Shell Pipeline C++ not ending command - c++

Question

I have a problem with Pipelines on a Shell that I want to create in C++. I'm trying to pipeline firstly between two commands and when I run the program, it pipelines as it should be, but it does not end my file or program to come back to the command line. It's an infinite loop of entering everything without showing an error.
The code for the pipeline is the following:
if(countpipes!=0)
{
pid_t pid;
int pipefd[2];
pipe(pipefd);
for(int j=0; j<commands.size(); j++)
{ //cout<<endl<<"IT of "<<j<<" "<<commands[j]<<endl;
vector<string> all_commands = split(commands[j]);
vector<string> next_commands;
if(j!=commands.size()-1)
{ //cout<<"Piped once"<<endl;
pipe(pipefd);
}
if(j+1<commands.size())
{
next_commands = split(commands[j+1]);
}
char* arguments[all_commands.size()+1];
arguments[all_commands.size()] = NULL;
for (int k = 0; k < all_commands.size(); k++)
{
arguments[k] = (char*)all_commands[k].c_str();
}
char* next_arguments[next_commands.size()+1];
next_arguments[next_commands.size()] = NULL;
for (int l = 0; l < next_commands.size(); l++)
{
next_arguments[l] = (char*)next_commands[l].c_str();
}
pid = fork();
//cout<<"Child: "<<pid<<endl;
if (pid < 0) {
// Showing an error
perror("fork");
exit(1);
}
if (pid == 0) {
// Child Process
if(j==0)
{
close(pipefd[0]);
dup2(pipefd[1], STDOUT_FILENO);
close(pipefd[1]);
int result2;
result2 = execvp(arguments[0], arguments);
if(result2 < 0)
{
perror("execvp");
exit(1);
}
}
else
{
close(pipefd[1]);
dup2(pipefd[0], STDIN_FILENO);
close(pipefd[0]);
int result;
result = execvp(arguments[0], arguments);
if (result < 0) {
perror("execvp");
exit(1);
}
}
}
else {
//cout<<"Wait pid"<<endl;
wait(NULL);
//cout<<"Wait after pid"<<endl;
}
}
}
I'm using a vector of strings that takes each of the commands. The final output for the pipeline looks like this and lets me introduce any text I want forever: https://imgur.com/LBvQRjF

Answer 1

The way you have your fork/exec set up, each child will only be connected to one pipe -- the first child connected (stdout) to the write end of a pipe, and each subsequent child connected (stdin) to the read end of a pipe. This will not work for 3+ children in a pipeline, as the middle children need to be connected to two (different) pipes -- stdin to the read end of one pipe and stdout to the write end of another.
You are not closing the pipe ends in the parent, so any child that is reading from a pipe will never see an EOF -- even when the previous child that is writing to the pipe exits, the parent will still have the write end open.
You also seem to be creating a next_agruments argv that you then never use, which is useless and probably indicates some confusion in your design.
pseudo-code for what you want:
int prev_out = -1, pipefds[2];
for (each command in the pipeline) {
if (not last command)
pipe(pipefds);
if ((child[i++] = fork()) == 0) {
// child
if (prev_out >= 0)
dup2(prev_out, 0);
if (not last command) {
dup2(pipefd[1], 1);
close(pipefd[0]); }
execvp(...
} else {
// parent
if (prev_out >= 0)
close(prev_out);
if (not last command) {
close(pipefd[1]);
prev_out = pipe_fd[0]; }
}
}
// now wait for the children.
You need a pipe between each pair of children connecting them. You need to close ALL the pipe fds in the parent after forking the children. You don't want to wait for any of the children until you've created all of them.