lets label your forks :-
main()
{
int ret;
ret=fork(); ------ 1
ret=fork(); ------ 2
ret=fork(); ------- 3
ret=fork(); ------- 4
if(!ret)
printf("one");
else
printf("two");
}
Now
original main() is suppose named M
then after encountering fork() 1st then
M
/ \
/ \
/ \
M C1
now after fork() -2
M
/ \
/ \
/ \
M C1
/
\ / \
M C2 C1 C3
after fork()- 4
it will be
M
C1 C2 C3 ...... C15
now we have half of them include main() have ret!=0 and
rest of them ret=0
i hope its clear now...
On Mon, Aug 8, 2011 at 12:53 PM, Shachindra A C <[email protected]>wrote:
> At the point of execution of the 4th fork(), there are 8 processes i.e, the
> 4th fork will get executed 8 times. The final value of ret will depend on
> this fork. the fork will return 0 in the 8 child processes created and
> returns pid of the child in the parent processes.
>
>
> On Mon, Aug 8, 2011 at 12:49 PM, Kamakshii Aggarwal <[email protected]
> > wrote:
>
>> then please elaborate?
>>
>>
>> On Mon, Aug 8, 2011 at 12:34 PM, Pradex <[email protected]> wrote:
>>
>>> get it..!! :) :)
>>>
>>> On Aug 7, 10:49 pm, Shachindra A C <[email protected]> wrote:
>>> > 8 one's and 8 two's. The order in which they get printed might vary.
>>> >
>>> > On Mon, Aug 8, 2011 at 11:11 AM, Kamakshii Aggarwal
>>> > <[email protected]>wrote:
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> >
>>> > > what will be the o/p of the following program:
>>> >
>>> > > main()
>>> > > {
>>> > > int ret;
>>> > > ret=fork();
>>> > > ret=fork();
>>> > > ret=fork();
>>> > > ret=fork();
>>> >
>>> > > if(!ret)
>>> > > printf("one");
>>> > > else
>>> > > printf("two");
>>> > > }
>>> >
>>> > > --
>>> > > Regards,
>>> > > Kamakshi
>>> > > [email protected]
>>> >
>>> > > --
>>> > > You received this message because you are subscribed to the Google
>>> Groups
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>>> >
>>> > --
>>> > Regards,
>>> > Shachindra A C
>>>
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>>
>>
>> --
>> Regards,
>> Kamakshi
>> [email protected]
>>
>> --
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>
>
>
> --
> Regards,
> Shachindra A C
>
> --
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--
**Regards
SAGAR PAREEK
COMPUTER SCIENCE AND ENGINEERING
NIT ALLAHABAD
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