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copied!<p>You are hitting a small quirk in the syntax (not that C++ has few of them...). The default way to access a member variable is through direct use of the name or through <code>this-></code>. That is, the simpler spelling of your show function would be:</p> <pre><code>void B::show() { std::cout << a << std::endl; // alternatively this->a std::cout << &a << std::endl; // &(this->a) } </code></pre> <p>Which has a simple and consistent syntax. Now the language allows you to drop in extra qualifiers to access members of a base when accessing a member:</p> <pre><code>std::cout << A::a << std::endl; // Extra qualification </code></pre> <p>This is really equivalent to <code>this->A::a</code> and the main uses of the extra qualification are to disambiguate (if two bases had a member <code>a</code>, select the one in <code>A</code>) and in the case of virtual function disabling dynamic dispatch.</p> <p>Now you can do the same with pointers, as in <code>&this->A::a</code>, which will take the address of the member <code>a</code> in the subobject <code>A</code> of the current object. The problem in this case is that you cannot drop the <code>this-></code> qualifier, as the syntax <code>&A::a</code> is reserved to obtain a pointer to member, although by adding an extra set of parenthesis, as in <code>&(A::a)</code> the parser can no longer interpret that as obtaining a pointer to member but rather as taking the address of the object represented by <code>A::a</code>, which as seen before is the member in the base.</p>

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