please check sleep code.

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

int main(int agrc, char* argv[])
{

printf("HELLO");
for( ; ; )
{
printf("Thank you");
sleep(20);
}
return 0;
}
[335937@oracleclient ~]$ gcc slep.c
[335937@oracleclient ~]$ ./a.out

No error but the output is not displayed.

... its infinite loop ... its sleepin infinitely......

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

int main(int agrc, char* argv[])
{
int i;
printf("HELLO");
for(i=0 ;i<5 ;i++ )
{
printf("Thank you");
sleep(1);
}
return 0;
}


this ll work.......

@anand.

for( ; ; ) means only the loop will run infinitely and not sleep.

we have sleep command in infinite loop ... i mean it .....

Code:


#include<fcntl.h>
#include<sys/types.h>
#include<unistd.h>
#include<sys/stat.h>



int main(int argc, char* argv[])
{
        char buff[512];
        int n;

        n = read(0, buff, sizeof buff);
        for(;;)
        {
                write(1,buff,n);
                sleep(10);


        }
return 0;
}


This works fine with infinite loop..

yeah it works fine..

ya the last code is working ..

it gives more priority to signals.

@ dinesh
in ur example, read,write,sleep are interrupts. so it gives equal priority to all the calls. but printf is not a interrupt. tatsy sleep is executed first.

Christopher wrote:it gives more priority to signals.

@ dinesh
in ur example, read,write,sleep are interrupts. so it gives equal priority to all the calls. but printf is not a interrupt. tatsy sleep is executed first.

No, thats not correct.

Navitha's program works fine if u just include "\n" at the end of the printf statements.
Equal priority to printf and sleep.

Code:

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

int main(int agrc, char* argv[])
{

printf("HELLO\n");
for( ; ; )
{
printf("Thank you\n");
sleep(20);
}
return 0;
}

its k.. but how??? why??

Dint know \n has this much of significance :-P

as far as i hope!
there are 2 modes!
character by character updation and line by line updation!

first one does not use buffer and later uses buffer

buffer is displayed once it encounters a newline character or at the end of program .

so in the code of navitha the output produced will be HELLOThankyouHELLOThankyou..... and so on!
so the output is not displayed as the program does not end!

but displays HELLO & ThankYou if u add \n at the end as it encounters newline char

stdio is buffered dats why its not displaying nything. if it encounters ny escape character it display all d buffered content.

if u don want to print a new line, add "fflush(stdout)" after printf stmt.

Code:

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

int main(int agrc, char* argv[])
{

printf("HELLO");
for( ; ; )
{
sleep(6);
printf("Thank you\n");
sleep(20);
printf("second line");
}
return 0;
}

hey ppl
use this code!

u will see that it will not print HELLO immediately but will wait for 6 sec then instantly will display HELLOThankYou
bcoz newline char is present in printf("Thankyou\n"); command

similerly the second lineThankyou in loop

@urveshi

But is it the case for a code without sleep()?

Code with sleep :

int main()
{
printf("hello");
printf("jksad \n");
printf("hsagdh");
sleep(20);
}


The output for the code above would be
hellojksad ---->as it encountered a \n character.

And it didn't buffer out the contents of third printf().


Consider this code now without sleep and /n :
int main()
{
printf("dsf");
printf("sad");
}

If stdio doesn't buffer out until it recognizes an escape sequence.,this program shouldn't display anything but we know wat it would.

So there is something else with sleep() and \n we need to crack it out.

The "fflush(stdout)" command is not working.
Again no display.

Sorry,I was wrong.

Printf buffers out wen it encounters a \n character. If not,it buffers out when the program terminates.

The "fflush(stdout)" command is not working.
Again no display.

include appropriate header files.

ARVINTH B.J wrote:@urveshi

But is it the case for a code without sleep()?

Code with sleep :

int main()
{
printf("hello");
printf("jksad \n");
printf("hsagdh");
sleep(20);
}


The output for the code above would be
hellojksad ---->as it encountered a \n character.

And it didn't buffer out the contents of third printf().

look at this:
int main()
{
printf("hello");
printf("jksad \n");
printf("hsagdh");
sleep(5);
printf("hi back again");
}

[335802@oracleclient ~]$ ./a.out
hellojksad .....(after delay of 5sec it prints the second line)
hsagdhhi back again