
/* Write something in this function so that the output of printf in
main
function should give 5 . Do not change the main function */
}

void main()
{
int i=5;
change();
i=10;
printf("%d",i);
}

IMHO: Impossible.
There is no way to get your hands at variable i in main().

--
Karl Heinz Buchegger
kb******@gascad.at

[Rolf Magnus]

sojin wrote:

#include<stdio.h>
void change()
{
/* Write something in this function so that the output of printf in
main
function should give 5 . Do not change the main function */
}

void main()
main() must return int.
{
int i=5;
change();
i=10;
printf("%d",i);
}

That's impossible in standard C++. I also fail to see what that would be
good for.

[Victor Bazarov]

sojin wrote:

#include<stdio.h>
void change()
{
/* Write something in this function so that the output of printf in
main
function should give 5 . Do not change the main function */
}

void main()
int main() // C++ does NOT have 'void main'
{
int i=5;
change();
i=10;
printf("%d",i);
}

There is no solution to your homework problem in the Standard C++, AFAICS.

Since the value that printf outputs is a local [to 'main'] variable,
nothing that another function can do would access that variable if you
don't pass its address or a reference to it into that function. Passing
that reference (or the address) would require changing the 'main'
function, which would mean not satisfying the requirement not to change
the 'main' function.

There is one way to make the _program_ to output 5 (instead of 10) but
the output will not be generated by the call to 'printf' in 'main'. You
could write

void change()
{
printf("5");
exit(0);
}

which basically causes the program to quit before even getting of the
'change' function. However, that goes against the "output of printf
in main function should give 5" because the output is from a different
call to printf.

Using non-standard means you can tweak the stack to make the 'change'
function return to 'main' in the place _after_ the "i=10" statement,
but that wouldn't be done in C++.

V

[John Carson]

"sojin" <ju*****@gmail.com> wrote in message
news:11**********************@g49g2000cwa.googlegr oups.com

#include<stdio.h>
void change()
{
/* Write something in this function so that the output of printf in
main
function should give 5 . Do not change the main function */
}

void main()
{
int i=5;
change();
i=10;
printf("%d",i);
}

Not sure if it answers the question, but:

void change()
{
printf("5");
exit(0);
}

--
John Carson

[Alf P. Steinbach]

* sojin:

#include<stdio.h>
void change()
{
/* Write something in this function so that the output of printf in
main
function should give 5 . Do not change the main function */
Here you don't have access to the values in the 'main' function, so the only
thing you can do, in a semi-portable way, is to redefine 'printf'.

And that's trivial, so I won't spoil it for you, but in your answer,
remember to include a comment in LARGE LETTERS to your instructor stating
that this technique is prohibited by the standard (i.e. formally invalid);
for full marks point out the relevant paragraph(s) in the standard.

Hopefully the children will not spoil it for you, but they always do...

}

void main()
{
As already remarked on, 'void main' is non-standard. Use 'int main'.
Include also this information in a LARGE LETTERS comment to your instructor.
int i=5;
change();
i=10;
printf("%d",i);
}

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
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[sojin]

Hi experts,

I have been asked this question by one of my friend... nd claiming that
he's having the answer (not funny one..)
i dont think so..any way i just want u good peoples suggetion also...

I'm sorry to post this 'c' stuff is this highly acclaimed standard c++
group...
Thnx,
soj

[Default User]

sojin wrote:
Please quote. From Google, click "show options" and use the Reply in
the expanded header.

I have been asked this question by one of my friend... nd claiming
that he's having the answer (not funny one..)
i dont think so..any way i just want u good peoples suggetion also...
As others of said, there's no standard way.
I'm sorry to post this 'c' stuff is this highly acclaimed standard c++
group...

printf() is C++ as well. The answer is the same in either language, you
can't do what you want in any sort of portable or standard manner.
PS, if this was a homework assignment, see if you can drop the class
and get into a real one some other time.


Brian

[Rolf Magnus]

sojin wrote:

Hi experts,

I have been asked this question by one of my friend... nd claiming that
he's having the answer (not funny one..) i dont think so..any way i just
want u good peoples suggetion also...
That's what you got. We all agree with you.
I'm sorry to post this 'c' stuff is this highly acclaimed standard c++
group...

In C, the answer would be the same.

[SerGioGio]

It's probably not really useful to even try to resolve such puzzles but
anyway here's one "standard" solution:

#include<stdio.h>
void change()
{
/* Write something in this function so that the output of printf in
main
function should give 5 . Do not change the main function */ #define int int dummy;
#define i 5, dummy }

void main()
{
int i=5;
change();
i=10;
printf("%d",i);
}

The printf *does* print "5" doesn't it ;)

SerGioGioGio