Your Own Shell - forks() and concurrent programs

[sourlemon]

Hi guys,

I'm currently having problem editing my code that a child of a process can run concurrently with its parent.

This is the exact statement from my homework.

Add functionality to the shell from Problem A so that a user can use the "&" operator as a command terminator. A command terminated with "&" should be executed concurrently with the shell (rather than the shell's waiting for the child to terminate before it prompts the user for another command).

I don't exactly understand how the fork () and shell works together. But after days of browsing, this is what I concluded. When I call fork, it runs another process (supposedly, my shell?). But I'm lost with the parent and the child. Who are they? Why is my child calling the execution? If only the child can call execv, does that mean I have to create another fork to have the process running concurrently? Or can I edit the wait statement?

This is my current execution function:

void exeCmd(char *cmd){
int i = 0;

if(fork() == 0) {
printf("%s\n", cmd);
i = execve(cmd, margv, menvp);
printf("Error: %d\n", errno);
if(i < 0) {
printf("%s: Command not found.\n", cmd);
exit(1);
}
} else {
wait(NULL);
}
}

My teacher gave me another waiting statement, but it doesn't work correctly when I do it like that.

int status = 0;
wait(&status);
printf("Child exited with status of %i.\n", status);

I know that's a lot I'm asking, but if anyone can help me, I'll be greatly appreciated. THANK YOU!!!

[dargueta]

First of all, I wouldn't call fork() like you did. This is the way I do it:

int exit_code;
pid_t pid = fork();

switch(pid)
{
case 0:
    //child code - execute exec() here
    break;
case -1:
    //failed to fork
    break;
default:
    //parent code
    waitpid(pid,&exit_code,0);
    break;
}

//check to see if child program exited immediately.
//if so, it failed.
if(WIFEXITED(exit_code))
{
    //exec() failed
    printf("Failed with code %d\n",WEXITSTATUS(exit_code));
}

The reason is that if fork() fails, your code won't catch it. Hence my use of a switch statement.

[kiddies]

thanks....

[dargueta]

That ellipsis leads me to believe it didn't exactly work.

[sourlemon]

thanks dargueta. I forgot about that. I think I solved the background problem by making it an if statement. If the bg is false, I make the parent wait. It looks messy, but I think that's how it is. Thank you for your help.

[dargueta]

Anytime.

[h4x]

int exit_code;
pid_t pid = fork();

switch(pid)
{
case 0:
    //child code - execute exec() here
    break;
case -1:
    //failed to fork
    break;
default:
    //parent code
    waitpid(pid,&exit_code,0);
    break;
}

//check to see if child program exited immediately.
//if so, it failed.
if(WIFEXITED(exit_code))
{
    //exec() failed
    printf("Failed with code %d\n",WEXITSTATUS(exit_code));
}

this code is flawed.

case 0:
    //child code - execute exec() here
    break;

what if break will be executed? you assume exec() wont fail.

case -1:
    //failed to fork
    break;

and here we jump right to

//check to see if child program exited immediately.
	//if so, it failed.
	if(WIFEXITED(exit_code))

wich is uninitilized

waitpid(pid,&exit_code,0);

cant it fail?
it might be pedantic, but always check return value unless it return void...

[dargueta]

H4x...I only put comments because kiddies is supposed to put his/her own code in there. It's called a template. Trust me, I know what to do.

[h4x]

oh well, perhaps i was misguided by your statment:

First of all, I wouldn't call fork() like you did. This is the way I do it:

[dargueta]

Should I have put this, then?

//insert your code here

I thought it was fairly obvious, as did the OP...