从 boost::shared_ptr 到 std::shared_ptr 的转换?

2021-12-24 00:00:00 c++ boost shared-ptr std

我有一个库,它在内部使用 Boost 的 shared_ptr 版本,并且只公开那些.对于我的应用程序,我想尽可能使用 std::shared_ptr .遗憾的是,这两种类型之间没有直接转换,因为引用计数的内容取决于实现.

I got a library that internally uses Boost's version of shared_ptr and exposes only those. For my application, I'd like to use std::shared_ptr whenever possible though. Sadly, there is no direct conversion between the two types, as the ref counting stuff is implementation dependent.

有没有办法让 boost::shared_ptrstd::shared_ptr 共享同一个引用计数对象?或者至少从 Boost 版本中窃取 ref-count 而只让 stdlib 版本来处理它?

Is there any way to have both a boost::shared_ptr and a std::shared_ptr share the same ref-count-object? Or at least steal the ref-count from the Boost version and only let the stdlib version take care of it?

推荐答案

您可以通过使用析构函数携带引用来将 boost::shared_ptr 内部"携带到 std::shared_ptr 中:

You can carry the boost::shared_ptr "inside" the std::shared_ptr by using the destructor to carry the reference around:

template<typename T>
void do_release(typename boost::shared_ptr<T> const&, T*)
{
}

template<typename T>
typename std::shared_ptr<T> to_std(typename boost::shared_ptr<T> const& p)
{
    return
        std::shared_ptr<T>(
                p.get(),
                boost::bind(&do_release<T>, p, _1));

}

这样做的唯一真正原因是,如果您有一堆需要 std::shared_ptr<T> 的代码.

The only real reason to do this is if you have a bunch of code that expects std::shared_ptr<T>.

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