C++ - 提升获取问题
有人知道 boost::variant 的 boost::get 是性能消耗操作还是不是。
现在，我正在性能关键部分重构一些旧代码，其中“变体”是由容器为每种可能的类型和相应的枚举实现的。
显然，这很快，但很难看，现在当我必须重构代码以便它可以与另一种类型一起使用时，我想删除旧的代码部分并将其替换为boost::variant。
此外，我不能简单地“分析两个变体并进行比较”，因为这种重构非常痛苦并且非常耗时。
因此，如果有人知道如何boost:: get 与通用的基于枚举的类型调度进行比较，如果您分享这些知识，我将不胜感激。
还有另一种将 boost::variant 与自定义访问者一起使用的变体（如 boost::variant 文档中所述） - 这可能比 boost::variant 更快code>boost::get 就我而言？
谢谢。
Does someone know if the boost::get for the boost::variant is a performance-consuming operation or not.
Right now I am refactoring some old code in a performance-critical part, where "varianting" was implementing by containers for each possible types and corresponding enum.
Obviously, this is fast, but ugly and right now when I have to refactor the code so that it would work with one more type, I want to get rid of that old part of code and replace it with boost::variant.
Also, I can't simple "profile both variants and compare" because this refactoring is a pain in the ass and would be pretty time consuming.
So, if someone knows how does boost::get<x> performs comparing to generic enum-based type dispatching, I would appreciate if you share this knowledge.
There is another variant of using boost::variant<types> with custom visitor (as described in boost::variant documentation) - may this be faster than boost::get in my case?
Thank you.
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您仍然可以编写一个简单的测试应用程序来比较两者，它不必是生产环境。
我的一位同事最近也遇到了与此类似的问题。在他的场景中存在不同类型的对象，但他总是事先知道他期望哪种类型。而且他的数据结构很大，所以内存是一个问题。他通过使用
void *和reinterpret_cast解决了这个问题。这可以防止多态性的内存开销，并且速度非常快。但你必须绝对确定自己在做什么，否则事情就会爆炸。You could still write a simple test-application to compare the two, it doesn't have to be the production environment.
A colleague of mine had a similar problem to this one recently. In his scenario there where objects of different types, but he always knew beforehand which type he expected. Also his data-structure was huge, so memory was an issue. He solved the problem by using
void *andreinterpret_cast. This prevents the memory-overhead of polymorphism and is very fast. You have to be absolutely sure of what you are doing though, otherwise things will explode.看代码，
get<>是使用boost::variant的内部访问者机制实现的。反过来，访问者机制依赖于mpl序列，并且是一步步执行的。这意味着在 n 类型变体上最多执行 n 步，但循环（递归调用）是存在的。同样，正如太空牛仔所建议的那样，进行一次小型性能测试会很有帮助。Looking at the code,
get<>is implemented using the internal visitor mechanism of theboost::variant. In turn, the visitor mechanism relies onmplsequences, and it is executed step by step. That means at most n steps on an n type variant, but the loop (recursive call) is there. Again, as Space Cowboy suggests, a small performance test would be helpful.