QUAINT !!! is this a gcc bug ? pls check out this simple C program.()

cryincold · 02-09-2008, 01:31 AM

Quote:

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>

int main(int argc, char* argv[])
{
int n = 0;
if ( fork() == 0 )
{
while (n++)
printf("Child ...\n");
}

while (n++)
printf("Parent ...\n");
}

As expected, both in Child and Parent will evaluate n++ in while(n++) and get false. So, nothing will be printed ,whatever the process is Parent or Child.

How strange it is. The output is infinite:
...
Parent ...
Parent ...
Parent ...
...

I am totally puzzled ! Could anybody straighten me out ?
TKS in advance.

cryincold · 02-09-2008, 01:35 AM

and, when i use gdb to set a break point at：

while (n++)
printf("Parent ...\n");

But gdb cannot stop there !!!

Nylex · 02-09-2008, 01:36 AM

You're incrementing n every time in the loop, so of course it will run infinitely (since every non-zero value means "true").

cryincold · 02-09-2008, 01:59 AM

Hi, Nylex.

Im sure u made a small mistake.
That's n++, not ++n, and n is initialized as 0. so, the first time the program evaluate n++, it get 0(FALSE).

I change the code slightly as below, then the output is nothing. There is no substantially difference between the 2 versions.

int main(int argc, char* argv[])
{
if ( fork() == 0 )
{

int n = 0;
while (n++)
printf("Child ...\n");
}

int n = 0;
while (n++)
printf("Parent ...\n");
exit(0);
}

jeetu · 02-09-2008, 02:00 AM

hi, myself jitender. what i think that's wrong in ur program is that the statement is being executed infinte times
while(n++)
printf(parent...);
see the compiler is not executing the statement above it
if(fork()==0)
{
while(n++)
printf(child...);
}
bcoz it's not checkin the condition fork()==0;
nd the statement below it is executing infinte times bcoz the condition u have mentioned
while(n++)
is true every time the compiler checks it
so mention the limits of n nd the try it out it will definitely goona work.
nd plz do reply back that whether u the program runs or not.

cryincold · 02-09-2008, 02:16 AM

Hi, jeetu.
tku for your reply ...
Im sorry for my English. I can't follow your mean .
Im puzzled, after i change the position of the sentence "int n = 0;" to the new position as showed at #4.
Then output is nothing.
...
frustrated ...

cryincold · 02-09-2008, 02:32 AM

Tku Tku ....

I got it !!!!

IBall · 02-09-2008, 02:38 AM

jeetu: Can you try posting in proper sentences with correct spelling and punctuation please? It would make your post easier to read and therefore easier to understand.

cryincold: You can't assume that a C compiler will always initialise variables!
It is always best to explicitly initialise variables before using them, as it ensures that you will get the expected results.

Out of curiousity, what is the point of the infinite loops? Why not

Code:

if(n++)

? In fact, why not:

Code:

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <unistd.h>

int main(int argc, char* argv[])
{
int n = 0;

if ( fork() == 0 )
   {
      printf("Child ...\n");
   }
else
   printf("Parent ...\n");
}

In this code, the child process will print "Child ..." and the parent will print "Parent ...". Surely this would be what you want?

Having said that, I always thought that the child got an exact copy of all local variables that the parent has when it is created. So the child process should get n=0 and the parent should have n=0....

--Ian

cryincold · 02-09-2008, 02:38 AM

Feel sooooooooooo embarrassed ...

sooooooooo awkward ...

I ever thought child process will not achieve the next while loop.

cryincold · 02-09-2008, 02:43 AM

Tku iBall.
it's so kind of u to help me. ^_^
I made a mistake.
Newbie to Unix programming. TKU.