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    <p>First off, if you make the base class destructor <code>virtual</code>, all derived classes will automatically get a <code>virtual</code> destructor if you declare them as <code>virtual</code> or not. This is generally true for matching signatures: if a base class has a <code>virtual</code> function with the same signature as a function in a derived class, the function in the derived class is an <code>override</code> and is <code>virtual</code> (although in C++ 2011 you can prevent further overriding using the <code>final</code> keyword in which case another override would create an error).</p> <p>That said, destructors are special: when you make a destructor <code>virtual</code> it will still be called even if there is another overriding destructor! The only impact of a destructor being <code>virtual</code> is what happens if you <code>delete</code> an object using a pointer to a base class when the object actually happens to be of a derived type: If the destructor isn't <code>virtual</code> you get undefined behavior while the Right Thing happens if the destructor is <code>virtual</code>. For example:</p> <pre><code>class not_a_base {}; class bad_idea: public not_a_base {}; class a_base { public: virtual ~a_base() {} }; class ok: public a_base {}; int main() { a_base* ab = new ok; delete ab; // &lt;---- all is good here! not_a_base* nab = new bad_idea; delete nab; // &lt;---- results in undefined behavior } </code></pre> <p>The reason destructors are not <code>virtual</code> by default is simply that this would mean that object size is always increased by a word size which is unacceptable in general.</p>
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    1. USDietmar Kühl
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    1. COThanks for clarifying exactly when issues could occur.
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