komly

openssl req -x509 -newkey rsa:2048 -nodes -keyout=test_key.pem -out test.pem -days=365 -subj='/CN=test.ru'  -reqexts SAN     -extensions SAN     -config <(cat /usr/lib/ssl/openssl.cnf \
        <(printf '[SAN]\nsubjectAltName=@alt_names\n[alt_names]\nDNS.1 = test.ru\nDNS.2 = *.test.ru'))     -sha256

komly

protected void setScrollingEnabled(boolean isEnabled) {
    AppBarLayout.LayoutParams params = (AppBarLayout.LayoutParams) toolbar.getLayoutParams();
    params.setScrollFlags(isEnabled ? (AppBarLayout.LayoutParams.SCROLL_FLAG_SCROLL | AppBarLayout.LayoutParams.SCROLL_FLAG_ENTER_ALWAYS) : 0);
}

komly

def get_char(c, i):
    return c if i == 1 else '%s%d' % (c, i)

def encode(s):
    if s == '':
        return ''
        
    res = []
    last_char = s[0]
    i = 1

komly

http://stackoverflow.com/q/10236953/1333025

I quite enjoyed this exercise. I tried to do it without looking at the answers,
and it was worth it. It took me considerable time, but the result is
surprisingly close to two of the other answers, as well as to
[monad-coroutine](http://hackage.haskell.org/package/monad-coroutine) library.
So I guess this is somewhat natural solution to this problem. Without this
exercise, I wouldn't understand how _monad-coroutine_ really works.

To add some additional value, I'll explain the steps that eventually led me to

komly

https://medium.com/@mondaini/assembling-a-continuous-integration-service-for-a-django-project-on-jenkins-5f979d4c4184

http://www.deploypython.com/source/theme/static/pdf/table-of-contents.pdf

komly

https://medium.com/@mondaini/assembling-a-continuous-integration-service-for-a-django-project-on-jenkins-5f979d4c4184

komly

#sudo sysctl -w net.inet.ip.portrange.first=32768
sudo sysctl -w net.inet.ip.portrange.first=12000
sudo sysctl -w net.inet.tcp.msl=1000
sudo sysctl -w kern.maxfiles=1000000 kern.maxfilesperproc=1000000

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#include <sys/types.h>
#include <sys/socket.h>
#include <netinet/in.h>
#include <signal.h>

#include <iostream>

using namespace std;

int main() {

komly

# Меняет первое вхождение a на b в arr
def replace_first(a, b, arr):
    arr = arr[:] 
    for i, x in enumerate(arr):
        if x == a:
            arr[i] = b
            return arr
    return arr

komly

#!/usr/bin/env python3

import os
import sys
import binascii


def find_duplicates(path):
    checksums = {}