我在C面临一个问题：
#include <iostream>
class A
{
protected:
void some_func(const unsigned int& param1)
{
std::cout << "A::some_func(" << param1 << ")" << std::endl;
}
public:
virtual ~A() {}
virtual void some_func(const unsigned int& param1,const char*)
{
some_func(param1);
}
};
class B : public A
{
public:
virtual ~B() {}
virtual void some_func(const unsigned int& param1,const char*)
{
some_func(param1);
}
};
int main(int,char**)
{
A* t = new B();
t->some_func(21,"some char*");
return 0;
}
我正在使用g 4.0.1和编译错误：
$g++ -W -Wall -Werror test.cc test.cc: In member function ‘virtual void B::some_func(const unsigned int&,const char*)’: test.cc:24: error: no matching function for call to ‘B::some_func(const unsigned int&)’ test.cc:22: note: candidates are: virtual void B::some_func(const unsigned int&,const char*)
为什么我必须指定B类中some_func(param1)的调用是A :: some_func(param1)？它是g的bug还是来自g的随机消息,以防止我看不到的特殊情况？
解决方法
问题是在派生类中,您将隐藏受保护的方法在基类中.您可以执行以下操作：要么完全限定派生对象中的受保护方法,要么使用using指令将该方法带入范围：
class B : public A
{
protected:
using A::some_func; // bring A::some_func overloads into B
public:
virtual ~B() {}
virtual void some_func(const unsigned int& param1,const char*)
{
A::some_func(param1); // or fully qualify the call
}
};
