习惯的方式来声明C ++不可变类 [英] Idiomatic Way to declare C++ Immutable Classes

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本文介绍了习惯的方式来声明C ++不可变类的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

所以我有一些相当广泛的功能代码,其中主要的数据类型是不可变的structs /类。我一直声明不变性的方式是几乎不可变的通过成员变量和任何方法const。

So I have some pretty extensive functional code where the main data type is immutable structs/classes. The way I have been declaring immutability is "practically immutable" by making member variables and any methods const.

struct RockSolid {
   const float x;
   const float y;
   float MakeHarderConcrete() const { return x + y; }
}

这是C ++中的或者有更好的方法吗?

Is this actually the way "we should do it" in C++? Or is there a better way?

推荐答案

你提出的方法是完全正常的,除非在你的代码中,的RockSolid变量,如下所示:

The way you proposed is perfectly fine, except if in your code you need to make assignment of RockSolid variables, like this:

RockSolid a(0,1);
RockSolid b(0,1);
a = b;

这将无法正常工作,因为编译器会删除复制赋值操作符。

This would not work as the copy assignment operator would have been deleted by the compiler.

所以一个替代方法是重写你的struct作为一个私有数据成员的类,只有public const函数。

So an alternative is to rewrite your struct as a class with private data members, and only public const functions.

class RockSolid {
  private:
    float x;
    float y;

  public:
    RockSolid(float _x, float _y) : x(_x), y(_y) {
    }
    float MakeHarderConcrete() const { return x + y; }
    float getX() const { return x; }
    float getY() const { return y; }
 }

这样,你的RockSolid对象是(pseudo-)不可变的,您仍然可以进行分配。

In this way, your RockSolid objects are (pseudo-)immutables, but you are still able to make assignments.

这篇关于习惯的方式来声明C ++不可变类的文章就介绍到这了,希望我们推荐的答案对大家有所帮助,也希望大家多多支持IT屋!

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