在 C++ 中确定是 Linux 还是 Windows
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Determine if Linux or Windows in C++
提问by Elpezmuerto
I am writing a cross-platform compatible function in C++ that creates directories based on input filenames. I need to know if the machine is Linux or windows and use the appropriate forward or back slash. For the following code below, if the machine is Linux then isLinux = true. How do I determine the OS?
我正在用 C++ 编写一个跨平台兼容函数，它根据输入文件名创建目录。我需要知道机器是 Linux 还是 windows 并使用适当的正斜杠或反斜杠。对于下面的代码，如果机器是 Linux 则isLinux = true. 如何确定操作系统？
bool isLinux;
std::string slash;
std::string directoryName;
if isLinux
slash = "/";
else
slash = "\";
end
boost::filesystem::create_directory (full_path.native_directory_string() + slash + directoryName);
回答by Artyom
Use:
用：
#if defined(WIN32) || defined(_WIN32) || defined(__WIN32) && !defined(__CYGWIN__)
static const std::string slash="\";
#else
static const std::string slash="/";
#endif
BTW, you can still safely use this slash "/" on Windows as windows understands this perfectly. So just sticking with "/" slash would solve problems for all OSes even like OpenVMS where path is foo:[bar.bee]test.extcan be represented as /foo/bar/bee/test.ext.
顺便说一句，您仍然可以安全地在 Windows 上使用这个斜杠“/”，因为 Windows 完全理解这一点。因此，坚持使用“/”斜线可以解决所有操作系统的问题，甚至像 OpenVMS 一样，其中路径foo:[bar.bee]test.ext可以表示为/foo/bar/bee/test.ext.
回答by Billy ONeal
Generally speaking, you'd have do do this with conditional compilation.
一般来说，您必须使用条件编译来完成此操作。
That said, if you're using boost::filesystemyou should be using the portable generic path formatso that you can forget about things like this.
也就是说，如果你正在使用boost::filesystem你应该使用可移植的通用路径格式，这样你就可以忘记这样的事情。
回答by bobobobo
By default, Visual Studio #defines _WIN32in the preprocessor project settings.
默认情况下，Visual Studio中#defineŞ_WIN32预处理器中的项目设置。
So you can use
所以你可以使用
// _WIN32 = we're in windows
#ifdef _WIN32
// Windows
#else
// Not windows
#endif
回答by Fanatic23
Look into http://en.wikipedia.org/wiki/Uname
查看http://en.wikipedia.org/wiki/Uname
If you are using g++ as your compiler/GNU then you could try the code below. POSIX compliant environments support this:
如果您使用 g++ 作为编译器/GNU，那么您可以尝试下面的代码。POSIX 兼容环境支持：
#include <stdio.h>
#include <sys/utsname.h>
#include <stdlib.h>
int main()
{
struct utsname sysinfo;
if(uname(&sysinfo)) exit(9);
printf("os name: %s\n", sysinfo.sysname);
return 0;
}
回答by nathan
One of the most used methods to do this is with a pre-processor directive. The link is for C but they're used the same way in C++. Fair warning, each compiler & OS can have their own set of directives.
最常用的方法之一是使用预处理器指令。该链接适用于 C，但它们在 C++ 中的使用方式相同。公平警告，每个编译器和操作系统都可以有自己的一组指令。
回答by Ronny Brendel
predef.sourceforge.netis a comprehensive collection of all kinds of MACROs that identify compilers/operating systems and more. (linking directly to the operating system sub-site)
predef.sourceforge.net是各种MACRO的综合集合，用于识别编译器/操作系统等。（直接链接到操作系统子站点）
