segmentation fault in recursion

[asit]

#include <stdio.h>

int main()
{
int i;
for(i=1;i<=10;i ++)
main();
printf("C is urs..");
return 0;
}

Why it shows error segmentation fault ??

[Willem]

Keith wrote:
) CBFalconer <cb********@yah oo.comwrites:
)However a smart enough compiler could optimize it to:
)>
)#include <stdio.h>
)int main(void) {
) while (1) printf("C is urs..");
) return 0;
)}
)>
)(ignoring the failure to output any '\n's). The result should
)progress considerably faster.
)
) No, that would be an invalid optimization. In the original program,
) even assuming infinite resources, the printf call will never be
) reached.
)
) It could be optimized to:
)
) int main(void) {
) while (1);
) }
)
) The only difference in behavior would be the lack of memory exhaustion
) (a "stack overflow" in some implementations ).

I agree. A smart compiler could prove that the main() call never
returns, and optimize accordingly (it would first throw away all the
unreachable code after the first main-call, and then optimize away
the tail recursion).
SaSW, Willem
--
Disclaimer: I am in no way responsible for any of the statements
made in the above text. For all I know I might be
drugged or something..
No I'm not paranoid. You all think I'm paranoid, don't you !
#EOT

[user923005]

On Jan 15, 1:49*pm, Eric Sosman <Eric.Sos...@su n.comwrote:

vipps...@gmail. com wrote:

On Jan 15, 10:09 pm, asit <lipu...@gmail. comwrote:

#include <stdio.h>

int main()
{
* * * * int i;
* * * * for(i=1;i<=10;i ++)
* * * * * * * * main();
* * * * printf("C is urs..");
* * * * return 0;

}

Why it shows error segmentation fault ??

Because you're a troll.

* * *You may be right about that.

As far as C is concerned, that is perfectly valid code.
perhaps a system limitation. C (or c.l.c) has nothing to do with that.

* * *The code is "valid" in that it expresses a computation.
Unfortunately, the computation will not complete in finite
time, nor will a finite machine have an easy time keeping
track of its intermediate states. *The inconvenient word
"finite" does in fact express a system limitation, but it's
a limitation that can be removed only with great difficulty
and at some expense.

I see no condition that will lead to termination.
This program will not terminate on an infinite machine with infinite
memory and an infinite number of CPUs.
It exhausts resources exponentially.

[Army1987]

vippstar wrote:

On Jan 15, 10:09 pm, asit <lipu...@gmail. comwrote:

>#include <stdio.h>

int main()
{
int i;
for(i=1;i<=10;i ++)
main();
printf("C is urs..");
return 0;

}

Why it shows error segmentation fault ??

Because you're a troll.
As far as C is concerned, that is perfectly valid code.
perhaps a system limitation. C (or c.l.c) has nothing to do with that.

I remembered that there was a limit on the number of nested function calls
an implementation was required to support, but re-reading 5.2.4.1 shows
that it was a false memory. (It does say "127 nesting levels of blocks",
but I'm not sure whether that applies here.)
--
Army1987 (Replace "NOSPAM" with "email")

[CBFalconer]

Keith Thompson wrote:

CBFalconer <cb********@yah oo.comwrites:
[...]

>asit <lipu...@gmail. comwrote:

>>#include <stdio.h>
int main() {
int i;
for(i=1;i<=10;i ++)
main();
printf("C is urs..");
return 0;
}

[...]

>However a smart enough compiler could optimize it to:

#include <stdio.h>
int main(void) {
while (1) printf("C is urs..");
return 0;
}

(ignoring the failure to output any '\n's). The result should
progress considerably faster.

No, that would be an invalid optimization. In the original
program, even assuming infinite resources, the printf call will
never be reached.

It could be optimized to:

int main(void) {
while (1);
}

I looked again, and you are right. That should also be a faster
optimization, in terms of loops per unit time.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

--
Posted via a free Usenet account from http://www.teranews.com

[JosephSlomka]

On Jan 15, 12:09 pm, asit <lipu...@gmail. comwrote:

#include <stdio.h>

int main()
{
int i;
for(i=1;i<=10;i ++)
main();
printf("C is urs..");
return 0;

}

Why it shows error segmentation fault ??

I'll bite.

Like others were saying here the program you have written has infinite
recursion. Each time you call the main function it spawns ten more
copies of the main(), thus you will eventually consume all of your
machines memory or process. While technically you should not get a seg
fault it's actually a 'graceful' solution to this block of code.

When you write recursive algorithms you need to include a way for the
recursion to end.
The following code is one way to limit recursion.

void func(int limit)
{
int i;
for(i=1;i<=limi t;i++)
func(limit-1);

}

int main()
{
int i;
for(i=1;i<=10;i ++)
func(10);
printf("C is urs..");
return 0;
}

limit -1 is the key to end the recursion. The first instance of func()
call func() 10 times, the each of these call func()9 times all the way
down to 0 times. This example shows how to do recursion but, doesn't
do anything with it.

Hope this helps.