> Why would an allocation require an atomic write for a reference count?

It won't always require it, but it usually will because you have to ensure the memory containing the reference count is correctly set before handing off a pointer to the item. This has to be done almost first thing in the construction of the item.

It's not impossible that a smart compiler could see and remove that initialization and destruction if it can determine that the item never escapes the current scope. But if it does escape it by, for example, being added to a list or returned from a function, then those two atomic writes are required.