So I don\'t know If this is the right place to ask this. I am new to programming

Question

So I don't know If this is the right place to ask this. I am new to programming, I am not very in to it (I know classes,arrays,functions..etc(Basic Things)).

I was programming with Visual Studio 2013 , so I decided to change my OS to Linux.

First things I notice is that the code is not exactly the same also the including headers. For example: I can't use conio.h and with it I can't end my main function with _getch(); function.

My question is: What's the difference between windows and linux programming? Can someone recommend me where can I look/read for this changes(whats the different and what can I use with linux).

Any help will be appreciated! Thank you in advance !! :)

Explanation / Answer

ou could also be interested in the system programming aspect of your task. Then read Advanced Linux Programming, read the intro(2) & syscalls(2) man pages. Be aware of the POSIX standard.

For thread programming, read a good Posix thread tutorial or a C++11 thread tutorial.

Regarding C++ programming, the Qt and POCO projects should interest you (and also perhaps Boost): they provide you with a cross-platform framework.

Regarding tools and programming habits, see these hints.

I recommend using at least C++11, not some earlier C++ standard (see documentation on cppreference & on cplusplus, etc...). So use recent versions of compilers (e.g. GCC 4.9 or Clang/LLVM 3.5 at least, in spring 2015). Don't forget to enable all warnings and debug info.

Study the source code of some existing free software (see e.g. sourceforge or github, etc...), you'll learn a big lot. Try even to contribute to it.

BTW, learning other programming languages (e.g. Scheme with SICP, Haskell, Ocaml, Prolog, Python, ...) will also teach you a big lot (even improve your way of thinking in C++).

Related Questions

Navigate

• Browse (All)
• Browse S
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.