Programming

hello everybody,
here is my problem:
________________________________________
#include <stdio.h>

int main()
{
int i=10;
printf("value is %i",i);
return 0;
}
_________________________________________

when i compile and execute, nothing appears on screen!!
but if i replace the printf line with this one:
printf("value is %i\n",i); // i just added a \n
then i can see the line displayed on my screen.
ok, that's allright, but what if i want to display something _without_ the \n ?

someone told me it was a classic buffer problem and adviced me to use fflush()
but when i write:
printf("value is %i",i); fflush(stdout);
nothing happens. i also tried to put fflush(stdout) before the printf, because i don't know how to use this function correcly, but nothing happened.

can someone explain me what's the matter?!!
im using Linux Mandrake 8.0
thanks

My guess is that your shell is overwriting the data with its prompt. Try this:
printf("value is %d", i);
fflush(stdout);
sleep(10);

that's great, you were right!

nice guess. now i finally understand what's happening, thank you!


anyway im disapointed with this bash shell...
is there a way to set the bash so that it doesn't overwrite the data?