c++ - 当数据可以在类对象超出范围之前被删除时，类允许访问其数据(通过 ptr/it)是不是糟糕的设计？

Question

经典的例子是迭代器失效:

std::string test("A");
auto it = test.insert(test.begin()+1,'B');
test.erase();
...
std::cout << *it;

您是否认为拥有这种 API 是糟糕的设计，并且对于初学者来说会难以学习/使用？

在这种情况下，一种昂贵的、性能/内存方面的解决方案是，在使用 clear 方法时将指针/迭代器分配给空字符串(或 nullptr，但这不是很有用)。

一些精度

我正在考虑这种设计，用于返回可以在内部修改的 const chars*(也许它们存储在可以清除的 std::vector 中)。我不想返回 std::string (二进制兼容性)并且我不想要 get(char*,std::size_t) 方法，因为需要获取大小参数(太慢)。我也不想围绕 std::string 或我自己的字符串类创建包装器。

Answer 1

我建议继续阅读 Stepanov's design philosophy (第 9-11 页):

[This example] is written in a clear object-oriented style with getters and setters. The proponents of this style say that the advantage of having such functions is that it allows programmers later on to change the implementation. What they forget to mention is that sometimes it is awfully good to expose the implementation. Let us see what I mean. It is hard for me to imagine an evolution of a system that would let you keep the interface of get and set, but be able to change the implementation. I could imagine that the implementation outgrows int and you need to switch to long. But that is a different interface. I can imagine that you decide to switch from an array to a list but that also will force you to change the interface, since it is really not a very good idea to index into a linked list.

Now let us see why it is really good to expose the implementation. Let us assume that tomorrow you decide to sort your integers. How can you do it? Could you use the C library qsort? No, since it knows nothing about your getters and setters. Could you use the STL sort? The answer is the same. While you design your class to survive some hypothetical change in the implementation, you did not design it for the very common task of sorting. Of course, the proponents of getters and setters will suggest that you extend your interface with a member function sort. After you do that, you will discover that you need binary search and median, etc. Very soon your class will have 30 member functions but, of course, it will be hiding the implementation. And that could be done only if you are the owner of the class. Otherwise, you need to implement a decent sorting algorithm on top of the setter-getter interface from scratch and that is a far more difficult and dangerous activity than one can imagine. ...

Setters and getters make our daily programming hard but promise huge rewards in the future when we discover better ways to store arrays of integers in memory. But I do not know a single realistic scenario when hiding memory locations inside our data structure helps and exposure hurts; it is, therefore, my obligation to expose a much more convenient interface that also happens to be consistent with the familiar interface to the C arrays. When we program in C++ we should not be ashamed of its C heritage, but make full use of it. The only problems with C++, and even the only problems with C, arise when they themselves are not consistent with their own logic. ...

My remark about exposing the address locations of consecutive integers is not facetious. It took a major effort to convince the standard committee that such a requirement is an essential property of vectors; they would not, however, agree that vector iterators should be pointers and, therefore, on several major platforms – including the Microsoft one – it is faster to sort your vector by saying the unbelievably ugly

if (!v.empty()) {
    sort(&*v.begin(), &*v.begin() + v.size());
}

than the intended

sort(v.begin(), v.end());

Attempts to impose pseudo-abstractness at the cost of efficiency can be defeated, but at a terrible cost.

Stepanov has a lot of other interesting documents available ，尤其是在“类笔记”部分。

是的，有几个关于 OOP 的经验法则。不，我不相信它们真的是最好的做事方式。当您使用 STL 时，以 STL 兼容的方式做事很有意义。当您的抽象级别较低时(例如 std::vector，它专门用于使动态分配的数组更容易工作；即，它应该像具有一些附加功能的数组一样可用)，那么其中一些 OOP 经验法则就毫无意义了。

回答最初的问题:即使是初学者最终也需要了解迭代器、对象生命周期以及我称之为对象的使用生命周期(即“对象没有超出范围，但不再有效使用，就像一个无效的迭代器”)。我看不出有任何理由试图向用户隐藏这些生活事实，所以我个人不会基于这些理由排除基于迭代器的 API。真正的问题是您的 API 意味着抽象什么以及它意味着公开什么(类似于 vector 是一个更好的数组并且旨在公开其数组性质的事实)。如果您回答这个问题，您应该对基于迭代器的 API 是否有意义有更好的了解。