给定类型的转换运算符与构造函数。哪个更好？ [英] Conversion operator vs constructor from given type. Which is preferable?

问题描述

我正在为我的容器定义迭代器类型，当然，我希望 iterator 可转换为 const_iterator 。但我不确定哪个更好/更可取：

I'm defining iterator types for my container and of course I want iterator to be convertible to const_iterator. But I'm not sure which is better/preferable:

iterator

class iterator
{
    operator const_iterator();
};

或 const_iterator <中的非显式构造函数/ p>

or non-explicit constructor in const_iterator

class iterator
{
    // implementation
    friend class iterator; // hard to avoid this
};

class const_iterator
{
    const_iterator(iterator const &);
};

有没有什么指南是更好的方法？

Are there any guidelines which way is better?

推荐答案

仅当可以在不构造新对象的情况下返回期望的类型，或者返回的是简单数据类型时，才应编写转换运算符。例如，如果要返回对类内某个属性的const引用，则应将其编写为强制转换运算符。

You should write a casting operator only if it is possible to return the desired type without constructing a new object, or if its return is a simple data type. For example, if it were to return a const reference to a property within the class, then you should write it as a casting operator.

在所有其他情况下，都应只需编写适当的构造函数即可。

In all other cases, you should just write the appropriate constructor.

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