Reputation: 776
Pipes between child processes
I wrote a C program that is supposed to create a certain number of child processes, each child process having to change 1 letter from a string. The string and the number of child processes are read from the keyboard.
I want to do it using pipes. It should work like this: The parent changes one letter, then the first child takes the string modified by the parent and changes one more letter. The second child takes the string modified by the first one (2 letters are already changed) and changes one more and so on. I am new to C and am not quite sure how it all works, especially pipes.
Also can the children be linked between them through the pipe, or can they only be linked to the parent and it has to be something like: first child changes a letter, gives the string back to the parent and then the second child reads from there, modifies letter and gives back. If it's like that, is there any way to make sure that this doesn't happen: Apples becomes AppleD and then AppleX and then AppleQ?
For example:
input:
3 Apples
output:
Applex Appldx Apqldx
My problem is: I don't get any output from the children. Unsure what I'm doing wrong. Help would be much appreciated, thanks in advance!
Here's my code:
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<sys/types.h>
#include<unistd.h>
#include<sys/wait.h>
void error(char* msg)
{
fprintf(stderr, "%s\n", msg);
exit(1);
}
char* modify(char msg[])
{
srand(time(NULL));
int pos1=rand()%((int)strlen(msg));
srand(time(NULL));
int pos2=rand()%26;
srand(time(NULL));
int big=rand()%2;
if(big==1)
{
msg[pos1]=(char)(((int)'A')+pos2);
}
else
{
msg[pos1]=(char)(((int)'a')+pos2);
}
return msg;
}
int main(int argc, char *argv[])
{
if(argc!=3)
{
error("Wrong number of arguments\n");
}
int nrch;
nrch=atoi(argv[1]);
char* msg=argv[2];
printf("Parent: erhalten: %s\n", msg);
int i=0;
msg=modify(argv[2]);
printf("Parent: weiter: %s\n", msg);
pid_t pids[10];
int fd[2];
if(pipe(fd) == -1)
{
error("Can't create the pipe");
}
dup2(fd[1], 1);
close(fd[0]);
fprintf(stdout, msg);
/* Start children. */
for (i = 0; i < nrch; ++i)
{
if ((pids[i] = fork()) < 0)
{
error("Can't fork process");
}
else if (pids[i] == 0)
{
dup2(fd[0], 0);
close(fd[1]);
fgets(msg,255,stdin);
printf("child%d: erhalten: %s\n", (i+1), msg);
modify(msg);
printf("child%d: weiter: %s\n", (i+1), msg);
if (pipe(fd) == -1)
{
error("Can’t create the pipe");
}
fprintf(stdout, msg);
dup2(fd[1], 1);
close(fd[0]);
exit(0);
}
}
/* Wait for children to exit. */
int status;
pid_t pid;
while (nrch > 0)
{
pid = wait(&status);
printf("Child with PID %ld exited with status 0x%x.\n", (long)pid, status);
--nrch;
}
}
Upvotes: 3
Views: 3266
Answers (2)
Reputation: 754900
One reason you see no output from the children is that you hook their standard output to the write end of the pipe, so when they write to standard output, it goes into the pipe, not to the screen (or wherever you sent the standard output of the program to originally).
Where the children are not going to execute a program that needs standard input and standard output going to the pipe, don't use I/O redirection. Just write to and read from the correct ends of the pipe.
If you've got multiple children, you probably need a pipe per child, but the parent process will need to do the creating. Your code creates a pipe in the child; that pipe is no use because only the child knows about it. You probably can do it all with one pipe, but it becomes indeterminate which sequence the children will run in. If determinacy is important, use multiple pipe() calls, and at least twice as many close() calls.
Single pipe solution
#include <assert.h>
#include <stdarg.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <sys/wait.h>
#include <time.h>
#include <unistd.h>
static void error(const char *fmt, ...)
{
va_list args;
va_start(args, fmt);
vfprintf(stderr, fmt, args);
va_end(args);
putc('\n', stderr);
exit(1);
}
static void modify(char msg[])
{
int pos1 = rand() % ((int)strlen(msg));
int pos2 = rand() % 26;
int big = rand() % 2;
if (big == 1)
msg[pos1] = (char)(((int)'A') + pos2);
else
msg[pos1] = (char)(((int)'a') + pos2);
}
static int read_pipe(int fd, char *buffer, size_t buflen)
{
int nbytes = read(fd, buffer, buflen);
if (nbytes <= 0)
error("Unexpected EOF or error reading pipe");
assert((size_t)nbytes < buflen);
buffer[nbytes] = '\0';
return nbytes;
}
int main(int argc, char *argv[])
{
if (argc != 3)
error("Usage: %s number 'message'", argv[0]);
srand(time(NULL));
int nrch = atoi(argv[1]);
char *msg = argv[2];
size_t len = strlen(msg);
printf("Parent: erhalten: %s\n", msg);
modify(msg);
printf("Parent: weiter: %s\n", msg);
int fd[2];
if (pipe(fd) == -1)
error("Can't create the pipe");
if (write(fd[1], msg, len) != (ssize_t)len)
error("Failed to write to pipe");
/* Start children. */
for (int i = 0; i < nrch; ++i)
{
int pid;
if ((pid = fork()) < 0)
error("Can't fork process");
else if (pid == 0)
{
char buffer[255];
int nbytes = read_pipe(fd[0], buffer, sizeof(buffer));
printf("child%d: erhalten (%d): %s\n", (i + 1), nbytes, buffer);
modify(buffer);
printf("child%d: weiter (%d): %s\n", (i + 1), nbytes, buffer);
write(fd[1], buffer, nbytes);
exit(0);
}
else
printf("Random: %d\n", rand());
}
/* Wait for children to exit. */
while (nrch > 0)
{
int status;
pid_t pid = wait(&status);
printf("Child with PID %ld exited with status 0x%.4X.\n", (long)pid, status);
--nrch;
}
char buffer[255];
int nbytes = read_pipe(fd[0], buffer, sizeof(buffer));
printf("Parent: weiter (%d): %s\n", nbytes, buffer);
return 0;
}
Example output
Code in file p1.c:
$ make p1 && ./p1 4 "Absolutely nothing to do with me"
gcc -O3 -g -std=c11 -Wall -Wextra -Wmissing-prototypes -Wstrict-prototypes -Wold-style-definition -Werror p1.c -o p1
Parent: erhalten: Absolutely nothing to do with me
Parent: weiter: AbsolutEly nothing to do with me
Random: 1120753102
child1: erhalten (32): AbsolutEly nothing to do with me
Random: 918317477
child1: weiter (32): AbsolutEly notzing to do with me
child2: erhalten (32): AbsolutEly notzing to do with me
child2: weiter (32): AbsolwtEly notzing to do with me
Random: 196864950
child3: erhalten (32): AbsolwtEly notzing to do with me
child3: weiter (32): AbsolwtEly notzing to ao with me
Random: 1584398270
Child with PID 42928 exited with status 0x0000.
Child with PID 42927 exited with status 0x0000.
Child with PID 42926 exited with status 0x0000.
child4: erhalten (32): AbsolwtEly notzing to ao with me
child4: weiter (32): AbsolwtEly notzing to ao with Ue
Child with PID 42929 exited with status 0x0000.
Parent: weiter (32): AbsolwtEly notzing to ao with Ue
$
Note the stray use of rand() in the loop. It makes sure the children change different letters in the message. Without that, they all end up changing the same 'random' letter in the same 'random' position in the message.
You can create a multi-pipe solution if you wish. I got what appeared to be deterministic behaviour from the single-pipe solution, though there is no guarantee of the sequencing. If, for example, each child waited for a random delay using nanosleep() or equivalent:
struct timespec nap = { .tv_sec = 0, .tv_nsec = (rand() % 1000) * 1000000 };
nanosleep(&nap, 0);
then you get an arbitrary sequence in the child processing. For example:
Parent: erhalten: Absolutely nothing to do with me
Parent: weiter: Absolutely nothinglto do with me
Random: 2028074573
Random: 988903227
Random: 1120592056
Random: 359101002
child4: erhalten (32): Absolutely nothinglto do with me
child4: weiter (32): vbsolutely nothinglto do with me
Child with PID 43008 exited with status 0x0000.
child3: erhalten (32): vbsolutely nothinglto do with me
child3: weiter (32): vbsolutelyGnothinglto do with me
Child with PID 43007 exited with status 0x0000.
child2: erhalten (32): vbsolutelyGnothinglto do with me
child2: weiter (32): vbsolutelyGnothinglto doawith me
Child with PID 43006 exited with status 0x0000.
child1: erhalten (32): vbsolutelyGnothinglto doawith me
child1: weiter (32): vbsolutelyGnothinglto doawimh me
Child with PID 43005 exited with status 0x0000.
Parent: weiter (32): vbsolutelyGnothinglto doawimh me
Upvotes: 1
Reputation:
Tried to change your code acc to below, not sure if this is exectly what you want. Anyhow the children are running...
int main(int argc, char *argv[])
{
if (argc != 3)
{
error("Wrong number of arguments\n");
}
int nrch;
nrch = atoi(argv[1]);
char *msg = argv[2];
printf("Parent: erhalten: %s\n", msg);
int i = 0;
msg = modify(argv[2]);
printf("Parent: weiter: %s\n", msg);
pid_t pids[10];
int fd[2];
/* Start children. */
for (i = 0; i < nrch; ++i)
{
if ((pids[i] = fork()) < 0)
{
error("Can't fork process");
}
if (pipe(fd) == -1)
{
error("Can't create the pipe");
}
// printf ( " pids[i] %d , i %d \n", pids[i] , i);
if (pids[i] == 0)
{
if (dup2(fd[0], 0) == -1)
{
error("Can't dup2 (A)");
}
close(fd[1]);
fgets(msg, 255, stdin);
printf("child%d: erhalten: %s\n", (i + 1), msg);
modify(msg);
printf("child%d: weiter: %s\n", (i + 1), msg);
fprintf(stdout, msg);
}
else
{
// printf("in else i= %d \n",i);
if (dup2(fd[1], 0) == -1)
{
error("Can't dup2 (B)");
}
close(fd[0])
exit(0);
}
}
/* Wait for children to exit. */
int status;
pid_t pid;
while (nrch > 0)
{
pid = wait(&status);
if (pid > -1)
printf("Child with PID %ld exited with status 0x%x.\n", (long)pid, status);
--nrch;
}
return 0;
}
Upvotes: 0
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