用多个签名覆盖C ++虚拟方法 [英] Overriding C++ virtual methods with multiple signatures

问题描述

我有一个C ++基类，它声明了一个具有两个不同签名的虚方法。

只要覆盖派生类中的某个虚方法签名，编译器（g ++ 4.6.3和g ++ 4.7）不再能够将该方法与同一派生类中的第二个签名进行匹配。

下面的示例代码不会编译如果我只将SPECIALIZE_ONEARG定义为1.为了重新编译它，我还必须将PASSTHRU_TWOARG定义为1.使用PASSTHRU方法并不理想，因为它的效率很高，而且真正的类层次结构要深得多，而且我会

这种行为是特定于g ++还是我只是想做一些C ++不支持的东西？

  #define SPECIALIZE_ONEARG（0）
 #define PASSTHRU_TWOARG（0）
 
 class base 
 {
 public：
 virtual int myMethod（char a）{return 1; } 
 virtual int myMethod（char a，int b）{return 2; } 
}; 
 
 class derived：public base 
 {
 public：
 #if SPECIALIZE_ONEARG 
 virtual int myMethod（char a）{return 3; } 
 #endif // SPECIALIZE_ONEARG 
 
 #if PASSTHRU_TWOARG 
 virtual int myMethod（char a，int b）{return base :: myMethod（a，b）; } 
 #endif // PASSTHRU_TWOARG 
}; 
 
 int main（int argc，char * argv []）
 {
派生myObj; 
 
返回myObj.myMethod（'a'）* 10 + myObj.myMethod（'b'，0）; 
 
  

解决方案

隐藏来自基类的定义。为了让这个定义在派生范围内可见，您需要使用base :: myMethod 。

 派生类：public base 
 {
 public：
 using base :: myMethod; //< --- here 
 
 #if SPECIALIZE_ONEARG 
 virtual int myMethod（char a）{return 3; } 
 #endif // SPECIALIZE_ONEARG 
 
 #if PASSTHRU_TWOARG 
 virtual int myMethod（char a，int b）{return base :: myMethod（a，b）; } 
 #endif // PASSTHRU_TWOARG 
}; 
  

I have a C++ base class which declares a virtual method with two different signatures.

As soon as I override one of the virtual method signatures in a derived class the compiler (g++ 4.6.3 and g++ 4.7) is no longer able to match the method with the second signature in the same derived class.

The example code below will not compile if I only define SPECIALIZE_ONEARG to 1. In order to get it to compile again I also have to define PASSTHRU_TWOARG to 1. Using the "PASSTHRU" method isn't ideal because of efficiency and because the real class hierarchy is much deeper and I would prefer not to hardwire in the call to the base class.

Is this behavior specific to g++ or am I just trying to do something which isn't supported in C++?

#define SPECIALIZE_ONEARG ( 0 )
#define PASSTHRU_TWOARG   ( 0 )

class base
{
public:
  virtual int myMethod( char a )         { return 1; }
  virtual int myMethod( char a, int b )  { return 2; }
};

class derived : public base
{
public:
#if SPECIALIZE_ONEARG
  virtual int myMethod( char a )         { return 3; }
#endif // SPECIALIZE_ONEARG

#if PASSTHRU_TWOARG
  virtual int myMethod( char a, int b )  { return base::myMethod( a, b ); }
#endif // PASSTHRU_TWOARG
};

int main( int argc, char* argv[])
{
  derived myObj;

  return myObj.myMethod( 'a' ) * 10 + myObj.myMethod( 'b', 0 );
}

解决方案

Your definition is hiding the definition from the base class. In order for that definition to be visible at your derived scope, you need using base::myMethod.

class derived : public base
{
public:
  using base::myMethod; // <--- here

#if SPECIALIZE_ONEARG
  virtual int myMethod( char a )         { return 3; }
#endif // SPECIALIZE_ONEARG

#if PASSTHRU_TWOARG
  virtual int myMethod( char a, int b )  { return base::myMethod( a, b ); }
#endif // PASSTHRU_TWOARG
};

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