why are there so many edgy kids willing to learn how to hack
Likai
Hey folks.
Likai
Just poking around some C resources right now, trying to see who may be interested in working on a reverse engineeri g / code cleanup project for long term reward.
Likai
Need to do this with antminer firmware. I have some existing resources and code for this, but am looking to make improvements. The team will get a percent fee from all machines running the firmware.
that didnt help my code lol i just figure out myself
Sukesh
anyone knows how to make use of uhttpd in c++ with openWRT
Anonymous
Hmmmm I want some one among of you that we meet privatly to explaining well about c++ programming languege
Dima
porgraming lenguege is good
Dima
my frend
Dima
c plas plas one loev
Anonymous
Thanks sir
Dima
Jk it's not possible to tell most of it in one chat
Anonymous
Don't worry sir we do it gradually gradually
Mike
/help@GroupButler_bot
Anonymous
/help@GroupButler_bot
Group Butler
Start me to get the list of commands
Anonymous
Hello everybody ..
C++ and C are great fast programming language
But there are something I think missing in there standard libraries .. GUI
The whole loud of using extensions like Qt & WxWidget ... lablablablaaa is the best choice to develop GUI application but
I want to make my own special for GUI api how to make that..
Let's say I will use css & html to recognize the style or behavier
How to make all these thing from scratch and as beginner
Any help 🤔
Anonymous
/help@GroupButler_bot
Group Butler
Start me to get the list of commands
Anonymous
/help@GroupButler_bot
Group Butler
Start me to get the list of commands
Vivid
So worth it
𝑠𝑎𝑛𝑖
correctmaninwrongplace
Do you want the homework done?
or just a clue about how to start?
Anonymous
take number as array, run
for
to sum up digits and build the number ( like multiply by 10, 100, 1000 etc ) and then see if its divisible by
%
operator
you need to divide by 10 the number to get the digits(the digits are the modulus), and save the value of the sum of the modulus in a variable,
them check if the modulus(%) of the original number and the sum of modulus is 0
correctmaninwrongplace
you need to know:
1. 1 type of loop from C++
2. operator %
3. operator /