I create a linked list dynamically and initialize the first node in main(), and I add to the list every time I spawn a worker process. Before the worker process exits, I print the list. Also, I print the list inside my sigchld signal handler.
in main():
head = NULL;
tail = NULL;
// linked list to keep track of worker process
dll_node_t *node;
node = (dll_node_t *) malloc(sizeof(dll_node_t)); // initialize list, allocate memory
append_node(node);
node->pid = mainPID; // the first node is the MAIN process
node->type = MAIN;
in a fork()’d process:
// add to list
dll_node_t *node;
node = (dll_node_t *) malloc(sizeof(dll_node_t));
append_node(node);
node->pid = mmapFileWorkerStats->childPID;
node->workerFileName = mmapFileWorkerStats->workerFileName;
node->type = WORK;
functions:
void append_node(dll_node_t *nodeToAppend) {
/*
* append param node to end of list
*/
// if the list is empty
if (head == NULL) {
// create the first/head node
head = nodeToAppend;
nodeToAppend->prev = NULL;
} else {
tail->next = nodeToAppend;
nodeToAppend->prev = tail;
}
// fix the tail to point to the new node
tail = nodeToAppend;
nodeToAppend->next = NULL;
}
finally… the signal handler:
void chld_signalHandler() {
dll_node_t *temp1 = head;
while (temp1 != NULL) {
printf("2. node's pid: %d\n", temp1->pid);
temp1 = temp1->next;
}
int termChildPID = waitpid(-1, NULL, WNOHANG);
dll_node_t *temp = head;
while (temp != NULL) {
if (temp->pid == termChildPID) {
printf("found process: %d\n", temp->pid);
}
temp = temp->next;
}
return;
}
Is it true that upon the worker process exiting, the SIGCHLD signal handler is triggered? If so, that would mean that after I print the tree before exiting, the next thing I do is in the signal handler which is print the tree… which would mean i would print the tree twice?
But the tree isn’t the same. The node I add in the worker process doesn’t exist when I print in the signal handler or at the very end of main(). Any idea why?
Thanks,
Hristo
Presumably, you are calling
fork()to spawn the worker process, and adding to the linked list in the child process.After you’ve called
fork(), there are now two independent copies of your linked list – one belongs to the parent, and one belongs to the child. If the child adds a node, it’s adding it to its own linked list – the modification won’t be seen by the parent.You need to have the parent process add the node to its linked list.