My program has to create multiple children, i take the number of children from terminal. Then, I need to group them and every group has different tasks.
I created the children but the problem is that I can't give them separated tasks, I can't distinguish children. Only thing I can do is, every children works on the same part(like printing pids).
How can I separate them and give them specific tasks?
For example, first 4 children should call a function, other 3 should print something and other 3 should write to a file etc.
pid_t pid[10];
pid[0] = fork();
if(pid[0] > 0)
{
for(int i = 0; i < 9; i++)
{
if(pid[i] > 0)
{
pid[i + 1] = fork();
}
}
}
for(int i = 0; i < 10; i++)
{
if(pid[i] == 0)
{
printf("child %d, parent %d\n", getpid(), getppid());
exit(1);
}
}
I think you should take a look at how the fork() function works. Here is the man page, here a useful answer and here a useful tutorial.
When you use fork() in your code, know that the child process continue from where the parent was. So when you call fork() in the first for loop, all the child processes continue the loop that the parent has begun. I don't think this is the behavior you expect.
Anyway, here's a possible solution for your problem. In this way your processes do some stuff divided in small groups. Note the exit(0) function at the end of the work. It is important to be sure that every process does only its work and not also its parent's work:
pid_t pid[10];
for (int i = 0; i < 9; i++)
{
pid[i] = fork();
//First group
if (pid[i] == 0 && i < 4){
//Insert here the code for the first group or call a fuction
exit(0);
}
//Second group
if (pid[i] == 0 && i >=4 && i < 8){
//Insert here the code for the second group or call a fuction
exit(0);
}
//Third group
if (pid[i] == 0 && i >=8){
//Insert here the code for the third group or call a fuction
exit(0);
}
if (pid[i] < 0){
perror("Something wrong with the fork");
exit(1);
}
}
int flag = 0;
flag |= (fork() == 0) << 0;
flag |= (fork() == 0) << 1;
flag |= (fork() == 0) << 2;
printf("this process works on problem %d\n", flag);
Related
I'm trying to exec a file within another file for maximum 5 times and save results. I use a fork-exec-wait in a for loop, this is how it looks like:
for(i = 0; i< Number_of_Processes; i++){
pid = fork();
if(pid == 0)
{
execl(...);
exit(EXIT_Failure);
}
else if(pid > 0)
{
wait(&status);
result = WEXITSTATUS;
}
}
The problem is, it executes the second file for first time correctly and then reprints the first result for 5 times(or whatever the user entered)
I tried so many other ways, like break when it's pid == 0 but it didn't work. i couldn't find a proper example of how to use exec within a loop so I'm stuck for hours, trying moving functions up and down and get nothing.
Whatever i do, i can't see a second execution of the inner file.
Apparently it was a problem of timing, when i add a sleep(1) just before execl in child process, it fixed everything and execed needed processes one by one and printed true results
for(i = 0; i< Number_of_Processes; i++){
pid = fork();
if(pid == 0)
{
sleep(1);
execl(...);
exit(EXIT_Failure);
}
else if(pid > 0)
{
wait(&status);
result = WEXITSTATUS;
}
}
I'm trying to write a small program that forks processes from a single parent. Currently my code does this a few times but then the children create more child processes, which I want to eliminate.
int main() {
pid_t c;
for (int i = 0; i < 5; i++) {
c = fork();
if(c < 0) {
perror("fork");
exit(1);
}
else if( c > 0 ) {
printf("parentID = %d, childID = %d\n", getppid(i), getpid(i));
}
}
}
I'm not sure how to modify it so that fork is only forking from the parent though.
EDIT: thanks for the help, got the solution:
int main() {
pid_t c;
for (int i = 0; i < 5; i++) {
c = fork();
if(c < 0) {
perror("fork");
exit(1);
}
else if( c > 0 ) {
printf("parentID = %d, childID = %d\n", getppid(i), getpid(i));
}
else {
exit(0);
}
}
}
nothing in the posted code is recognizing the child (0 == pid)
so a child hits (and skips) the two 'if' statements.
hits the end of the loop,
branches back to the top of the loop, calls fork()....etc.
Suggest: adding
elseif( 0 == pid )
{ // then child ...
exit( EXIT_SUCCESS );
}
The child process does not enter any part of the if block, and just loops back to the top of the for loop creating more children. Also, the if (n > 0) block gets run for the parent, not the child, since fork returns 0 to the parent and the child's pid to the parent.
Change if (n > 0) to if (n == 0), and call exit() at the bottom of the block to prevent the child from continuing. Also, getpid() and getppid() don't take any arguments.
int main() {
pid_t c;
for (int i = 0; i < 5; i++) {
c = fork();
if(c < 0) {
perror("fork");
exit(1);
}
else if( c == 0 ) {
printf("parentID = %d, childID = %d\n", getppid(), getpid());
exit(0); // <-- here
}
}
}
I'm trying to use the multiple fork() calls to create several children with different task
I found a code on
Multiple child process
Which is really close for what I want , yet I couldn't fully understand it
pid_t firstChild, secondChild;
firstChild = fork();
if(firstChild != 0)
{
// In parent
secondChild = fork();
if(secondChild != 0)
{
// In parent
}
else
{
// In secondChild
}
}
else
{
// In firstChild
}
My questions are:
How many process have been created (I assume that we have 4 since it's 2 forks!)?
In this part of the code
firstChild = fork();
if(firstChild != 0)
{
// In parent
secondChild = fork();
if(secondChild != 0)
{
// In parent
}
Does "//in parent" mean both of them are the same process (they have the same PID when I tried to test it).
How can I create 3 children using 2 forks?( I can draw the tree that ends with 4 leaves 3 of them are children and 1 parent)
Thank you (please feel free to tell me if I'm not totally getting the Fork concept)
How many process have been created (I assume that we have 4 since it's 2 forks!)?
Depending on the result of your forks it should be 0 to 2. Probably 2 if nothing goes wrong. There's a parent process that forks 2 children processes.
Does "//in parent" mean both of them are the same process (they have the same PID when I tried to test it).
Yes. In your case the code is checking for a return value of fork being non zero. That's not a very good idea since it covers 2 distinct cases:
It could be less than zero indicating an error, or ...
It could be greater than zero indicating to the parent the pid of the newly spawned process
Anyway ... considering all goes well and both the forks succeed, you will end up with a parent process having 2 different children.
How can I create 3 children using 2 forks?( I can draw the tree that ends with 4 leaves 3 of them are children and 1 parent
Something like this should do the trick:
firstChild = fork();
if (firstChild < 0) {
exit(EXIT_FAILURE);
perror("fork");
}
secondChild = fork();
Notice that by not checking the return value of fork() any more I'm getting a child process continuing execution at the same place as the parent. So the next fork will actually be executed by both the parent and the children each spawning a new process. So I'll get something like this ...
parent─┬─child1───(child1's child)
└─child2
I can't think of any way you can get this with only 2 forks though:
parent─┬─child1
├─child3
└─child2
Note: It's customary on stackoverflow to only limit yourself to one question per topic.
The following code shows how to create 4 process (1 parent 3 children) with only 2 forks
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
#include <sys/types.h>
int *list;
void calculate_average(int );
void calculate_maximum(int);
void calculate_minimum(int);
void calculate_average(int count)
{
int i, total = 0;
for (i = 0; i < count; i++)
total += list[i];
double average = total / count;
printf("average is %f\n",average);
}
void calculate_maximum(int count)
{
int i;
int maximum = list[0];
for (i = 1; i < count; i++)
if (list[i] > maximum)
maximum = list[i];
printf("maximum is %d\n",maximum);
}
void calculate_minimum(int count)
{
int i;
int minimum = list[0];
for (i = 1; i < count; i++)
if (list[i] < minimum)
minimum = list[i];
printf("minimum is %d\n",minimum);
}
int main(int argc, char *argv[])
{
pid_t pid, pid1;
int num_of_args = argc-1;
int i;
/* allocate memory to hold array of integers */
list = malloc(sizeof(int)*num_of_args);
for (i = 0; i < num_of_args; i++)
list[i] = atoi(argv[i+1]);
printf("The %d number of input ingeters are\n",num_of_args);
for (i = 0; i < num_of_args; i++)
printf("%d\n",list[i]);
/* fork a child process */
pid = fork();
if (pid < 0) { /* error occurred */
fprintf(stderr, "Fork Failed\n");
return 1;
}
else if (pid == 0) { /* P2 */
pid1=getppid();
calculate_average(num_of_args);
}
else { /* P1 */
pid1=getpid();
wait(NULL);
}
pid = fork();
if (pid < 0) { /* error occurred */
fprintf(stderr, "Fork Failed\n");
return 1;
}
else if (pid == 0) { /* could be either P3 or P4 */
if (getppid() == pid1) { /* P3 */
calculate_minimum(num_of_args);
}
else { /* P4 */
calculate_maximum(num_of_args);
}
}
else {
wait(NULL);
}
return 0;
}
Note that one of the children will be a parent for a grandchild
This is a homework question. I have to write a program forking itself 20 times. Each new process is adding +1 to a variable (integer) which is shared between all of them. The thing is, I have to use semaphores (IPC). This piece of code is 'working' - giving the value of 20 in the end.
*buf = 0;
for(i=1; i<=20; ++i)
{
if(fork()!=0)
{
*buf += 1;
exit(0);
}
}
EDIT:
Based on this code I am trying to get output like :
I am child 1...
I am child 2...
.
.
.
I am child 20...
It worked once (first time), and then the order became random. But I did not change any code. What am I doing wrong?
Well your major problem is this:
if (fork()!=0) //<-- this
fork() will return -1 on error, the parent pid, OR ZERO for the child. So you are actually doing everything in the parent. Change to (fork() ==0) and it does what you want.
Also you should wait on your children and detach shared memory. (I added some output of the process ids to make it a little clearer.)
printf("I AM THE PARENT pid = %d\n", getpid());
*buf = 0;
for(i=1; i<=20; ++i)
{
if((pid = fork()) == -1)
{
perror("fork");
exit(1);
}
if (pid == 0)
{
v(semid, 0);
*buf += 1;
p(semid, 0);
printf("I am child %d with pid = %d\n", i, getpid());
shmdt(buf);
exit(0);
}
}
for (i = 1; i <= 20; ++i)
{
pid = wait(&status);
printf("child pid = %d reaped\n", pid);
}
printf("buf: %p\n", buf);
printf("*buf: %d\n", *buf);
shmdt(buf);
return 0;
I have found examples of how to fork multiple children by having something like this:
if ( fork() = 0 ) {
//In child
} else {
if ( fork() = 0 ) {
//in second child
But if I don't know how many children I am going to need, how might I do this?
For instance, if I have a linked list of commands, and I want to fork and exec for each of them... So I guess I need to know which child it is as well...
Taking you at your word that you need to do this for a linked list:
linked_list_of_commands_t *node = root;
while (node != NULL) {
int pid = fork();
if (pid == -1) {
break; // handle error
} else if (pid == 0) {
// child
execv(node->command, node->argv);
exit(1); // execv should not return, but just in case the execv call fails
} else {
node = node->next;
}
}
This will launch a separate process for every item in the list.
But the number of routines would have to be fixed, even if the execution over those branches is unbounded. So what about a while loop with some sort of switch statement logic to each routine?
How about
for (i=0; i< 1000; i++) {
pid = fork();
if (pid) {
// Error handling for pid==-1
break;
}
// Some logic dependent on value of 'i'
}
for(i = 0; i < num_children_to_spawn(); ++i) {
pid_t pid = fork();
if (pid == -1) {
exit(-1); /* error */
} else if (pid == 0) {
/* child */
do_child_things();
break;
} else {
/* parent */
}
}
Note that I didn't use a switch() because it would make it more cumbersome to break out of the loop.
pid_t children_pids[MAX_CHILDREN];
int last_child_index = 0;
for (int i=0; i < num_wanted_children; i++) {
pid_t pid = fork();
if (pid == 0)
// in child
else
children_pids[last_child_index++] = pid;
}