如何在派生类构造函数中初始化基类成员变量？

问题描述：

为什么我不能这样做？

class A
{
public:
    int a, b;
}

class B : public A
{
    B() : A(), a(0), b(0)
    {
    }

}

I meant for the A member variables to be public.

答

您不能初始化 a 中的和和 b ，因为它们不是 B 。它们是 A 的成员，因此只有 A 可以初始化它们。您可以将它们设为公开，然后在 B 中进行赋值，但这不是推荐的选项，因为它会破坏封装。而是在 A 中创建一个构造函数以允许 B （或 A ）初始化它们：

You can't initialize a and b in B because they are not members of B. They are members of A, therefore only A can initialize them. You can make them public, then do assignment in B, but that is not a recommended option since it would destroy encapsulation. Instead, create a constructor in A to allow B (or any subclass of A) to initialize them:

class A 
{
protected:
    A(int a, int b) : a(a), b(b) {} // Accessible to derived classes
    // Change "protected" to "public" to allow others to instantiate A.
private:
    int a, b; // Keep these variables private in A
};

class B : public A 
{
public:
    B() : A(0, 0) // Calls A's constructor, initializing a and b in A to 0.
    {
    } 
};

如何在派生类构造函数中初始化基类成员变量？

相关推荐