Fork: Resource temporarily unavailable when running shell with one arg - c++

Question

I am trying to write a microshell in C++ that will take in 1 or 2 args and run them in UNIX. My shell takes two args split by || fine, but when I run only one I get a massive fork error. My shell will look for || as a pipe instead of just |. Thank you in advance!
Some Functional commands are:
cat filename || sort
ls -l || less
Code:
#include <iomanip>
#include <iostream>
#include <string.h>
#include <unistd.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <errno.h>
#include <stdio.h>
using namespace std;
void getParms (char[], char* [], char* []);
int main()
{
char command[160];
pid_t pid1 = 1, pid2 = 1;
cout << "myshell> ";
cin.getline(command, 160);
while (strcmp(command, "q") != 0 && strcmp(command, "quit") != 0 && pid1 > 0 && pid2 > 0)
{
char* arg1[6];
char* arg2[6];
char path1[21], path2[21];
int pipefd[2];
arg1[0]=NULL;
arg2[0]=NULL;
getParms(command, arg1, arg2);
if (pipe(pipefd) < 0)
{
perror ("Pipe");
exit (-1);
}
//cerr <<"This is arg2"<<arg2[0]<<endl;
pid1 = fork();
if (pid1 < 0)
{
perror ("Fork");
exit (-1);
}
if (pid1 == 0)
{
//cout<<"Child 1"<<endl;
//cerr<<arg1[0]<<endl;
if(arg2[0] != NULL)
{
close(pipefd[0]);
close(1);
dup(pipefd[1]);
close(pipefd[1]);
}
strcpy(path1, "/bin/");
strcat(path1, arg1[0]);
if (execvp(path1, arg1) < 0)
{
strcpy(path1, "/usr/bin/");
strncat(path1, arg1[0], strlen(arg1[0]));
if (execvp(path1, arg1) < 0)
{
cout<<"Couldn't execute "<<arg1[0]<<endl;
exit (127);
}
}
if(arg2[0]== NULL)
{ // Parent process
close (pipefd[0]); //read
close (pipefd[1]); //write
waitpid(pid1, NULL, 0); // Waits for child2
cout << "myshell> ";
cin.getline(command, 160);
}
}
else if(arg2[0] != NULL)
{
//cerr<<"Child 2"<<endl;
pid2 = fork();
if (pid2 < 0)
{
perror ("Fork");
exit (-1);
}
if (pid2 == 0)
{
close(pipefd[1]);
close(0);
dup(pipefd[0]);
close(pipefd[0]);
strcpy(path2, "/bin/");
strncat(path2, arg2[0], strlen(arg2[0]));
if (execvp(path2, arg2) < 0)
{
strcpy(path2, "/usr/bin/");
strncat(path2, arg2[0], strlen(arg2[0]));
if (execvp(path2, arg2) < 0)
{
cout<<"Couldn't execute "<<arg2[0]<<endl;
exit (127);
}
}
}
else
{ // Parent process
//cerr<<"in last 2 else"<<endl;
close (pipefd[0]); //read
close (pipefd[1]); //write
waitpid(pid2, NULL, 0); // Waits for child2
cout << "myshell> ";
cin.getline(command, 160);
}
}
}
return 0;
}
/****************************************************************
FUNCTION: void getParms (char [], char* [], char* [])
ARGUMENTS: char str[] which holds full command
char* args[] args2[] which will hold the individual commands
RETURNS: N/A
****************************************************************/
void getParms(char str[], char* args[], char* args2[])
{
char* index;
int i= 0;
int j= 0;
index = strtok(str, " ");
//cerr<<"before first while"<<endl;
// While the token isn't NULL or pipe
while (index != NULL && strstr(index,"||") == NULL)
{
args[i] = index;
index = strtok(NULL, " ");
i++;
}
args[i] = (char*) NULL; // makes last element Null
//cerr<<" getParms before ||"<<endl;
if(index != NULL && strcmp(index,"||") != 0)
{
//cerr<<"after checking for ||"<<endl;
index = strtok(NULL," ");
while (index != NULL)
{
args2[j] = index;
index = strtok(NULL," ");
j++;
}
}
//cerr<<"After second IF"<<endl;
args2[j] = (char*) NULL; // makes last element Null
}

Answer 1

Your problem is that the main while loop is not going to any of the if-else statements in which you have the prompt for another command - the same statement is executed over and over. When you use the double pipe it goes to else if(arg2[0] != NULL) and the parent process shows a new prompt.
Try removing both prompts for a command from the main while loop in your if-else statement and move the prompt to the beginning of the loop like this:
//Move these two below into the while loop
//cout << "myshell> ";
//cin.getline(command, 160);
while (strcmp(command, "q") != 0 && strcmp(command, "quit") != 0 && pid1 > 0 && pid2 > 0)
{
cout << "myshell> ";
cin.getline(command, 160);
//...
}
Try not to make such redundant calls of the same thing. If you have a couple of those and you need to change something it can get messy.