CppCoreGuidelines C.147 使用 dynamic_cast 转换成引用类型的时候，如果无法转换，就会认为出错了，抛出异常

04 January 2023

“Use dynamic_cast to a reference type when failure to find the required class is considered an error”

理由

把一个对象转换成引用，标明你希望最终得到一个有效的对象，所以类型转换必须成功。如果失败， dynamic_cast 就应该抛出一个异常。

例子

class Base {
  public:
    virtual string to_string() {
        return "Derived";
    }
};

class Derived : public Base {
  public:
    string to_string() {
        return "Derived";
    }
};

std::string f(Base& b)
{
    return dynamic_cast<Derived&>(b).to_string();
}

以下代码，在 f 函数中，d 被成功转换成了 Derived 本身，所以正确执行。

class Base {
  public:
    virtual string to_string() {
        return "Derived";
    }
};

class Derived : public Base {
  public:
    string to_string() {
        return "Derived";
    }
};

std::string f(Base& b)
{
    return dynamic_cast<Derived&>(b).to_string();
}
int main()
{
    Derived d;
    cout << f(d);
}

Derived

以下代码，因为无法将 Base 转换成 Derived，所以抛出异常 std::bad_cast

class Base {
  public:
    virtual string to_string() {
        return "Derived";
    }
};

class Derived : public Base {
  public:
    string to_string() {
        return "Derived";
    }
};

std::string f(Base& b)
{
    return dynamic_cast<Derived&>(b).to_string();
}
int main()
{
    Base b;
    cout << f(b);
}

terminate called after throwing an instance of 'std::bad_cast'

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