Hey there, I got homework (related to C), which asks to read different .txt files with different processes.
At the end of each child process, we must print .txt file's characters number
I tried this approach:
#include <stdio.h>
#include <stdlib.h>
void main(int argc, char* argv[])
{
int i, j, FILE *fp;
char c;
for(j=1;j<argc;j++)
{
if (fork() == 0) // child process
{
if ((fp = fopen(argv[j],"r")) == NULL)
{
printf("Error when trying to open file");
exit(-1);
}
i = 0;
while((c=fgetc(fp)) != EOF) // tried while(!eof(fp)) but it didn't work
{ i++; }
printf("The characters number from %d text file is %d",j,i);
fclose(fp);
// even tried kill(getpid()) after closing the file, but it also didn't work
}
}
exit(0);
}
As predicted, if we run the program using
2 files as parameters, we get
4 prints..
The thing is, child processes aren't stopping, which is why I tried to kill each one of them after closing the file, but it didn't work either.
What am I missing here? I know exactly why it is wrong, but I don't know how to make each process read only 1 file.
Thanks,
Apidcloud
Answer:#include <stdio.h>
#include <stdlib.h>
void main(int argc, char* argv[])
{
int i, j, FILE *fp;
char c;
for(j=1;j<argc;j++)
{
if (fork() == 0) // child process
{
if ((fp = fopen(argv[j],"r")) == NULL)
{
printf("Error when trying to open file");
exit(-1);
}
fseek(fp, 0, SEEK_END); // this and the line below have better performance, and are also cleaner than a loop
i = ftell(fp);
printf("The characters number from %d text file is %d",j,i);
fclose(fp);
exit(0); // The reason why it wasn't working normally
}
}
exit(0);
}
That's weird. fork() should spawn a child process which automatically exits when you exit the parent process.
Are you saying that theoretically the code should work?
when i first thought about it, i wondered if, in fact, the second iteration of the loop would be run by first child process..when running the program, i got 4 prints instead of 2, which means something is wrong :wacko:
Yeah, that's definitely unusual. There's also a thing which you can use to make the code cleaner/faster.
// after opening the file
fseek(fp, 0, SEEK_END);
i = ftell(fp);
// now printing stuff
What do your 4 prints say?
Thanks for pointing a cleaner/faster way to count files characters.
I keep getting 4 prints:
The characters number from 2 text file is 18
The characters number from 1 text file is 48
The characters number from 2 text file is 18
The characters number from 1 text file is 48
Print out the PID's as well. Let's see what happens.
prints:
pid<21654> Number of characters from file 2 -> 18
pid<21653> Number of characters from file 1 -> 48
pid<21653> Number of characters from file 1 -> 48
pid<21655> Number of characters from file 2 -> 18
I don't understand why it prints 4 times. I would understand 3 prints, but not 4..
Edit: I think I need to kill each child process using exit or kill
Likely. Your prints indicate that child process 1 might be spawning an own child process. Try this:
#include <stdio.h>
#include <stdlib.h>
void main(int argc, char* argv[])
{
int i, j, FILE *fp;
char c;
for(j=1;j<argc;j++)
{
if (fork() == 0) // child process
{
if ((fp = fopen(argv[j],"r")) == NULL)
{
printf("Error when trying to open file");
exit(-1);
}
fseek(fp, 0, SEEK_END);
i = ftell(fp);
printf("The characters number from %d text file is %d",j,i);
fclose(fp);
exit(0);
}
}
exit(0);
}
Yes it works perfectly, thanks
The thing is, even if the first child process creates a child process on its own, the output would be 3 prints instead of 4 right(theoretically) ?
I can't think of a reason to have 4 prints in my output...
Thanks again :)
Actually it may make sense after all. The first child process not only spawns another child, but also executes its own code twice (first run and after spawning its own child). Not that it makes much sense, but it would explain why the prints of the first file would be attributed to the same PID.
The part of executing the code twice doesn't make much sense since it would normally increment the iteration and "would" normally exit after creating its own child but it doesn't seem that way ehehe
Thanks for helping, Blizzard :shy:
(http://4.bp.blogspot.com/-EMRr2coYbuY/TttAFNdNLuI/AAAAAAAAAqA/ushPxLzprW4/s1600/Neil%252BPatrick%252BHarris%252B2008brocodekopiekopie.jpg)
Edit:
Should I edit first post and add the answer?
Grateful :)
Quote from: Apidcloud on November 15, 2012, 02:44:34 pm
The part of executing the code twice doesn't make much sense since it would normally increment the iteration and "would" normally exit after creating its own child but it doesn't seem that way ehehe
I think it fell through in the second iteration of fork() == 0 or something like that.
Quote from: Apidcloud on November 15, 2012, 02:44:34 pm
Should I edit first post and add the answer?
Sure, why not.
*Adds the answer to the main post*
I tried to use spoiler to contain the code, but it doesn't work well :P