为什么Boost中有两个变体类实现? [英] Why are there two variant class implementations in Boost?
问题描述
Boost似乎具有变体类模板的两种实现:
Boost seems to have two implementations of a variant class template:
Boost很少有(虽然不是闻所未闻)两个包含两个相同概念的东西.为什么变体会发生这种情况?这些变体有何不同?
It is rare (though not unheard of) for Boost two include two takes on the same concept. Why has this happened for variants? How do these variants differ?
推荐答案
第一种变体类模板 boost :: variant 早于C ++ 17的 std :: variant .有关与 std :: variant 的比较，请参见此问题.主要是，差异涉及在变体中构造一个值时引发异常时的处理方式.
The first kind of variant class template, boost::variant, predates C++17's std::variant. See this question for its comparison to std::variant. Mainly, the difference regards what to do when an exception is thrown on construction of a value within the variant.
std :: variant 选择是允许无值状态； boost :: variant 选择是在堆上而不是就地构造新对象，并存储指向该位置的指针.
The std::variant choice is to allow a valueless state; the boost::variant choice is to construct the new object on the heap, not in-place, and store a pointer to that location.
std :: variant2 是后来的版本，它一方面希望遵守C ++ 17 API，另一方面却表达了对其选择的不满在这个问题上.
std::variant2 is a later addition, which, on the one hand, wishes to adhere to the C++17 API, but on the other hand is an expression of dissatisfaction with its choice on this matter.
std :: variant2 选择第三个选项，与之前的两个实现不同:双缓冲.它占据了有问题类型的两倍大小；在未使用的一半中构造一个新值，一旦构造成功-则破坏另一半中的旧值.如果所有类型都不都是可构造的，则这不是必需的，并且 boost :: variant2 实际上不会具有双倍缓冲区.
std::variant2 chooses a third option, different than both previous implementations: double-buffering. It takes up twice the size of the problematic type; constructs a new value in the unused half, and once the construction succeeds - destroys the old value in the other half. If all types are nothrow-constructible, this is not necessary and boost::variant2 will not actually have the double buffer.
此选择意味着 boost :: variant2 不可能毫无价值；实际上，其文档标题强调了这一事实.
This choice means that boost::variant2 can never be valueless; and indeed, its documentation title emphasizes this fact.
这篇关于为什么Boost中有两个变体类实现?的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！
