From: "Iñaki Baz Castillo" <ibc@...>
Date: 2009-12-07T01:41:11+09:00
Subject: Re: Why doesn't Ruby "compile" strings?

El Domingo, 6 de Diciembre de 2009, Robert Klemme escribió:
> On 06.12.2009 11:17, Brian Candler wrote:
> > Iñaki Baz Castillo wrote:
> >>> When you are using the string literals, the interpreter doesn't know
> >>> what you are going to do with that string literal, so it's not really
> >>> safe for it to assume that it can use a single ruby object to represent
> >>> all instances of it.
> >>
> >> Why not? It's obviously a string writen in the script, with no variables
> >> into
> >> it and so...
> >
> > Suppose you do:
> >
> >   10.times do
> >     puts "hello"
> >   end
> >
> > The interpreter has no way of knowing that puts does not mutate the
> > argument passed to it. This is a silly example, but you might have done:
> >
> >   alias :old_puts :puts
> >   def puts(x)
> >     old_puts x
> >     x.replace "rubbish"
> >   end
> >
> > So it is forced to create a new string object each time round the loop.
> >
> > It's a shame that Ruby doesn't have immutable strings. Symbols are the
> > closest, but they have different semantics to strings.
> >
> > If the literal syntax "xxx" gave a *frozen* String, then it would be
> > safe to re-use it. But then if you wanted to append to a string, you'd
> > have to write:
> >
> >   a = "".dup
> >   a << "stuff"
> >
> > or perhaps
> >
> >   a = String.new
> >   a << "stuff"
> 
> Another thing you could not do with the auto interning (as with Java
> String constants):
> 
> ...each do |whatever|
>    s = "intro " << whatever << " outro"
>    store_away(s)
> end
> 
> Ruby leaves the decision up to you where you want to optimize while
> still keeping things nice for other use cases.  The code above would
> have to look like this if "" and '' would not construct new objects:
> 
> ...each do |whatever|
>    s = "intro ".dup << whatever << " outro"
>    store_away(s)
> end
> 
> Now, that looks worse IMHO.
> 
> Kind regards
> 
> 	robert
> 
> 
> PS: Note also that all strings created via "" and '' do share the
> internal character buffer until one of them is modified (copy on write)
> so it could be more inefficient as it actually is. :-)

Ok, thanks a lot to all for so good explanations. 100% understood now :) 


-- 
Iñaki Baz Castillo <ibc@aliax.net>