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std :: basic_fstream和std :: unique_lock的接口设计

[英]Interface designs of std::basic_fstream and std::unique_lock

原文 2015-11-26 07:09:41 8 1 c++/ c++11/ move/ lazy-initialization

I'm comparing the two classes, because they both associate with something else. 我正在比较这两个类，因为它们都与其他东西相关联。 std::basic_fstream has to associate with a file, while std::unique_lock has to associate with a mutex. std::basic_fstream必须与文件关联，而std::unique_lock必须与互斥体关联。 As such, providing an open() method seems reasonable. 因此，提供open()方法似乎是合理的。 Still, std::unique_lock provides no such methods. 尽管如此， std::unique_lock提供此类方法。 Lazy initialization can be performed anyway via move assignment. 可以通过移动分配执行延迟初始化。 It may thus seem redundant to provide an open() method. 因此，提供open()方法似乎是多余的。 std::basic_fstream , on the other hand, does provide an open() method. 另一方面， std::basic_fstream确实提供了open()方法。 std::basic_fstream exists before C++11, and it is the only way to perform lazy initialization then. std::basic_fstream在C ++ 11之前存在，并且这是执行延迟初始化的唯一方法。 Putting aside backward compatibility considerations, would std::basic_fstream::open() just be removed? 抛开向后兼容性考虑，是否只会删除std::basic_fstream::open() ？ Or it still needs to be there for the operation may fail actually? 还是仍然需要在那里进行操作可能实际上会失败？ Note that, the open (associate) operation always succeeds with std::unique_lock (not to be confused with the lock operation). 请注意，打开（关联）操作始终以std::unique_lock成功（不要与lock操作混淆）。

1 个解决方案

Your last sentence actually contains a hint: " open (associate) operation always succeeds with std::unique_lock ". 您的最后一句话实际上包含一个提示：“ open （关联）操作始终以std::unique_lock成功”。 This makes it convenient to declare the open-ness as class invariant, establish it in the constructor and throw exceptions on rare failures. 这样可以方便地将开放性声明为类不变，在构造函数中建立开放性，并在极少数失败时引发异常。 Compared that to files: Opening them can and does fail, so declaring open-ness as class invariant doesn't work as well there. 与文件相比：打开文件可能并且确实会失败，因此将开放性声明为类不变性在那儿行不通。