c++ - boost::variant 单一存储保证

Question

我的目标是保证所有变体类型的单一存储:根据 'never empty' guarantee from Boost::variant ，我们需要覆盖 boost::has_nothrow_copy 每个有界类型。但过了一会儿 该文档提到了一些关于 'boost::blank' 的内容，如果 类型已绑定(bind)，变体将设置该值而不是尝试不抛出 默认复制构造函数。

不清楚的是如果在有界类型列表中添加 boost::blank 将避免覆盖/专门化 has_nothrow_copy 的要求 与其他类型？

Answer 1

我相信这已经很明确了。以下是 boost 文档中的相关部分:

Accordingly, variant is designed to enable the following optimizations once the following criteria on its bounded types are met:

For each bounded type T that is nothrow copy-constructible (as indicated by boost::has_nothrow_copy), the library guarantees variant will use only single storage and in-place construction for T.

If any bounded type is nothrow default-constructible (as indicated by boost::has_nothrow_constructor), the library guarantees variant will use only single storage and in-place construction for every bounded type in the variant. Note, however, that in the event of assignment failure, an unspecified nothrow default-constructible bounded type will be default-constructed in the left-hand side operand so as to preserve the never-empty guarantee.

因为 boost::blank 是不可抛出默认构造的，所以第二个子句适用。听起来 Boost 对这种特殊类型进行了特殊处理，以支持所有其他类型，因此与其未指定将实例化哪个默认可构造类型，不如保证该类型是 boost::blank 如果这是一个选项。