Pipe communication in C problems - c

Question

I'm trying to write some code which uses pipes to communicate between a parent process and it's children. However, my pipe seems to give up after the first time I use it (that is, it stops working after the first use of the pipe). I'm not really sure how to fix this problem, and any help will be greatly appreciated. I also know that some of the coding practice I am using in this are not really ideal (mainly use of sleep).
const int READ = 0;
const int WRITE = 1;
char* COOP = "Criminal cooperates\n";
char* SIL = "Criminal doesn't talk\n";
char* reader(int);
void writer(int, char *c);
int main()
{
int c1pipe1[2];
int c1pipe2[2];
int c2pipe1[2];
int c2pipe2[2];
int c1sentence = 0;
int c2sentence = 0;
int r;
int c;
pipe(c1pipe1);
pipe(c1pipe2);
pipe(c2pipe1);
pipe(c2pipe2);
int C2;
int C1 = fork();
if(C1 > 0)
C2 = fork();
if(C1 < 0 || C2 < 0) //error
{
perror("fork() failed");
exit(1);
}
else if(C1 == 0)
{
close(c1pipe1[WRITE]);
close(c1pipe2[READ]);
for(c = 0; c < 10; c++)
{
r = rand();
//printf("C1 rand = %d\n", r%2);
if(r % 2 == 1)
writer(c1pipe2[WRITE], "1");
else
writer(c1pipe2[WRITE], "0");
sleep(1);
}
exit(0);
}
else if(C2 == 0)
{
close(c2pipe1[WRITE]);
close(c2pipe2[READ]);
for(c = 0; c < 10; c++)
{
r = rand();
//printf("C2 rand = %d\n", r%2);
if(r % 2 == 1)
writer(c2pipe2[WRITE], "1");
else
writer(c2pipe2[WRITE], "0");
sleep(1);
}
exit(0);
}
else //parent
{
int buff1; //stores choice of c1
int buff2; //stores choice of c2
close(c1pipe1[READ]);
close(c1pipe2[WRITE]);
close(c2pipe1[READ]);
close(c2pipe2[WRITE]);
for(c = 0; c< 10; c++)
{
buff1 = atoi(reader(c1pipe2[READ]));
buff2 = atoi(reader(c2pipe2[READ]));
printf("C1's \(%d)\ choice trial %d : %d\n", C1, c+1, buff1);
printf("C2's \(%d)\ choice trial %d : %d\n", C2, c+1, buff2);
if(buff1 && buff2) //c1 and c2 cooperate with police
{
c1sentence = c1sentence + 6;
c2sentence = c2sentence + 6;
}
else if(buff1 || buff2) // one cooperates, one is silent
{
if(buff1) // if c1 cooperates and c2 is silent
{
c1sentence = c1sentence + 0;
c2sentence = c2sentence + 10;
}
else // if c2 cooperates and c1 is silent
{
c1sentence = c1sentence + 10;
c2sentence = c2sentence + 0;
}
}
else if(!(buff1 && buff2)) //both c1 and c2 are silent
{
c1sentence = c1sentence + 1;
c2sentence = c2sentence + 1;
}
sleep(1);
}
printf("C1 is in jail for %d years total\n", c1sentence);
printf("C2 is in jail for %d years total\n", c2sentence);
exit(0);
}
exit(0);
}
void writer(int pipe_write_fd, char *c)
{
open(pipe_write_fd);
char* choice = c;
// Write to the pipe
write(pipe_write_fd, choice, strlen(choice));
// Close the pipe
// (Sends 'end of file' to reader)
close(pipe_write_fd);
}
char* reader(int pipe_read_fd)
{
open(pipe_read_fd);
// Allocate buffer to store
// result of read
int buffer_size = 1024;
char buffer[buffer_size];
// Keep reading until we exhaust
// buffer or reach end of file
int i = 0;
while (i < buffer_size
&& read(pipe_read_fd, &buffer[i], 1) > 0)
{ i++; }
if (i < buffer_size) {
// Add null termination
buffer[i] = '\0';
} else {
// We exhausted buffer
fprintf(stderr, "Warning: buffer full.\n");
buffer[buffer_size-1] = '\0';
}
//printf("%s", buffer);
// Close the pipe
close(pipe_read_fd);
return buffer;
}

Answer 1

You need to close more of the pipes. The child processes must close every pipe file descriptor that they are not using. You have 8 pipe file descriptors; each child process has to close 6 of those - at least! You would be very well advised not to create all the pipes up front as you have done - it is complicated to control things and get all the right descriptors closed.
Looking at the code more closely, the parent does not write messages to the child processes, so you have twice as many pipes as you need - you only need one pipe for each child process to write back to the parent with.
You also do not open() already open file descriptors to the pipes...but how did you get the code to compile? You must be missing the correct header (#include <fcntl.h>) for open() and compiling without enough warning options enabled.
Your variables COOP and SIL are unused in the code presented.
Your writer() function not only mistakenly tries to open an already closed file descriptor, it also closes it, which means that there is no way to send back the extra messages after the first. You should only close the file descriptor once finished - after the loop in the main program for each child. This is why you only see one message.
It is also worth getting into the habit of error-checking the return from every system call that can fail. There are a few that can't fail - getpid() is one such. But I/O operations are notorious for failing for reasons outside the direct control of the program (or, in this case, within the control of the program), so you should check that writes succeed. When you get back an EBADF - bad file descriptor - error, you know something is up.
You have similar problems with close() (and open()) in reader(), plus the additional problem that you attempt to return a pointer to a local automatic variable - which is not a good idea, ever. Again, a decent compiler (like GCC) with warnings enabled will tell you about such things. I used this command to compile your program:
gcc -O -std=c99 -Wall -Wextra -Wmissing-prototypes -Wstrict-prototypes \
pipe.c -o pipe
Your child processes are always going to generate the same sequence of (pseudo-)random numbers, which isn't very exciting. You should probably use something like:
srand(getpid());
to ensure they get different sequences.
Your reader() function is both not enthusiastic enough and too enthusiastic about reading the data. You read a single byte at a time, but you then loop to accumulate single bytes, so the code waits around for all 10 results to be known, and then spits everything out at once. Since a 32-bit integer can store a number up to 1,111,111,111 without problem, you would get just one number back from your call to atoi() on the first iteration, which isn't quite what you wanted.
Reads and writes on pipes are atomic - in the sense that if the writing process writes 6 bytes and the reading process attempts to read more than 6 bytes, then the packet of 6 bytes will be returned by a single read, even if there are other bytes in the pipe waiting to be read; those extra bytes will be returned on subsequent calls to read().
So, your reader() function should be passed in a buffer to use, along with its size; the code should attempt to read that buffer size; it should null terminate what it does receive; it can return the pointer to the buffer it was passed; it should error check the returned value from read().
The code for the two child processes is essentially the same - you should use an appropriately parameterized function rather than writing out the code twice.
Putting it all together, you end up with something like this (which works fine for me on MacOS X 10.6.6 with GCC 4.5.2):
#include <errno.h>
#include <string.h>
#include <fcntl.h>
#include <stdio.h>
#include <unistd.h>
#include <stdarg.h>
#include <stdlib.h>
const int READ = 0;
const int WRITE = 1;
static char* reader(int fd, char *buffer, size_t bufsiz);
static void writer(int fd, const char *c);
static void child_process(int *my_pipe, int *his_pipe);
static void err_exit(const char *fmt, ...)
{
va_list args;
int errnum = errno;
va_start(args, fmt);
vfprintf(stderr, fmt, args);
va_end(args);
if (errnum != 0)
fprintf(stderr, "%d: %s\n", errnum, strerror(errnum));
exit(1);
}
int main(void)
{
int c1pipe[2];
int c2pipe[2];
int c1sentence = 0;
int c2sentence = 0;
int c;
if (pipe(c1pipe) != 0 || pipe(c2pipe) != 0)
err_exit("Failed to open a pipe\n");
int C2 = 0;
int C1 = fork();
if (C1 > 0)
C2 = fork();
if (C1 < 0 || C2 < 0) //error
err_exit("fork() failed\n");
else if (C1 == 0)
child_process(c1pipe, c2pipe);
else if (C2 == 0)
child_process(c2pipe, c1pipe);
else //parent
{
int choice1; //stores choice of c1
int choice2; //stores choice of c2
char buffer1[BUFSIZ];
char buffer2[BUFSIZ];
close(c1pipe[WRITE]);
close(c2pipe[WRITE]);
for (c = 0; c< 10; c++)
{
choice1 = atoi(reader(c1pipe[READ], buffer1, sizeof(buffer1)));
choice2 = atoi(reader(c2pipe[READ], buffer2, sizeof(buffer1)));
printf("C1's (%d) choice trial %d : %d\n", C1, c+1, choice1);
printf("C2's (%d) choice trial %d : %d\n", C2, c+1, choice2);
if (choice1 && choice2) //c1 and c2 cooperate with police
{
c1sentence = c1sentence + 6;
c2sentence = c2sentence + 6;
}
else if (!(choice1 && choice2)) //both c1 and c2 are silent
{
c1sentence = c1sentence + 1;
c2sentence = c2sentence + 1;
}
else if (choice1) // if c1 cooperates and c2 is silent
{
c1sentence = c1sentence + 0;
c2sentence = c2sentence + 10;
}
else // if c2 cooperates and c1 is silent
{
c1sentence = c1sentence + 10;
c2sentence = c2sentence + 0;
}
}
printf("C1 is in jail for %d years total\n", c1sentence);
printf("C2 is in jail for %d years total\n", c2sentence);
}
return(0);
}
static void writer(int pipe_write_fd, const char *c)
{
int len = strlen(c);
if (write(pipe_write_fd, c, len) != len)
err_exit("Write failed\n");
}
static char* reader(int pipe_read_fd, char *buffer, size_t bufsiz)
{
int i = read(pipe_read_fd, buffer, bufsiz-1);
if (i < 0)
err_exit("Read failed\n");
buffer[i] = '\0';
return buffer;
}
static void child_process(int *my_pipe, int *his_pipe)
{
int c;
srand(getpid());
close(my_pipe[READ]);
close(his_pipe[READ]);
close(his_pipe[WRITE]);
for (c = 0; c < 10; c++)
{
writer(my_pipe[WRITE], ((rand() % 2) == 1) ? "1" : "0");
sleep(1);
}
close(my_pipe[WRITE]);
}
Note how the error routine captures errno early - to avoid damaging it. It is one of the perils of using global variables; they may change when you call a function. Don't use them when you can avoid them (but note that you can't avoid using errno completely, in general).

Answer 2

void writer(int pipe_write_fd, char *c)
{
open(pipe_write_fd);
char* choice = c;
// Write to the pipe
write(pipe_write_fd, choice, strlen(choice));
// Close the pipe
// (Sends 'end of file' to reader)
close(pipe_write_fd);
}
I'm not sure which function open you are trying yo use but the usual one takes a filename and returns a file descriptor. In any case you are discarding the return value so I suppose that this doesn't matter.
What is clear is that you close the pipe immediately after the first write so it is "correct" that the next write will fail; the pipe has been closed.
If you fix this problem then the next problem is that reader will, one byte at a time, all of the available input - up to 1024 bytes - before closing the read pipe. As reader is called in a loop, the read attempt in the second iteration will fail.