在 C++0x 中转发所有构造函数

2021-12-13 00:00:00 constructor templates c++ c++11

在 C++0x 中转发所有父级构造函数的正确方法是什么?

What is the correct way to forward all of the parent's constructors in C++0x?

我一直在这样做:

class X: public Super {
    template<typename... Args>
        X(Args&&... args): Super(args...) {}
};

推荐答案

C++0x 中有一个更好的方法来解决这个问题

There is a better way in C++0x for this

class X: public Super {
  using Super::Super;
};

如果你声明一个完美转发模板，你的类型在重载解析中会表现得很糟糕.假设您的基类可以从 int 转换，并且有两个函数可以打印出类

If you declare a perfect-forwarding template, your type will behave badly in overload resolution. Imagine your base class is convertible from int and there exist two functions to print out classes

class Base {
public:
  Base(int n);
};

class Specific: public Base {
public:
  template<typename... Args>
    Specific(Args&&... args);
};

void printOut(Specific const& b);
void printOut(std::string const& s);

你用

printOut("hello");

会叫什么?这是模棱两可的，因为 Specific 可以转换任何参数，包括字符数组.它这样做不考虑现有的基类构造函数.使用 using 声明继承构造函数仅声明使此工作所需的构造函数.

What will be called? It's ambiguous, because Specific can convert any argument, including character arrays. It does so without regard of existing base class constructors. Inheriting constructors with using declarations only declare the constructors that are needed to make this work.

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