Having trouble with fork(), pipe(), dup2() and exec() in C

Question

Here\'s my code:

#include 
#include 
#include 
#include 
#include 

#         


        

Answer 1

Even after the first command of your pipeline exits (and thust closes stdout=~fdPipe[1]), the parent still has fdPipe[1] open.

Thus, the second command of the pipeline has a stdin=~fdPipe[0] that never gets an EOF, because the other endpoint of the pipe is still open.

You need to create a new pipe(fdPipe) for each |, and make sure to close both endpoints in the parent; i.e.

for cmd in cmds
    if there is a next cmd
        pipe(new_fds)
    fork
    if child
        if there is a previous cmd
            dup2(old_fds[0], 0)
            close(old_fds[0])
            close(old_fds[1])
        if there is a next cmd
            close(new_fds[0])
            dup2(new_fds[1], 1)
            close(new_fds[1])
        exec cmd || die
    else
        if there is a previous cmd
            close(old_fds[0])
            close(old_fds[1])
        if there is a next cmd
            old_fds = new_fds
if there are multiple cmds
    close(old_fds[0])
    close(old_fds[1])

Also, to be safer, you should handle the case of fdPipe and {STDIN_FILENO,STDOUT_FILENO} overlapping before performing any of the close and dup2 operations. This may happen if somebody has managed to start your shell with stdin or stdout closed, and will result in great confusion with the code here.

Edit

   fdPipe1           fdPipe3
      v                 v
cmd1  |  cmd2  |  cmd3  |  cmd4  |  cmd5
               ^                 ^
            fdPipe2           fdPipe4

In addition to making sure you close the pipe's endpoints in the parent, I was trying to make the point that fdPipe1, fdPipe2, etc. cannot be the same pipe().

/* suppose stdin and stdout have been closed...
 * for example, if your program was started with "./a.out <&- >&-" */
close(0), close(1);

/* then the result you get back from pipe() is {0, 1} or {1, 0}, since
 * fd numbers are always allocated from the lowest available */
pipe(fdPipe);

close(0);
dup2(fdPipe[0], 0);

I know you don't use close(0) in your present code, but the last paragraph is warning you to watch out for this case.

Edit

The following minimal change to your code makes it work in the specific failing case you mentioned:

@@ -12,6 +12,4 @@
     pid_t pid;

-    pipe(fdPipe);
-
     while(1) {
         bBuffer = readline("Shell> ");
@@ -29,4 +27,6 @@
         } while(aPtr);

+        pipe(fdPipe);
+
         for(i = 0; i < pCount; i++) {
                 aCount = -1;
@@ -72,4 +72,7 @@
         }

+        close(fdPipe[0]);
+        close(fdPipe[1]);
+
         for(i = 0; i < pCount; i++) {
                 waitpid(lPids[i], &status, 0);

This won't work for more than one command in the pipeline; for that, you'd need something like this: (untested, as you have to fix other things as well)

@@ -9,9 +9,7 @@
 int main(int argc, char *argv[]) {
     char *bBuffer, *sPtr, *aPtr = NULL, *pipeComms[NUMPIPES], *cmdArgs[10];
-    int fdPipe[2], pCount, aCount, i, status, lPids[NUMPIPES];
+    int fdPipe[2], fdPipe2[2], pCount, aCount, i, status, lPids[NUMPIPES];
     pid_t pid;

-    pipe(fdPipe);
-
     while(1) {
         bBuffer = readline("Shell> ");
@@ -32,4 +30,7 @@
                 aCount = -1;

+                if (i + 1 < pCount)
+                    pipe(fdPipe2);
+
                 do {
                         aPtr = strsep(&pipeComms[i], " ");
@@ -43,11 +44,12 @@

                         if(pid == 0) {
-                                if(i == 0) {
-                                        close(fdPipe[0]);
+                                if(i + 1 < pCount) {
+                                        close(fdPipe2[0]);

-                                        dup2(fdPipe[1], STDOUT_FILENO);
+                                        dup2(fdPipe2[1], STDOUT_FILENO);

-                                        close(fdPipe[1]);
-                                } else if(i == 1) {
+                                        close(fdPipe2[1]);
+                                }
+                                if(i != 0) {
                                         close(fdPipe[1]);

@@ -70,4 +72,17 @@
                         }
                 }
+
+                if (i != 0) {
+                    close(fdPipe[0]);
+                    close(fdPipe[1]);
+                }
+
+                fdPipe[0] = fdPipe2[0];
+                fdPipe[1] = fdPipe2[1];
+        }
+
+        if (pCount) {
+            close(fdPipe[0]);
+            close(fdPipe[1]);
         }

Answer 2

You should have an error exit after execvp() - it will fail sometime.

exit(EXIT_FAILURE);

As @uncleo points out, the argument list must have a null pointer to indicate the end:

cmdArgs[aCount] = 0;

It is not clear to me that you let both programs run free - it appears that you require the first program in the pipeline to finish before starting the second, which is not a recipe for success if the first program blocks because the pipe is full.

Answer 3

One potential problem is that cmdargs may have garbage at the end of it. You're supposed to terminate that array with a null pointer before passing it to execvp().

It looks like grep is accepting STDIN, though, so that might not be causing any problems (yet).

Answer 4

Jonathan has the right idea. You rely on the first process to fork all the others. Each one has to run to completion before the next one is forked.

Instead, fork the processes in a loop like you are doing, but wait for them outside the inner loop, (at the bottom of the big loop for the shell prompt).

loop //for prompt
    next prompt
    loop //to fork tasks, store the pids
        if pid == 0 run command
        else store the pid
    end loop
    loop // on pids
        wait
    end loop
end loop

Answer 5

I think your forked processes will continue executing.

Try either:

• Changing it to 'return execvp'
• Add 'exit(1);' after execvp

Answer 6

the file descriptors from the pipe are reference counted, and incremented with each fork. for every fork, you have to issue a close on both descriptors in order to reduce the reference count to zero and allow the pipe to close. I'm guessing.