任何人都可以解释这个opencv c ++代码的含义

Question

string getFilename（string s）{

char sep = '/';
char sepExt='.';

#ifdef _WIN32
    sep = '\\';
#endif

size_t i = s.rfind(sep, s.length( ));
if (i != string::npos) {
    string fn= (s.substr(i+1, s.length( ) - i));
    size_t j = fn.rfind(sepExt, fn.length( ));
    if (i != string::npos) {
        return fn.substr(0,j);
    }else{
        return fn;
    }
}else{
    return "";
}

}

一个的getFileName =（文件名）; //文件名是图像

Answer 1

看起来它提取文件的名称没有它的扩展名和路径：

"/home/user/Documents/someimage.jpg" -> "someimage"

size_t i = s.rfind(sep, s.length( )); // find location of the "/"
if (i != string::npos) {
  string fn= (s.substr(i+1, s.length( ) - i)); // extract filename with extension -> "someimage.jpg"
  size_t j = fn.rfind(sepExt, fn.length( )); // find location of the extension by looking for "."
  if (i != string::npos) {
    return fn.substr(0,j); // extract filename -> "someimage"
  }else{
    return fn;
  }
}else{
  return "";

}