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copied!<p>It's possible with C++11 and <code>decltype</code>. For that, we'll exploit that a pointer-to-member is not a pointer into the derived class when the member is inherited from a base class.</p> <p>For example:</p> <pre><code>struct base{ void f(){} }; struct derived : base{}; </code></pre> <p>The type of <code>&derived::f</code> will be <code>void (base::*)()</code>, not <code>void (derived::*)()</code>. This was already true in C++03, but it was impossible to get the base class type without actually specifying it. With <code>decltype</code>, it's easy and only needs this little function:</p> <pre><code>// unimplemented to make sure it's only used // in unevaluated contexts (sizeof, decltype, alignof) template<class T, class U> T base_of(U T::*); </code></pre> <p>Usage:</p> <pre><code>#include <iostream> // unimplemented to make sure it's only used // in unevaluated contexts (sizeof, decltype, alignof) template<class T, class R> T base_of(R T::*); struct base{ void f(){} void name(){ std::cout << "base::name()\n"; } }; struct derived : base{ void name(){ std::cout << "derived::name()\n"; } }; struct not_deducible : base{ void f(){} void name(){ std::cout << "not_deducible::name()\n"; } }; int main(){ decltype(base_of(&derived::f)) a; decltype(base_of(&base::f)) b; decltype(base_of(&not_deducible::f)) c; a.name(); b.name(); c.name(); } </code></pre> <p>Output:</p> <pre><code>base::name() base::name() not_deducible::name() </code></pre> <p>As the last example shows, you need to use a member that is actually an inherited member of the base class you're interested in.</p> <p>There are more flaws, however: The member must also be unambiguously identify a base class member:</p> <pre><code>struct base2{ void f(){} }; struct not_deducible2 : base, base2{}; int main(){ decltype(base_of(&not_deducible2::f)) x; // error: 'f' is ambiguous } </code></pre> <p>That's the best you can get though, without compiler support.</p>

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