返回无效的引用 [英] Returning an invalid reference

问题描述

有时候，当我在C ++中编程时，我希望对每个变量，如Javascript！有一个 undefined 值。
例如，当我为数组的out-of-bounds元素返回一个值时，返回一个 undefined 而不是抛出异常是有用的，或：

Sometimes when I'm programming in C++ I wish there was an undefined value for every variable something like Javascript!. For example when I'm returning a value for out-of-bounds element of an array, it was useful to return an undefined instead of throwing an exception, or:

template <typename T, int SIZE>
class MyArray
{
  T arr[SIZE];
  static T badref;
public:
  T &operator[](int i)
  {
    if (i >=0 && i < SIZE)
      return arr[i];
    else
      throw std::string("OUT-OF-BOUNDS"); // or: return badref; !!
  }
};

另一个脏（在我看来）选项返回一个预定义变量的引用为坏 - 参考变量。我知道我们不能将 null 或类似的东西赋给一个引用变量。

Another dirty(In my opinion) option is returning a reference of a pre-defind variable as a bad-reference variable. I know we can not assign null or something like that to a reference variable.

模式返回一个引用，其中调用者能够找到返回的值无效？

Is there an another well formed pattern to return a reference where caller has the ability to find out the returned value is not valid?

EDIT： 指针

推荐答案

在他的评论。它作为Boost libary的一部分。有关详情，请参见此页。

You can use boost::optional as @chris mentioned in his comment. It comes as a part of Boost libary. See this page for more details.

修改 MyArray class：

template <typename T, int SIZE>
class MyArray
{
  T arr[SIZE];
public:
  optional<T&> operator[](int i)
  {
    if (i >=0 && i < SIZE)
      return optional<T&>(arr[i]);
    else
      return optional<T&>();
  }
};

用法：

MyArray<int>() array;
// fill array with data

optional<int&> result = array[0];
if (result) {
    // item was found
} else {
    // index out of bounds
}

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