Reproduction of shell pipes
Pipex is a project that re-creates in C the way two commands are piped together via | in the shell
# ./pipex infile cmd1 cmd2 outfile
pipe()
|
|-- fork()
|
|-- child // cmd1
: |--dup2()
: |--close end[0]
: |--execve(cmd1)
:
|-- parent // cmd2
|--dup2()
|--close end[1]
|--execve(cmd2)
# pipe() sends the output of the first execve() as input to the second execve()
# fork() runs two processes (i.e. two commands) in one single program
# dup2() swaps our files with stdin and stdout
void pipex(int f1, int f2)
{
int end[2]; pipe(end);
}
# pipe() takes an array of two int, and links them together
# what is done in end[0] is visible to end[1], and vice versa
# pipe() assigns an fd to each end
# Fds are file descriptors
# since files can be read and written to, by getting an fd each, the two ends can communicate
# end[1] will write to the its own fd, and end[0] will read end[1]’s fd and write to its own
void pipex(int f1, int f2)
{
int end[2];
pid_t parent; pipe(end);
parent = fork();
if (parent < 0)
return (perror("Fork: "));
if (!parent) // if fork() returns 0, we are in the child process
child_process(f1, cmd1);
else
parent_process(f2, cmd2);
}
# fork() splits the process in two sub-processes -> parallel, simultaneous, happen at the same time
# it returns 0 for the child process, a non-zero for the parent process, -1 in case of error
end[1] is the child process, end[0] the parent process; the child writes, the parent reads
Since for something to be read, it must be written first, so cmd1 will be executed by the child, and cmd2 by the parent.
pipex is run like this ./pipex infile cmd1 cmd2 outfile
FDs 0, 1 and 2 are by default assigned to stdin, stdout and stderr
infile, outfile, the pipe, the stdin and stdout are all FDs
On linux, you can check your fds currently open with the command ls -la /proc/$$/fd
Our fd table right now looks like this:
-----------------
0 | stdin |
-----------------
1 | stdout |
-----------------
2 | stderr |
-----------------
3 | infile | // open()
-----------------
4 | outfile | // open()
-----------------
5 | end[0] |
-----------------
6 | end[1] |
-----------------
For the child process, we want infile to be our stdin (as input), and end[1] to be our stdout (we write to end[1] the output of cmd1)
In the parent process, we want end[0] to be our stdin (end[0] reads from end[1] the output of cmd1), and outfile to be our stdout (we write to it the output of cmd2)
Visually,
// each cmd needs a stdin (input) and returns an output (to stdout)
infile outfile
as stdin for cmd1 as stdout for cmd2
| PIPE ↑
| |---------------------------| |
↓ | | |
cmd1 --> end[1] ↔ end[0] --> cmd2
| |
cmd1 |---------------------------| end[0]
output reads end[1]
is written and sends cmd1
to end[1] output to cmd2
(end[1] becomes (end[0] becomes
cmd1 stdout) cmd2 stdin)
From the MAN,
int execve(const char *path, char *const argv[], char *envp[]);
# path: the path to our command
# type `which ls` and `which wc` in your terminal
# you'll see the exact path to the commands' binaries
# argv[]: the args the command needs, for ex. `ls -la`
# you can use your ft_split to obtain a char **
# like this { "ls", "-la", NULL }
# it must be null terminated
# envp: environmental variable -> retrieved from main (see below)
# in envp the line PATH contains all possible paths to the commands' binaries
# type env in the terminal to have a look
# split on : to retrieve all possible PATHs
int main(int ac, char **ag, char **envp)
{
int f1;
int f2;
f1 = open(ag[1], O_RDONLY);
f2 = open(ag[4], O_CREAT | O_RDWR | O_TRUNC, 0644);
if (f1 < 0 || f2 < 0)
return (-1);
pipex(f1, f2, ag, envp);
return (0);
}
execve() will try every possible path to the cmd until it finds the good one
If the command does not exist, execve() will do nothing and return -1;
else, it will execute the cmd and delete all ongoing processes (so no leaks)
void pipex(int f1, int f2, char *cmd1, char *cmd 2)
{
int end[2];
int status;
pid_t child1;
pid_t child2; pipe(end);
child1 = fork();
if (child1 < 0)
return (perror("Fork: "));
if (child1 == 0)
child_one(f1, cmd1);
child2 = fork();
if (child2 < 0)
return (perror("Fork: "));
if (child2 == 0)
child_two(f2, cmd2);
close(end[0]); // this is the parent
close(end[1]); // doing nothing
waitpid(child1, &status, 0); // supervising the children
waitpid(child2, &status, 0); // while they finish their tasks
}
If the command that does not exist, execve() will execute nothing without error messages
You need to check if the command exists before its execution with access(), else send an error pipex: weirdcmd: weirdcmd not found
[0] When splitting the env, print out the result of split. Add a / at the end for the path to work correctly.
[1] If the program gets stuck without executing anything, most probably the pipe ends are not closed correctly. Until one end is open, the other will be waiting for input and its process will not finish.
[2] Place perror("Error") in your code, especially right after fork() or execve() , to see what is going on in the pipe.
Inside the pipe, everything we do will go to one of its ends.
printf for ex. won’t print to the terminal, it will print to your outfile (because we swapped the stdout)
perror("Error") will work because it prints to stderr.
[3] Handle file rights when open()ing them.
Return error if the file cannot be opened, read or written.
Check how the shell treats infile and outfile when they do not exist, are not readable, writable etc. (chmod is your best friend).