问题描述
这称为Derived :: f(complex)。为什么?好吧,请记住 派生未声明“使用Base:f;”,因此很明显Base :: f(int) 和Base :: f(double)无法调用。
我决定尝试使用此代码
class Base {
public:
virtual void f( int ) {
cout << "Base::f(int)" << endl;
}
virtual void f( double ) {
cout << "Base::f(double)" << endl;
}
virtual void g( int i = 10 ) {
cout << i << endl;
}
};
class Derived: public Base {
using Base::f;
public:
void f( complex<double> ) {
cout << "Derived::f(complex)" << endl;
}
void g( int i = 20 ) {
cout << "Derived::g() " << i << endl;
}
};
int main() {
Derived d;
d.f(1.0);
}
我得到了错误 main.cpp:在函数'int main()'中:
main.cpp:43:16: error: 'virtual void Base::f(double)' is inaccessible within this context
43 | d.f(1.0);
我的问题是如何使用using Base::f;
以及如何解决此问题?
解决方法
简单。您应该在类的公共部分下编写“使用”声明,如下所示。如果将声明放在private部分中,则Base函数可用于Derived类,但它们将成为Derived类的私有成员。这就是为什么您的编译器会引发“无法访问”错误。
#include <iostream>
#include <complex>
using namespace std;
class Base {
public:
virtual void f( int ) {
cout << "Base::f(int)" << endl;
}
virtual void f( double ) {
cout << "Base::f(double)" << endl;
}
virtual void g( int i = 10 ) {
cout << i << endl;
}
};
class Derived: public Base {
public:
using Base::f;
void f( complex<double> ) {
cout << "Derived::f(complex)" << endl;
}
void g( int i = 20 ) {
cout << "Derived::g() " << i << endl;
}
};
int main() {
Derived d;
d.f(1.0);
}