From: Rich Kilmer <rich@...>
Date: 2001-10-29T23:29:46+09:00
Subject: [ruby-talk:23725] Re: ref. constructors and new (was: Re: class Foo does not call Class.new?)

The parser would have to know that when you call:

String.new("text") 

it is a special case because when you call:

Foo.new("text") => Foo.new(String.new("text"))

you don't want:

String.new("text") => String.new(String.new("text"))

I think ;-)

-Rich

> -----Original Message-----
> From: dave@thomases.com [mailto:dave@thomases.com]On Behalf Of Dave
> Thomas
> Sent: Monday, October 29, 2001 9:22 AM
> To: ruby-talk ML
> Subject: [ruby-talk:23724] Re: ref. constructors and new (was: Re: class
> Foo does not call Class.new?)
> 
> 
> matz@ruby-lang.org (Yukihiro Matsumoto) writes:
> 
> > |> They won't (at least String won't).  It will cause infinite loop.
> > |> 
> > |>   "abc" -> String.new("abc") -> String.new(String.new("abc")) ...
> > |
> > |But String.new("abc") would become String.new(aNewStringObject),
> > |wouldn't it?  The literal only gets converted once.
> > 
> > I'm not sure I get what you mean.  Then how literal strings are
> > created, without String.new?
> 
> You showed a chain:
> 
>   "abc" -> String.new("abc") -> String.new(String.new("abc")) ...
> 
> presumably to show that this would recurse indefinitely. But I don't
> think this would be the case.
> 
> When you evaluate a string literal, you call String.new, passing in
> the contents of the literal. This will allocate space for the string
> object, copy in the bytes from the string, set up the encoding, and
> then call #initialize(newString) (that is: the parameter would be
> equal to 'self' within initialize). I don't think there'd be a loop.
> 
> In the case where you explicitly call
> 
>    String.new("abc")
> 
> the literal string would be converted to a String object (as above)
> before being passed to String.new as a parameter. String.new already
> accepts String parameters, so again there'd be no infinite loop.
> 
> I'm probably missing something obvious, though  :)
> 
> 
> Dave