用相同的args声明两个构造函数的最理想的方法是什么？ [英] What's the most idotomatic way of declaring two constructors with the same args?

问题描述

作为一个例子，我有一个名为 File 的类。现在，文件可以二进制或文本形式打开。我的构造函数当前为 File（const char * filename）。让我们假设open的实现是完全不同的二进制和文本。我该怎么构造呢？

As an example lets say I have a class called File. Now file can be opened as binary or text. My constructor is currently File(const char*filename). Let's suppose the implementation of open is completely different binary and text. How the heck do I construct this?

我曾考虑使用静态函数，但不想返回指针。我可以传递一个指针，但我宁愿不允许在没有实际初始化的情况下构造一个类。

I thought about using a static function but I don't want to return a pointer. I could pass in a pointer but I rather not allow a class be constructed without actually initializing it.

我当时在考虑在构造函数中使用枚举或布尔值，但是感觉对我不对（以及我这样做的方式）。我可以为二进制和文本使用不同的类名称，并且都继承一个基本实现（或其他实现），即使唯一的区别是构造函数。

I was thinking about having an enum or bool in the constructor but it feels 'wrong' to me (and the way I may do this). I could have a different class name for binary and text and have both inherit a base implementation (or the other implementation) even though the only difference is the constuctor.

这是什么

推荐答案

两种惯用的方法是工厂函数（没有强制您返回指针的方法）？或标签分发（在标准库中使用，例如，在 std :: variant ）。

Two idiomatic ways are a factory function (nothing forces you to return a pointer), or tag dispatching (which is used in the standard library, for example in std::variant).

// Factory functions
struct File {
    static File openText(char const *filename);
    static File openBinary(char const *filename);
};



// Tag dispatching
struct open_as_binary_t {} constexpr open_as_binary;
struct open_as_text_t {} constexpr open_as_text;

struct File {
    File(char const *filename, open_as_binary_t);
    File(char const *filename, open_as_text_t);
};

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