谁能解释一下这个opencv c++；代码意味着

字符串getFilename（字符串s）{

}

---

a=getFilename（filename）；//文件名是一个图像

它似乎提取了文件名，但没有扩展名和路径：

"/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 "";

}

-1，因为它与opencv无关。这只是使用STD::string IN C++的简单路径名称争用

"/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 "";