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    <p>From Merriam Webster</p> <blockquote> <p>Full Definition of QUALIFY</p> <p>transitive verb 1 a : to reduce from a general to a particular or restricted form : modify b : to make less harsh or strict : moderate c : to alter the strength or flavor of d : to limit or modify the meaning of (as a noun) 2 : to characterize by naming an attribute : describe 3 a : to fit by training, skill, or ability for a special purpose b (1) : to declare competent or adequate : certify (2) : to invest with legal capacity : license</p> </blockquote> <p>Clauses 1 and 2 apply. Java and C++ both have scoping/namespaces, and "qualify" means to introduce sufficient scopes to distinguish between potential candidates.</p> <p>C.f.: If you have two classes with a member called "read".</p> <pre><code>class Foo { void read(); }; class Bar { void read(); }; </code></pre> <p>In your implementation file, you will implement both functions. But if you wrote (C++)</p> <pre><code>void read() {} </code></pre> <p>This is valid, but it creates a function in the global namespace, rather than implementing one of the two functions. The same code written inside the class Foo definition would implement Foo::read.</p> <p>So to implement our member functions outside the class definitions, we have to qualify - to reduce from general, to name the attribute of the container path - which read we are implementing.</p> <pre><code>void Foo::read() {} void Bar::read() {} </code></pre> <p>The global namespace is "::", so you can even be explicit if you are trying to use that:</p> <pre><code>void ::read() {} // note: this name is already taken by stdio's read() :) </code></pre>
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    1. POWhat does qualify mean?
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    1. CO"Given that the global namespace is anonymous", no, in C++ it isn't. "Anonymous namespaces" have a specific meaning, in particular they're distinct for each Translation Unit. The global namespace is the same namespace everywhere.
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    2. COGood point - editing to correct.
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