This is a quick cheat sheet guide for git users trying to migrate over to perforce. It does not include an exhaustive list of commands, this is just a simple cheat sheet to get people started.
The git status command isn’t a direct equivalent of p4 opened. However, p4 opened will show what files you have checked out that are not yet submitted.
The git checkout command is usually used for checking out a branch, there is no direct equivalent in git for p4 edit. All files in p4 must be “checked out” first before they can be checked in.
So I was recently commissioned to do some work on writing up a perforce installation script and also a script to create backups to install from s3 tarballs. This is the end result of the bash script I created. There are some issues with regards to chmod permissions on the system. However, the script is fully functional.
Here is a list of commands:
./p4d_install.sh -c 1 – Uninstalls everything that the script had installed previously
./p4d_install.sh -r -b s3://bucket-url -p superuserpassword – Installs perforce from an s3 bucket tarball, it will grab the latest tarball and use this for installation.
./p4d_install.sh -p superuserpassword – will install perforce without a backup, clean installation
#!/bin/bash
: ' @Author Daniel Gleason
@Email [email protected]
@Script_Name p4d_install.sh
@Date 12/11/2021
-- Developer Notes --
Most if not all of the permissions of the perforce files are owned by the perforce user
Use sudo -u perforce CMD_LINE for most commands
p4dctl commands:
sudo -u perforce p4dctl stop master
sudo -u perforce p4dctl start master
sudo -u perforce p4dctl status master
'
set -e
# Prints error to stderr in red
function print_error {
local RED='\033[0;31m'
local NC='\033[0m'
printf "${RED}$1${NC}\n" 1>&2;
}
# Prints out the usage of the script
function usage() {
local usage="Usage: $0 [-r <1|0> restore_from_s3 number] [-k aws_access_key str] [-s aws_secret_key str] [-d aws_region str] [-b s3_bucket_path str] [-p perforce_password str] [-c <1|0> clean_install]";
print_error $usage
exit 1;
}
# Removes any potential installation residue from previous runs of the script
function clean() {
sudo slay perforce 2> /dev/null
sudo rm -rf /perforce
sudo apt-get --purge remove helix-p4d -y
sudo apt-get remove awscli -y
sudo apt-get autoremove -y
sudo rm /etc/perforce/p4dctl.conf.d/master.conf 2> /dev/null
sudo userdel perforce -f 2> /dev/null
sudo delgroup perforce 2> /dev/null
}
# Get inputs
while getopts "r:k:s:d:b:p:c:" flag; do
case "${flag}" in
r)
export restore=${OPTARG}
;;
k)
export AWS_ACCESS_KEY_ID=${OPTARG}
;;
s)
export AWS_SECRET_ACCESS_KEY=${OPTARG}
;;
d)
export AWS_DEFAULT_REGION=${OPTARG}
;;
b)
export S3_BUCKET_PATH=${OPTARG}
;;
p)
export PERFORCE_PASSWORD=${OPTARG}
;;
c)
export CLEAN=1
clean
;;
*)
usage
;;
esac
done
function install_p4d() {
local ubuntu_distro=`lsb_release --codename --short` # focal, xenial, etc
wget -qO - https://package.perforce.com/perforce.pubkey | sudo apt-key add -
sudo rm /etc/apt/sources.list.d/perforce.list 2> /dev/null # Delete perforce apt file if it already eixsts, ignore the error output.
echo "deb http://package.perforce.com/apt/ubuntu $ubuntu_distro release" | sudo tee -a /etc/apt/sources.list.d/perforce.list
sudo apt-get update
sudo apt-get install helix-p4d -y
}
function configure_p4d() {
sudo mkdir /perforce 2> /dev/null
sudo /opt/perforce/sbin/configure-helix-p4d.sh -n master -p ssl:1666 -r /perforce -u perforce -P $PERFORCE_PASSWORD
sudo chown -R perforce.perforce /perforce
sudo usermod -aG perforce ubuntu # Add ubuntu user to perforce group
sudo -u perforce mkdir /perforce/depots
# All depot data will be located in /perforce/depots
p4 configure set master#server.depot.root=/perforce/depots
sudo chmod -R 777 /perforce/
sudo chmod -R 770 /perforce/depots
}
# Just some sueful tools I like to use
function install_prerequisites() {
sudo apt-get install awscli -y
sudo apt-get install slay -y
sudo apt-get install mlocate -y
}
# Produces credentials file in ~/.aws/credentials and config file in ~/.aws/config used by aws cli
function setup_aws_cli() {
aws configure set aws_access_key_id $AWS_ACCESS_KEY_ID
aws configure set aws_secret_access_key $AWS_SECRET_ACCESS_KEY
aws configure set default.region $AWS_DEFAULT_REGION
}
function restore_from_s3_backup() {
echo 'Restoring from s3 backup'
sudo mkdir /perforce/backups 2> /dev/null
sudo chown perforce.perforce /perforce/backups
sudo -u perforce p4dctl stop master
# Download from s3 into /perforce/backups/
pushd /perforce/backups/
local most_recent_checkpoint_s3=$(aws s3 ls $S3_BUCKET_PATH | sort | tail -n 1 | awk '{print $4}')
local backup_dir=`pwd`
sudo -u perforce aws s3 cp $S3_BUCKET_PATH/$most_recent_checkpoint_s3 .
sudo -u perforce tar -xf $most_recent_checkpoint_s3
sudo -u perforce rsync -a perforce/depots/ /perforce/depots/ # Move the depot files over
# Get checkpoint file we just extracted from the tarball
pushd perforce/backups/
local most_recent_checkpoint_file=`ls -t master.ckp.*.gz | head -1`
popd
# Restore the checkpoint
pushd /perforce/root/
sudo rm -rf /perforce/root/db.*
sudo -u perforce p4d -r . -jr $backup_dir/perforce/backups/$most_recent_checkpoint_file
popd
# Remove extracted files to keep things clean
sudo rm -rf perforce/
popd
# Add P4PORT to /etc/perforce/p4dctl.conf.d/master.conf
sudo sed -i '/P4SSLDIR\s.*=\s.*ssl/a \\tP4PORT=ssl:1666' /etc/perforce/p4dctl.conf.d/master.conf
# Change permissions of db files
sudo chmod -R 600 /perforce/root/db.*
# Generate new ssl certs
pushd /perforce/root/
sudo chmod -R 0700 ssl/
sudo rm ssl/privatekey.txt ssl/certificate.txt
P4SSLDIR=ssl sudo -Eu perforce p4d -Gc
popd
# Start p4d
sudo -u perforce p4dctl start master
# Run p4 trust on the new server setup.
p4 trust -f -y
}
function main() {
if [ ! $CLEAN ]
then
install_prerequisites
fi
if [ $PERFORCE_PASSWORD ]
then
install_p4d
configure_p4d
fi
if [ $AWS_ACCESS_KEY_ID ] && [ $AWS_SECRET_ACCESS_KEY ] && [ $AWS_DEFAULT_REGION ]
then
setup_aws_cli
fi
if [ $restore ] && [ $restore -eq 1 ]
then
if [ $S3_BUCKET_PATH ]
then
echo Bucket Path: $S3_BUCKET_PATH
restore_from_s3_backup
else
print_error "Error: restore flag must be 1 and s3_bucket_path must be set to restore from a backup."
usage
fi
fi
}
main
This next script is used in conjunction with the script above to create backups. You can run this script on a cron task and it will create a tarball and create a new perforce checkpoint and upload it to s3. It does not have functionality to setup an aws cli configuration. You will need to do that yourself, or give the aws instance the permissions it needs. I also hard coded the s3 backup location, you will need to change this to a variable and use getopts or just change the hard coded path for the bucket url.
#!/bin/bash
: ' @Author Daniel Gleason
@Email [email protected]
@Script_Name create_backup.sh
@Date 12/13/2021
'
function get_most_recent_checkpoint() {
pushd /perforce/backups/ > /dev/null
echo `ls -t master.ckp.*.gz | head -1`
popd > /dev/null
}
function make_backup() {
pushd /perforce > /dev/null
mkdir backups 2> /dev/null
sudo chown -R perforce.perforce backups/
pushd backups/
# Creates a checkpoint in /perforce/backups using the p4 database in /perforce/root and the journal in /perforce/root
sudo -u perforce p4d -r /perforce/root -J /perforce/root/journal -z -jc `pwd`/master
# Create tarball of depots and the checkpoint
local backup_name="$(date +"%m.%d.%Y.%I.%M.%p")_backup.tar.gz"
local most_recent_checkpoint=$(get_most_recent_checkpoint)
sudo -u perforce tar -Ppzcf $backup_name /perforce/depots/ /perforce/backups/$most_recent_checkpoint
aws s3 cp `pwd`/$backup_name " S3_LOCATION "
popd > /dev/null
popd > /dev/null
}
make_backup
int ar[] = {-2,3,4,-5,6,9,-1,-5};
int i = 0;
int j = sizeof(ar) / sizeof(ar[0]) - 1;
while(i <= j)
{
while(ar[i] < 0) { i++; }
while(ar[j] >= 0) { j--; }
if (i < j) {
std::swap(ar[i], ar[j]);
}
}
So this code is pretty simple. First we are creating an array of numbers, some negative, some positive, not in any order. Second we determine two index values, one at the beginning of the array and one at the end of the array. Then, as long as the starting index is less than or equal to the ending index, we loop.
After this we basically find the two indices to swap. We do two while loops. While the element in ar[i] is less than 0, which means it’s negative, increase i by 1. We stop once ar[i] is not negative. Then we do another loop where ar[j] is greater than or equal to zero, which means it is positive, and we keep doing this until a number that is not positive is found.
Finally, we just swap the values once we have found one that is positive and one that is negative. The final if statement if (i < j) is to make sure that we don’t swap it incorrectly after i is incremented by 1.
This is just a simple leetcode style question I found the solution to so I wanted to share it. The time complexity should be O(N).
Okay I just want to start out and say that this is outdated. The only reason why I am posting about this is because I find the history of technology very interesting and I want to share what I’ve read up on.
Alright so, if you’re a programmer you’re probably aware of multiple types of syntax. With Python we have:
int main(void)
{
printf("Hello World");
return 0;
}
But in the old days we had something called K & R syntax in C. It looked like this:
int Chicken(a, b, c)
int a;
float b;
char c;
{
return 0;
}
What’s interesting about this is that some modern compilers still support this syntax but will scream at you and tell you to not do it. However there’s some important things to note here. Look at the following code block.
int Chicken(a,b,c)
int a;
int b;
{
}
If you notice, I never declared the data type for c. By default, the compiler will use integer data types for anything that is undeclared. So the entire concept behind K & R syntax is that you define the function name, and it’s inputs. Then, you define the types of inputs afterwards. It’s very different from what I was taught in university. I just wanted to share this. Hopefully someone finds this interesting. 🙂
So I was doing a few leetcode questions today. Honestly, I just found this one and it seemed interesting because it was easy, and it had multiple solutions to the problem. You could take multiple jabs at it from different angles. For example, you could simply do math operations on it and get the modulo of 10 of the number which would give you the last digit, or you could do string manipulation.
You could also create a stack and a queue of each value and do the modulo of 10 for each number and then compare the stack and queue values because of the FIFO/LIFO rules of these data structures. However, this is not only inefficient in time complexity, but also in space because we are creating two unnecessary data structures.
We could also just convert the number to a string and then compare each value from the ends, for example:
bool isPalindrome(int x) {
string s = to_string(x);
int len = s.size() - 1;
int n = 0;
while(n < len)
{
if (s[n] != s[len])
return false;
len--;
n++;
}
return true;
}
However, this solution is not very good either because we are converting the integer to a string, which I believe is an O(N) time complexity, because each character has to be individually converted to a number.
So I found another solution which basically takes advantage of how math works.
bool isPalindrome(int n)
{
int original = n;
int reversed = 0;
while(n > 0)
{
reversed = reversed * 10 + n % 10;
n /= 10;
}
if (original == reversed)
return true;
else
return false;
}
This may not even be the most efficient solution as I may be ignorant on other mathematical concepts and how they work. However, given this, the math is simple. We know that n % 10 will give us the last digit in the number. So for example: 123 % 10 would give us 3.
Because we know this, we can simply loop through n, and say, while n is not 0, the reversed version of the number will be itself times 10 + the remainder (n % 10).
So given reversed = 0 for the first iteration of our loop, the math would be: 0 = 0 * 10 + 3
Obviously, this is assuming that n is 123. So after the first iteration, reversed will equate to 3. On the second iteration, after we divide x by 10, it would be:
3 = 3 * 10 + 2, which is 32, so reversed is now 32, which is obviously the first two digits in reverse of 123, it’s 32. If we do it again we will get the one. Let’s do it.
If we do n % 10 again, we get a remainder of 1, or our last digit. So then we take 32 = 32 * 10 + 1 which is 321. And there we go, we have the numbers in reverse. The last thing to do is to essentially check if the original value of n, is equal to this newly reversed value of n.
Return true if they are equal, return false otherwise.
Anyways, hopefully this helps other people who are trying to practice algorithms!
Today I had an interview and I incorrectly answered a question. It was silly because I remember reading about this a long time ago honestly in a C++ book written by IBM from the 1990s. It’s one of the first books I read growing up when learning to code. My mother bought it for me from Salvation Army. It had a yellowish-orange and red cover. Anyways, just as a personal note and so that I don’t forget this again.
The order is as follows:
Preprocessing – this is essentially the stage where the preprocessor parses the source files and looks for macros. This one is the one I’m most familiar with due to my heavy usage of Unreal Engine. Unreal Engine uses a lot of macros, and conditional compilation instructions.
Just as an example, in unreal engine there’s tons of conditional compilation instructions like this:
There’s more information about the visual studio C++ preprocessor here: https://docs.microsoft.com/en-us/cpp/preprocessor/c-cpp-preprocessor-reference?view=msvc-160
The preprocessor is really amazing honestly. There’s some token-pasting operations using the ## tag in a macro. I found this example on here: https://docs.microsoft.com/en-us/cpp/preprocessor/token-pasting-operator-hash-hash?view=msvc-160
#define paster( n ) printf_s( "token" #n " = %d", token##n )
int token10 = 15;
So essentially, if we call paster(10). It will essentially put token10 as the last parameter and it will fill in 15 where #n is at. So this will cause the console to print out “token10 = 15” There are other examples of preprocessor usage but those can be looked at on the reference I linked above.
Compilation – This is basically where the compiler goes over the code again (after it has been filled in with proper pre-processor changes) and then it begins to generate assembly source code.
Assembly – Takes assembly source code and builds an object file.
Linking – Takes libraries and object files to create the executable file.
So this is something we have all learned in a basic Computer Science course. However, I just wanted to write down a super simple implementation of Binary Search for people who are interested. As you should already know, Binary Search can only be done on a sorted array. This function will only give you the index of the target in the given array. When first calling it, pass 0 for left ptr and array.size() – 1 for the right ptr.
int binarySearch(vector<int>& array, int target, int leftPtr, int rightPtr)
{
int middle_index = (leftPtr + rightPtr) / 2;
int potentialMatch = array[middle_index];
if (target == potentialMatch)
{
return middle_index;
}
else if (target > potentialMatch)
{
return binarySearch(array, target, middle_index + 1, rightPtr);
}
else
{
return binarySearch(array, target, leftPtr, middle_index - 1);
}
}
So the question essentially asks you to write a boolean function that returns true if the input string is “balanced.” Balancing simply means that each open bracket has a closed bracket. You’re given a string like this:
string inString = "{[()]}";
In this example, this is a “balanced” string, because each opening bracket must have a closed bracket.
string inString = "[(])";
Now this one is not balanced. Why? Because it has overlapping brackets. Think of it like a venn diagram
Nothing can be in the center. So the question is, how do we implement this algorithm? This is my solution:
In my code I essentially start out by making two boolean methods for readability. One is called “IsOpenBracket” and the other is called “IsClosedBracket” both of these functions take in a character and then return true or false if they are an open or closed bracket respectively.
After this is our “balancedBrackets” method which takes in a string. At the beginning of the method, I define a stack of characters called “s”
After this I define a map called matching_brackets which essentially is used later to give us the opposite bracket of what we turn in. So if you pass in ‘)’ it gives you ‘(‘
The next step is to essentially loop through each character in the string one by one and then I define a character called currentCharacter, which I set as a local temp variable for each loop iteration.
The first step now is to essentially check if this currentCharacter is an open bracket or a closed bracket. If it’s an open bracket, we just push it to the stack. Why? Because at the end of our function, we will check to see if there is anything in the stack. If the stack is empty, we return true, if it’s not empty, we return false. The whole premise of this method is that we push when it’s an open character, and pop when it’s a matching closing character.
Alright so, if it’s not an open bracket, we then check to see if the size of the stack is 0. We do this because if the stack is empty and we are inside of the “IsClosedBracket” part of the if statement, then we essentially are adding a closed bracket because we have an open bracket, which would be an immediate failure case.
After this we essentially check if the top of the stack is equal to the opposite of our current closing bracket. This is essentially the main trick. Remember the map that was made earlier? This is why it was made. So essentially, because a stack is last in, first out, the top of the stack is essentially the last open bracket we pushed, and we are checking if the top of that stack is equal to the opposite of our current closing bracket.
The last part is that we return false. Why do we return false here? This is because if we are on a closing bracket, and the current top is not equal to the opposite of our closing bracket, then we are overlapping. This would not follow the rules, so we would return false.
Lastly, we just check if the current size is 0, which means we popped off all of the opening brackets, meaning there is no open bracket without a matching closing bracket. If it’s not zero, we just return false as to say that the string is not balanced.
If you want to read the entire rules of this algorithm, you can read it here on hackerrank: https://www.hackerrank.com/challenges/balanced-brackets/editorial
Alright so hopefully the DFS algorithm makes sense. It’s very simple once you got it. Once you got it, you got it!
Breadth First Search is a little bit trickier, but don’t worry. It’s still easy to learn. For BFS, it might be implemented like this:
vector<string> breadthFirstSearch(vector<string> *array) {
queue<Node*> q;
array->push_back(name);
q.push(this);
while (!q.empty())
{
Node* n = q.front();
q.pop();
for (int i = 0; i < n->children.size(); i++)
{
if (count(array->begin(), array->end(), n->children[i]->name) == 0)
{
array->push_back(n->children[i]->name);
q.push(n->children[i]);
}
}
}
return *array;
}
In the above sample, this method will essentially print out all of the children of a graph in breadth first search. That is, level by level. However, in a real situation, you would simply return the node when you find what you are looking for.
