[23] inheritance what your mother never told you, c++ faq lite.htm

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<P><A name=[23.3]></A>
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<H3>[23.3] 当基类构造函数调用虚函数时，为什么不调用派生类重写的该虚函数？</H3></DIV>
<P>当基类被构造时，对象还不是一个派生类的对象，所以如果&nbsp;<TT>Base::Base()</TT>调用了<A 
href="http://www.sunistudio.com/cppfaq/virtual-functions.html">虚函数</A> 
<TT>virt()</TT>，则&nbsp;<TT>Base::virt()</TT>&nbsp;将被调用，即使&nbsp;<TT>Derived::virt()</TT>（译注：即派生类重写的虚函数）存在。<BR><BR>同样，当基类被析构时，对象已经不再是一个派生类对象了，所以如果&nbsp;<TT>Base::~Base()</TT>调用了<TT>virt()</TT>，则&nbsp;<TT>Base::virt()</TT>得到控制权，而不是重写的&nbsp;<TT>Derived::virt()</TT> 
。<BR><BR>当你可以想象到如果&nbsp;<TT>Derived::virt()</TT>&nbsp;涉及到派生类的某个成员对象将造成的灾难的时候，你很快就能看到这种方法的明智。详细来说，如果&nbsp;<TT>Base::Base()</TT>调用了虚函数&nbsp;<TT>virt()</TT>，这个规则使得&nbsp;<TT>Base::virt()</TT>被调用。如果不按照这个规则，<TT>Derived::virt()</TT>将在派生对象的派生部分被构造之前被调用，此时属于派生对象的派生部分的某个成员对象还没有被构造，而&nbsp;<TT>Derived::virt()</TT>却能够访问它。这将是灾难。
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<P><A name=[23.4]></A>
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<H3>[23.4] 派生类可以重置（“覆盖”）基类的非虚函数吗？</H3></DIV>
<P>合法但不合理。 
<P>有经验的&nbsp;C++&nbsp;程序员有时会重新定义非虚函数（例如，派生类的实现可能可以更有效地利用派生类的资源），或者为了回避<A 
href="http://www.sunistudio.com/cppfaq/strange-inheritance.html#[23.5]">隐藏规则</A>。即使非虚函数的指派基于指针/引用的静态类型而不是指针/引用所指对象的动态类型，但其客户可见性必须是一致的。 

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<P><A name=[23.5]></A>
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<H3>[23.5] 
“<TT>Warning:&nbsp;Derived::f(float)&nbsp;hides&nbsp;Base::f(int)</TT>”&nbsp;是什么意思？</H3></DIV>
<P>意思是：你要完蛋了。
<P>你所处的困境是：如果基类声明了一个成员函数<TT>f(int)</TT>，并且派生类声明了一个成员函数&nbsp;<TT>f(float)</TT>（名称相同，但参数类型和/或数量不同），那么&nbsp;<TT>Base</TT>&nbsp;的&nbsp;<TT>f(int)</TT>被隐藏（hidden）而不是被重载（overloaded）或被重写（overridden）（即使&nbsp;基类的<TT>f(int)</TT>是<A 
href="http://www.sunistudio.com/cppfaq/virtual-functions.html">虚拟</A>的）
<P>以下是你如何摆脱困境：派生类必须有一个被隐藏成员函数的<TT>using</TT>&nbsp;声明，例如： 
<P>
<DIV 
class=CodeBlock><TT>&nbsp;class&nbsp;Base&nbsp;{<BR>&nbsp;public:<BR>&nbsp;&nbsp;&nbsp;void&nbsp;f(int);<BR>&nbsp;};<BR>&nbsp;<BR>&nbsp;class&nbsp;Derived&nbsp;:&nbsp;public&nbsp;Base&nbsp;{<BR>&nbsp;public:<BR>&nbsp;&nbsp;&nbsp;using&nbsp;Base::f;&nbsp;&nbsp;&nbsp;&nbsp;</TT><EM>//&nbsp;This&nbsp;un-hides&nbsp;<TT>Base::f(int)</TT></EM><TT><BR>&nbsp;&nbsp;&nbsp;void&nbsp;f(double);<BR>&nbsp;}; 
</TT></DIV>
<P>如果你的编译器不支持<TT>using</TT>语法，那么就重新定义基类的被隐藏的成员函数，<A 
href="http://www.sunistudio.com/cppfaq/strange-inheritance.html#[23.4]">即使它们是非虚的</A>。一般来说这种重定义只不过使用<TT>::</TT>语法调用了基类被隐藏的成员函数，如， 

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<DIV 
class=CodeBlock><TT>&nbsp;class&nbsp;Derived&nbsp;:&nbsp;public&nbsp;Base&nbsp;{<BR>&nbsp;public:<BR>&nbsp;&nbsp;&nbsp;void&nbsp;f(double);<BR>&nbsp;&nbsp;&nbsp;void&nbsp;f(int&nbsp;i)&nbsp;{&nbsp;Base::f(i);&nbsp;}&nbsp;&nbsp;</TT><EM>//&nbsp;The&nbsp;redefinition&nbsp;merely&nbsp;calls&nbsp;<TT>Base::f(int)</TT></EM><TT><BR>&nbsp;}; 
</TT></DIV>
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<P><A name=[23.6]></A>
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<H3>[23.6]&nbsp; "virtual table" is an unresolved external 是什么意思？</H3></DIV>
<P>如果你得到一个连接错误"<TT>Error:&nbsp;Unresolved&nbsp;or&nbsp;undefined&nbsp;symbols&nbsp;detected:&nbsp;virtual&nbsp;table&nbsp;for&nbsp;class&nbsp;Fred</TT>"，那么可能是你在&nbsp;<TT>Fred</TT>&nbsp;类中有一个未定义的<A 
href="http://www.sunistudio.com/cppfaq/virtual-functions.html">虚</A>成员函数。<BR><BR>编译器通常会为含有虚函数的类创建一个称为“虚函数表”的不可思议的数据结构（这就是它如何处理<A 
href="http://www.sunistudio.com/cppfaq/virtual-functions.html#[20.2]">动态绑定</A>的）。通常你根本不必知道它。但如果你忘了为<TT>Fred</TT>&nbsp;类定义一个虚函数，则有时会得到这个连接错误。<BR><BR>许多编译器将这个不可思议的“虚函数表”放进定义类的第一个非内联虚函数的编辑单元中。因此如果&nbsp;<TT>Fred</TT>&nbsp;类的第一个非内联虚函数是&nbsp;<TT>wilma()</TT>，那么编译器会将&nbsp;<TT>Fred</TT>&nbsp;的虚函数表放在&nbsp;<TT>Fred::wilma()</TT>&nbsp;所在的编辑单元里。不幸的是如果你意外的忘了定义&nbsp;<TT>Fred::wilma()</TT>，那么你会得到一个"<TT>Fred</TT>'s 
virtual table is undefined"（<TT>Fred</TT>的虚函数表未定义）的错误而不是“<TT>Fred::wilma()</TT> 
is undefined”（<TT>Fred::wilma()</TT>未定义）。
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