title: What is the meaning of Note 1 in the C++ class member name lookup rules?
body: <p>From <a href="http://eel.is/c++draft/class.member.lookup#1" rel="noreferrer">http://eel.is/c++draft/class.member.lookup#1</a> :</p>
  <blockquote>
  <p>A <em>search</em> in a scope <code>X</code> for a name <code>N</code> from a program point <em>P</em> is a single search in <code>X</code> for <code>N</code> from <code>P</code> unless <code>X</code> is the scope of a class or class template <code>T</code>, in which case the following steps define the result of the search.</p>
  <p>[<em>Note 1</em>: The result differs only if <code>N</code> is a <em>conversion-function-id</em> or if the single search would find nothing. — <em>end note</em>]</p>
  </blockquote>
  <p>I'm having a hard time making a sense of the Note. It seems that a &quot;single search&quot; from a class scope will find preceding declarations at namespace scope, since the namespace scope contains the class scope. But, as we know, if the name has also been declared as a member of a non-dependent base class, then the base class member takes precedence over the namespace member. <em>Note 1</em> seems to contradict this, since it's basically saying &quot;if <code>N</code> is not a conversion-function-id, then you can just do a normal single search, and only if you fail to find anything, then use the procedure in this section&quot;. But the single search will succeed by finding the namespace scope declaration, and the class member lookup will yield a different result.</p>
  <p>Where is the error in my understanding?</p>