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copied!<p>Here's how I understand this: During the construction of an object, each sub-object constructs its part. In the example, it means that <code>V::V()</code> initializes <code>V</code>'s members; <code>A</code> initializes <code>A</code>'s members, and so on. Since <code>V</code> is initialized before <code>A</code> and <code>B</code>, they can both rely on <code>V</code>'s members to be initialized.</p> <p>In the example, <code>B</code>'s constructor accepts two pointers to itself. Its <code>V</code> part is already constructed, so it's safe to call <code>v->g()</code>. However, at that point <code>D</code>'s <code>A</code> part has not been initialized yet. Therefore, the call <code>a->f()</code> accesses uninitialized memory, which is undefined behavior. </p> <p><strong>Edit:</strong></p> <p>In the <code>D</code> above, <code>A</code> is initialized before <code>B</code>, so there won't be any access to <code>A</code>'s uninitialized memory. On the other hand, once <code>A</code> has been fully constructed, its virtual functions are overridden by those of <code>D</code> (in practice: its vtable is set to <code>A</code>'s during construction, and to <code>D</code>'s once the construction is over). Therefore, the call to <code>a->f()</code> will invoke <code>D::f()</code>, before <code>D</code> has been initialized. So either way - <code>A</code> is constructed before <code>B</code> or after - you're going to call a method on an uninitialized object.</p> <p>The virtual functions part has already been discussed here, but for completeness: the call to <code>f()</code> uses <code>V::f</code> because <code>A</code> has not been initialized yet, and as far as <code>B</code> is concerned, that's the only implementation of <code>f</code>. <code>g()</code> calls <code>B::g</code> because <code>B</code> overrides <code>g</code>.</p>

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