DEV Community: MrSoUndso

Setting up a OpenTTD Server

MrSoUndso — Thu, 09 Jul 2020 15:04:33 +0000

Tl;Dr; at the end

OpenTTD is a business simulation game in which players try to earn money via transporting passengers and freight by road, rail, water and air. It is an open-source[3] remake and expansion of the 1994 Chris Sawyer video game Transport Tycoon Deluxe. (from: Wikipedia)

Setting up a server to play with your friends is a easy thing to do, but finding a guide has been really hard.
So I decided to write my own:

Prerequisites

You have a Linux server with Ubuntu
You have cli and sudo access to that server
You have your favorite editor installed (this guide uses nano)

Getting OpenTTD

Getting OpenTTD is easy. Simply run
sudo apt install openttd openttd-opengfx
and you're set!

Making OpenTTD run

To make sure OpenTTD keeps running even after you close your terminal window, it has to be a systemd service.
To create a service navigate to /etc/systemd/system (using cd /etc/systemd/system) and create a new file called openttd.service (sudo nano openttd.service).

In this file, we add the following lines:

[Unit]
Description=OpenTTD Server
Documentation=""

[Service]
Type=simple
ExecStart=/usr/games/openttd -D
Restart=on-failure
RestartSec=10s

[Install]
WantedBy=multi-user.target

Let's see whats going on here line by line:

The file is separated into three main parts, [Unit],[Service] and [Install]

[Unit] is for metadata like the title of the service and its relationship to other services
[Service] is the place for instructions on what the service should do
[Install] tells systemd how to enable and disable the service (this part stays the same most of the time)

A closer look at `[Service]`

First we define the service type with Type=simple. This is the default type and tells systemd that this service is going to be a long running one.
The next line is ExecStart=/usr/games/openttd -D. This tells systemd to start /usr/games/openttd with the -D command line flag (-D tells OpenTTD to start in server mode)
The next to lines Restart=on-failure and RestartSec=10s tell systemd to restart the service if if fails, but not before it has waited ten seconds.

(If you want more information on services, this is a good guide)

Starting the service

To start the service, you first have to enable it. Enable it with:
sudo systemctl enable openttd.service.
Afterwards, you can start it with
sudo systemctl start openttd.service
and view its status with
sudo systemctl status openttd.service.

Configuring the firewall

To access the server from anywhere in the world, you'll have to configure the firewall of your server. Thankfully, Ubuntu makes this easy!

Install the "Uncomplicated Fire Wall":
sudo apt install ufw
and allow connections on the ssh and the OpenTTD ports:
sudo ufw allow ssh
sudo ufw allow 3979
This makes your server available to the world, but it will not be visible on the public server list. If you want to add your server to the list, run
sudo ufw allow 3979/udp
Now you need to enable the firewall. You can do this by running:
sudo ufw enable

And that's it! Your OpenTTD server should be working fully now!

If your are searching for the openttd.cfg, it is located at /root/.openttd/openttd.cfg

TL;DR:

These are the commands you need to setup a server:

sudo apt install openttd openttd-opengfx ufw
sudo nano /etc/systemd/system/openttd.service

Paste this inside the file:

[Unit]
Description=OpenTTD Server
Documentation=""

[Service]
Type=simple
ExecStart=/usr/games/openttd -D
Restart=on-failure
RestartSec=10s

[Install]
WantedBy=multi-user.target

sudo systemctl enable openttd.service
sudo systemctl start openttd.service
sudo systemctl status openttd.service

sudo ufw allow ssh
sudo ufw allow 3979
sudo ufw enable

A strange Square

MrSoUndso — Tue, 29 Oct 2019 20:46:21 +0000

Hi everybody!

In the last post I showed you the the way I approached the "Print a rectangle test".
The next test I'll have to tackle is:

Print a crossed square to the console

So I started with the nearly the same code I used in my last project, but I quickly realized that the three for loops I used made everything way harder than it should be.

So I switched over to two nested for loops:

for (int i = 0; i < x; i++) {
    for(int b = 0; b < x; b++) {
      printf(%c,c);
    }
    printf("\n");
}

This resulted in a completely filled rectangle, just as I intended.
So I once again added a if check to make sure only the parts I wanted to be char where chars.

I started with the walls: the top line should be filled if i == 0, and the bottom if i = x - 1. The walls worked after a similar principle, just replace the i with a b. It's also important to add a else statement to print a space if none of the conditions on top are met.

The cross part was harder than I initially thought. While the line going from the top left corner to the bottom right one was simple as it simply follows the i == b condition, the one going the other way made me scratch my head a lot more.
The thing I came up with was this formula: (i + b) == x - 1
This condition is meet for every point on the line between the bottom left corner and the top right one.

In the end my complete code looked like this:

for (int i = 0; i < x; i++) {
  for(int b = 0; b < x; b++) {
    if(i == 0 || i == x -1 || b == 0 || b == x - 1 || i == b || i + b == x - 1) {
       printf("%c",c );
    } else {
      printf(" ");
    }
  }
printf("\n");

}

If you would try the same thing with 3 individual loops instead of 2 nested ones, you would run into really really unfunny issues with these simple if checks. For example, I had issues with the lines not meeting in the middle or having spaces everywhere but not where I wanted to have them.

Have a nice day!

P.S: I'm sorry for the strange formating. If you know how to fix it please leave me a comment!

P.P.S: Thanks to

.ltag__user__id__246551 .follow-action-button { background-color: #ee719e !important; color: #7f007f !important; border-color: #ee719e !important; }

Katie Nelson

Welcome tag moderator AKA Unofficial DEV cheerleader. While most of my friends are found on SnapChat or Tic-Toc, you can find me here. And I OOP, but I’m not a VSCO girl.

KatieGirlCoder KatNel20

for telling me how to fix my issues!

My first C Adventure

MrSoUndso — Tue, 29 Oct 2019 20:12:13 +0000

Hi everybody!

This is the first post of my new series: "Adventures in learning C".
First of all, please let my introduce myself: I'm Mr SoUndso and I currently study "Technische Informatik" (probably best translated as "technical computer science") in Germany. One of my lectures teaches me how to program in C. To pass the class, I'll have to complete different exercises. This series will be a write up on the different challenges I faced, and the ways I used to overcome them.

So lets start with the first test:

Write a program that prints a rectangle to the console

This sounds like a simple task, and it is. But i made a mistake along the way, that made my attempt at the next test way harder.

I started out with three different for loops, one for the top row, one for the side walls and one for the bottom.

for (int i = 0; i < y; i++) { //print first row } printf("/n"); for (int i = 1; i < x; i++) { //print the sides } for (int i = 0; i < y; i++) { //print last row }

The top and bottom rows were simple: I just added a printf(%c,c); into the loop and was set.

The sides were harder nuts to crack. I could not simply add a printf("%c %c \n",c, c); to the loop, as a requirement of the test was to make a the hight and width configurable.
So I added a second for loop into the mix.
My second loop now looked like this:
for (int i = 1; i < x; i++) { for (int b = 0; b < y; b++) { printf("%c", c); } printf("\n"); }

This would give me a completely filled rectangle. I realized I wasn't far from my goal, I only needed to fill the center with spaces instead of chars.

To do this, I added a if to check if the second iterator (b) is equal to 0 or y.
This resulted in a strange output:
xxxxx x x x x x xxxxx

I realized that b could never be equal to y, as the for loop would stop if
b == y .
So I changed the check from b == y to b == y - 1 and voilà: My code worked!

But wait: I told you earlier that I made a mistake making my next attempt way harder. Using 3 for loops instead of 2 nested ones made me scratch my head a lot when working on the next test. But this is a story for another post.

Have a nice day.