Comment by pat_shaughnessy | Hacker News Reader

Comment by pat_shaughnessy | Hacker News Reader

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pat_shaughnessy

Great questions! To be honest I don't have enough knowledge of the MRI internals yet to be able to answer these 100% correctly. Maybe I'll write a follow up post or an update to this one explaining these issues when I have time for more research.

But for today:

1. str2 = str doesn't actually create a new string. RString represents a string value, not a string object. So str2=str creates a new RString because that essentially defines what str2's value refers to… Think of RString as an internal pointer to the value that str or str2 is referring to. Sorry - not a good explanation :(

2. How are shared RStrings modified? Not sure yet, but I'm curious to find out. Will let you know on my site somehow.

3. Yes, simply calling str2 = str or in a variety of other ways will do it.

I'm 99% sure you're wrong about 1). The internal pointer is the VALUE that is pointing to the struct RString. `str2 = str` will make another VALUE point to the same struct RString. Consider

  str = "foo"
  str2 = str
  str.object_id == str2.object_id #=> true

I'm fairly sure that implies that str and str2 share everything including the RString object, so it couldn't really have one of str/str2 have a field that points to the other.

I am not sure about 2) and 3) but I'd expect some copy-on-write mechanism based on a flag field in the RBasic struct, and taking substrings might be O(1) thanks to sharing too.

Here's a better explanation for #1:

In Ruby, most classes (including String) are reference types. This is why `str2 = str` doesn't create a new string. It doesn't create a new RString either - str and str2 are both pointers to the same RString struct.

AFAIK str2 = str will just share the RVALUE-Pointer.

I think to actually see shared strings behavior, you would have to do str2 = str.clone or str2 = str[0..-1].

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