如何在c ++中读取Linux环境变量 [英] How to read Linux environment variables in c++
问题描述
在我的c ++程序中,我想将一些环境变量从shell加载到一些字符串中。
In my c++ program I want to load some environment variables from the shell into some strings. How can this be done?
推荐答案
使用getenv()函数 - 参见http://en.cppreference.com/w/cpp/utility/program/getenv 。我喜欢包装如下:
Use the getenv() function - see http://en.cppreference.com/w/cpp/utility/program/getenv. I like to wrap this as follows:
string GetEnv( const string & var ) {
const char * val = ::getenv( var.c_str() );
if ( val == 0 ) {
return "";
}
else {
return val;
}
}
这避免了环境变量不存在时的问题,并允许我使用C ++字符串轻松地查询环境。当然,它不允许我测试环境变量是否不存在,但一般来说,这不是我的代码中的问题。
which avoids problems when the environment variable does not exist, and allows me to use C++ strings easily to query the environment. Of course, it does not allow me to test if an environment variable does not exist, but in general that is not a problem in my code.
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