关于 c++ 的推导问题
为什么我无法访问基类 A 的 B 类初始化列表中的成员？
class A
{
public:
explicit A(int a1):a(a1)
{
}
explicit A()
{
}
public:
int a;
public:
virtual int GetA()
{
return a;
}
};
class B : public A
{
public:
explicit B(int a1):a(a1) // wrong!, I have to write a = a1 in {}. or use A(a1)
{
}
int GetA()
{
return a+1;
}
};
class C : public A
{
public:
explicit C(int a1):a(a1)
{
}
int GetA()
{
return a-1;
}
};
Why I can't access base class A's a member in class B initialization list?
class A
{
public:
explicit A(int a1):a(a1)
{
}
explicit A()
{
}
public:
int a;
public:
virtual int GetA()
{
return a;
}
};
class B : public A
{
public:
explicit B(int a1):a(a1) // wrong!, I have to write a = a1 in {}. or use A(a1)
{
}
int GetA()
{
return a+1;
}
};
class C : public A
{
public:
explicit C(int a1):a(a1)
{
}
int GetA()
{
return a-1;
}
};
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评论（3）
A 的构造函数在 B 的构造函数之前运行，并且隐式或显式地，前者构造了 A 的所有实例，包括
a成员。 因此 B 不能在a上使用构造函数，因为该字段已经构造完毕。 您尝试使用的符号准确地表明在a上使用构造函数，而此时这是不可能的。A's constructor runs before B's, and, implicitly or explicitly, the former construct all of A's instance, including the
amember. Therefore B cannot use a constructor ona, because that field is already constructed. The notation you're trying to use indicates exactly to use a constructor ona, and at that point it's just impossible.要以 Alex 的答案为基础，您可以通过控制其构造来初始化基类的 "a" 成员，如下所示：
请注意，我正在构造基类（大写 "A" ），而不是尝试直接初始化其继承的成员（小写“a”，取自您的示例）。
To build on Alex' answer, you can initialize the base class' "a" member by controlling its construction, like so:
Note that I'm constructing the base class (capital "A") above, rather than attempting to directly initialize its inherited member (lowercase "a", drawing from your example).
为了进一步构建 pilcrow 的答案，您可以通过在 B 类中覆盖它来轻松地初始化 A 成员，就像您想要的那样：
不过，我不一定推荐这样做；）
To build even further on pilcrow's answer, you could easily initialize the A member like you want by overriding it in your B class:
Though, I wouldn't necessarily recommend this ;)