I use libcurl to get webpage's source code, everything goes all right until to get the source code of "www.macys.com". 

My programme gets nothing without CURLOPT\_FOLLOWLOCATION option, but when with this option, the redirecting location is always '/' and the curl\_easy\_perform function will not return unless I set the CURLOPT\_MAXREDIRS option.

My question is how can I get this webpage's source code with libcurl ? 
Thanks a lot.

My libcurl version : 7.23.1, OS: CentOS 6.0（final) x64