Identify the platform linux or windows using C/C++ code [duplicate]

Question

I was asked in an interview to how would you identify the platform only by providing runnable C code?

The code must be runnable in both platform(i.e no special header files function used to check)?

Answer 1

If you considering only the Linux platform or the Windows platform, this code will be work as needed to you on C and C++.

#include <stdio.h>

#if defined(__linux__) // any linux distribution
    #define PLATFORM "linux"
#elif defined(_WIN32) // any windows system
    #define PLATFORM "windows"
#else
    #define PLATFORM "Is not linux or windows"
#endif


int main(int argc, char *argv[]) {
    puts(PLATFORM);
    return 0;
}

A full answer with an example see there https://stackoverflow.com/a/42040445/6003870

Answer 2

The question is wierd. You can't provide an executable which can run on multiple platforms. So the obvious answer is: compile it on one platform: if it runs, that's the platform you're on; if it doesn't you're on some other platform.

Supposing that you do have separate executables, there's really no real answer. For starters, what do you consider "a different platform"? Are Linux on a Sparc and Linux on a PC different platforms? What about two different versions of Linux? What about running under CygWin? Regardless of how you define it, though, you'll probably need a separate test for each platform. But the question is silly, because, except for considering different versions different platforms, you'll need a distinct executable for each platform, which means that you have to distinguish at compile time anyway

Just for the record: for starters, I'd use boost::filesystem, and try to read directories "c:\\" and "/usr". In theory, you could have a directory named "c:\\" in the current directory under Unix, or a directory named "/usr" in C: under Windows, but the odds are definitely against it. Otherwise, try various files in "/proc/myPID". The directory will almost certainly not exist under Windows, and even if it does, it won't have the dynamic structure it has under various Unix. And the structure varies from one Unix to the next, so you should even be able to distinguish between Solaris and Linux.

If the comparison is just between Windows and Linux on a PC, of course, the simplest is to compile in 32 bit mode, and take the address of a local variable. If it's greater than 0x8000000, you're under Linux; if it's less, Windows.

Answer 3

a quick google search shows this

#if defined(_WIN64)
    /* Microsoft Windows (64-bit). ------------------------------ */

#elif defined(_WIN32)
    /* Microsoft Windows (32-bit). ------------------------------ */

#endif

OS identification macros are predefined by all C/C++ compilers to enable #if/#endif sets to wrap OS-specific code. This is often necessary in cross-platform code that must use low-level library functions for fast disk I/O, inter-process communications, or threads. While differences between Windows and other OSes are acute, even differences among UNIX-style OSes can require #if/#endif constructs. This article surveys common compilers and shows how to use predefined macros to detect common OSes at compile time.

http://nadeausoftware.com/articles/2012/01/c_c_tip_how_use_compiler_predefined_macros_detect_operating_system