什么是返回引用一个恒定大小的指针数组的最好的方法？ [英] What is the best way to return reference to a constant sized array of pointers?

问题描述

我有一个声明为类成员的指针数组，如下所示：

I have an array of pointers declared as a class member like this:

class Bar
{
private:
  static constexpr int SIZE = 10;    
  Foo* m[SIZE];
}

在我的一个类方法中，我想返回一个指针优选地，参考）到该阵列。数组在编译时有一个已知的大小，但我跟踪我有多少项目我放在那里（它是一个缓冲区的东西）。

In one of my class methods, I would like to return a pointer (or preferably, a reference) to this array. The array has a known size at compile time, but I am keeping track of how many items I have put in there (it is a buffer of stuff).

在C ++ 11中返回此数组的引用的最佳方法是什么？

What is the best way to return a reference to this array in C++11 ?

我试过的东西：

GetArray(Foo* &f[], unsigned &size) const

我喜欢语法，因为它清楚地表明引用值是一个指针数组， ：声明为类型为Foo的引用数组*

I like the syntax because it makes it clear that the reference value is an array of pointers, but this gives a compiler error: Declared as array of references of type Foo*

GetArray(Foo** &f, unsigned &size) const
{
  f = m;
  size = mSize;
}

给我：错误：分配到 Foo ** 'from不兼容的类型Foo * const [10] 。将 mFoo 转换为（Foo **）可以缓解错误，但是IMHO不太优雅。

Gives me: Error: assigning to Foo **' from incompatible type Foo *const[10]. Casting mFoo to (Foo**) alleviates the error, but IMHO, this is not elegant.

推荐答案

没有人使用 std :: array 替换：

class Bar
{
    std::array<Foo *, 10>  m;
public:
    std::array<Foo *, 10> & getArray() { return m; }
    std::array<Foo *, 10> const & getArray() const { return m; }
};

在我看来，这比你要跳过的C数组版本。

This seems to me a lot simpler than the hoops you have to jump through to use your C-style array version.

为了避免代码重复，你可以 typedef std :: array< Foo *，10> FooArray; 。

To avoid code duplication you could typedef std::array<Foo *, 10> FooArray; .

同时具有 const code> const 实现是返回引用或指针的访问器函数的常见模式。 （如果你的访问器返回值是不需要的，当然）。

The technique of having both a const and a non-const implementation is a common pattern for accessor functions which return a reference or a pointer. (It's not required if your accessor returns by value, of course).

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