I'm new to system programming and curious how the exec function works. My first question is why does the child never print "I'm the child" after calling exec. How does exec replace the child process? My second question is why does the program (after calling exec) continue and ask for one more command line argument before it completely terminates. I'm not sure what's going on here. Anyone could explain what is going on would be very appreciated. Here is the code:
#include <stdio.h>
#include <unistd.h>
int main(void) {
if(fork() == 0){
printf("Hello from child!\n");
execl("/usr/bin/sort", "sort", "talk.c",NULL);
printf("I'm the child\n");
}
else{
printf("Hello from parent!\n");
printf("Iam the parent\n");
}
return 0;
}
You can read about execl in https://linux.die.net/man/3/execl
The exec() family of functions replaces the current process image with a new process image. The functions described in this manual page are front-ends for execve(2). (See the manual page for execve(2) for further details about the replacement of the current process image.)
exec family REPLACES the current process image with a new process image so nothing after execl happen.
Related
So I have this simple program that sleeps for 4 second if the value returned by fork is '0' meaning that the child process is executing, I've tried using sleep in child process but the program is blocked, and flushing standard output isn't working...
code:
#include <stdio.h>
#include <unistd.h>
int main(int argc, char const *argv[]) {
pid_t value = fork();
if (value == 0) {
sleep(4);
}
printf("Value returned by fork: %d\n", value);
printf("I'm the process N°%d\n", getpid());
return 0;
}
I'm running on Ubuntu 20.04.3 LTS.
Output:
Value returned by fork: 12618
I'm the process N°12617\
farouk#farouk-HP-Pavilion-Desktop-TP01-1xxx:~/sysexp$ Value returned by fork: 0
I'm the process N°12618
To allow this question to have an accepted answer.
The child process is not blocking the shell. The shell gave its prompt and the child wrote some output after the prompt, leaving the cursor at the start of a line without a shell prompt visible — because the shell prompt already appeared earlier.
There are a variety of ways around this.
The simplest is just to type a command such as ps and hit return, noting that the shell executes it, and that the ps output does not list the child process. If you type the ps command quick enough, you might see the child listed in the output before its output appears.
Another is to modify the program so that it waits for all child processes to exit before it exits — using wait() or waitpid(). The same code can be used in the child and the parent since the child will have no children of its own. The call to the wait function will return immediately with a 'no more children' status (error).
You can find extensive discussion of all this in the comments — I've chosen to make this a Community Wiki answer since there was a lot of activity in the comments that identified the gist of this answer.
My question is: after finishing the execution of the new process image, the function
execl()
would return the execution to the caller process or to the father process?
When using one of the exec family of functions, you do not expect the function to return at all. The program counter begins at the first instruction of the binary image that replaced the calling process.
From the Darwin man page:
If any of the exec() functions returns, an error will have occurred.
The return value is -1, and the global variable errno will be set to
indicate the error.
There was a comment asking about the following, but it was deleted:
If you are in a child process, and execl succeeds, then the child process is replaced by the new binary. If it fails, then control returns to that child process (the caller). There's no strict relationship between fork and exec, if that's what you're asking. If you are in a child process, and exec fails, then you have a "forked" child process, which is a copy of the original parent process. At this point you probably want to print some error message and exit from the child process.
If you want to know why it failed, you can use the following pattern:
if (execl(...)) {
perror(NULL);
exit(errno);
}
For example, try running this program, the error message will indicate how to fix the program:
#include <errno.h>
#include <stdio.h>
#include <stdlib.h>
#include <unistd.h>
int main (const int argc, char * const argv[]) {
if (execl("ls", "ls", "-la", NULL)) {
perror(NULL);
exit(errno);
}
return 0;
}
The solution, use execlp instead of execl in this case.
I have been asked this question for homework, and am having trouble figuring it out. If anyone can help me i would really appreciate it.
What Linux library function is like a fork(), but the parent process is terminated?
I'm fairly certain that whoever assigned you this homework is looking for the exec() family of functions, from the POSIX API header <unistd.h>, because there is nothing else that more closely resembles the sort of functionality you describe.
The exec() family of functions executes a new process and replaces the currently running process address space with the newly executed process.
From the man page:
The exec() family of functions replaces the current process image with
a new process image.
It's not exactly the same as "terminating" the parent process, but in practice it results in a similar situation where the parent process address space is erased (replaced) with the address space of the child process.
What Linux library function is like a fork(), but the parent process
is terminated?
The parent process should not terminate because , it must wait for the child processes to finish executing , after which they will be in a state called "zombie state", now it is the responsibility of the parent to clean up the leftovers of the child process. The parent process can terminate without cleaning up the child processes, but then, it is not a proper way to do it, as the exit status of the child processes should be collected and checked by the parent process.
Here is an example, to demonstrate , what i just said...
#include<stdio.h>
#include<unistd.h>
#include<sys/wait.h>
int main()
{
pid_t cpid = 1 ;
int status;
cpid = fork();
// Load a application to the child using execl() , and finish the job
printf("Parent waiting for child to terminate\n");
int wait_stat = waitpid(-1,&status,0); // Parent will hang here till all child processes finish executing..
if (wait_stat < 0)
{
perror("waitpid error");
exit(-1);
}
// WIFEXITED and WEXITSTATUS are macros to get exit status information, of the child process
if (WIFEXITED (status))
{
printf("Child of id %u cleaned up\n",wait_stat);
printf("Exit status of application = %u\n",WEXITSTATUS(status));
}
}
Consider the code:
#include <stdio.h>
#include <errno.h>
#include <sys/types.h>
#include <unistd.h>
/* main --- do the work */
int main(int argc, char **argv)
{
pid_t child;
if ((child = fork()) < 0) {
fprintf(stderr, "%s: fork of child failed: %s\n",
argv[0], strerror(errno));
exit(1);
} else if (child == 0) {
// do something in child
}
} else {
// do something in parent
}
}
My question is from where does in the code the child process starts executing, i.e. which line is executed first??
If it executes the whole code, it will also create its own child process and thing will go on happening continuously which does not happen for sure!!!
If it starts after the fork() command, how does it goes in if statement at first??
It starts the execution of the child in the return of the fork function. Not in the start of the code. The fork returns the pid of the child in the parent process, and return 0 in the child process.
When you execute a fork() the thread is duplicated into memory.
So what effectively happens is that you will have two threads that executes the snippet you posted but their fork() return values will be different.
For the child thread fork() will return 0, so the other branch of the if won't be executed, same thing happens for the father thread.
When fork() is called the operating system assigns a new address space to the new thread that is going to spawn, then starts it, they will both share the same code segment but since the return value will be different they'll execute different parts of the code (if correctly split, like in your example)
The child starts by executing the next instruction (not line) after fork. So in your case it is the assignment of the fork's return value to the child variable.
Well, if i understand your question correctly, i can say to you that your code will run as a process already.When you run a code,it is already a process , so that this process goes if statement anyway. After fork(), you will have another process(child process).
In Unix, a process can create another process, that's why that happens.
Code execution in a child process starts from the next instruction following the fork() system call.
fork() system call just creates a seperate address space for the child process therefore it is a cloned copy of the parent process and the child process has all the memory elements of it's parent's process.
Thus, after spawning a child process through fork(), both processes (the parent process and the child process) resumes the execution right from the next instruction following the fork() system call.
I'm new to c language and Linux. I have a problem related to fork(),getpid()and exec()function.
I wrote a c program using fork() call the code of my program is following"
code:
#include <stdio.h>
#include <sys/types.h>
#include <unistd.h>
#include <stdlib.h>
void fun()
{
printf("\n this is trial for child process");
}
int main (int argc, char const *argv[])
{
int i,status,pid,t;
if(pid=fork()<0)
{
printf("\nfailed to create the process\n");
}
if(pid=fork()==0)
{
printf("\n the child process is created");
fun();
exit(1);
}
while(wait(&status)!=pid);
return 0;
}
The out put of this program is following:
the child process is created
this is trial for child process
the child process is created
this is trial for child process
Now my questions are as follows:
Why the output of program showing same thing twice? The output supposed to be "child process is created this is trial for child process"
Why the output is not according to code ?
Can we have a program which has 4 processes and all the processes perform different task for example one process print "my name". One process print "my age", the other process print "my address ?
How to make multiple process in main function ?
How to control the execution of multiple process ?
what does the exec() function do? Can anyone please explain me the working of exec(), fork(), getpid() with a source code?
Please help this novice fellow.
Your code calls fork() multiple times:
if(pid=fork()<0) /* calls fork() */
{
...
}
if(pid=fork()==0) /* also calls fork() */
{
...
}
Each successful fork() creates a new child process. To make matters worse, the second fork() is called by both the parent and the first child.
If you're trying to create a single child process, you should call fork() just once:
pid_t pid; /* note the correct return type of fork() */
...
pid = fork();
if (pid < 0)
{
...
}
else if (pid == 0)
{
...
}
If you want to create multiple child processes, you can have the parent process call fork() in a loop.
As to questions like "what does exec do?", my advice is to learn how to use man and then come back with specific questions if there's something in the manpages that remains unclear.
In this code you are creating Three process not including your main process.
pid=fork()
is itself a statement , which forks a new process even though it is inside an if statement condition. After the first fork() call the remaining codes will be executed twice. so next fork call will be called twice. You have already created a new process.
fork returns zero to itself and its
process id to its parent
That is consider a process A forks B (not from your code)
pid = fork();
printf("pid is : %d",pid);
printf statement executes twice(one for A and one for B). For A it prints(A is parent)
pid is : 512 //some integer value
process id
and B prints
pid is : 0
So in your question
if(pid=fork()==0)
{
printf("\n the child process is created");
fun();
exit(1);
}
this is the second fork which is already executing twice. So each of this execution creates a new child process. For both childs pid value is 0. So your print statement executes, which is what you see in the output. But for both parents a pid value will be there and your if condition fails, so it wont print. These two childs are your second and third processes..So in short you create 3 processes along with the main process
The output is generated twice because you are forking twice:
if(pid=fork()<0) // Fork #1
{
printf("\nfailed to create the process\n");
}
if(pid=fork()==0) // Fork #2