From: nainar <vnainar@...>
Date: 2003-12-07T18:41:26+09:00
Subject: Re: newbie question

On Sun, Dec 07, 2003 at 03:14:28PM +0900, Dmitry V. Sabanin wrote:
> On Sunday 07 December 2003 12:43, nainar wrote:
> >  If function are params are only copies ,  why do  the two  functions below
> >   behave differently:
> >
> > 	def f1(a)
> > 	 a=[9,8,7]
> > 	end
> Because in this case, not object that 'a' refers to is changed, but variable 'a' itself assigned to a new object [9,8,7].
> > 	def f2(b)
> > 	 b[0]=[9,8,7]
> > 	end
> But now, object that 'b' refers to is changed, not variable 'b'.
> 
> > 	v=[1,2,3]
> > 	f1(v)
> > 	 p v    		# prints [1,2,3]  :  'v' not changed
> When variable assigned to a new object, old object still exists. And v holds reference to it, no matter what f1 did.
> 
> > 	f2(v)
> > 	 p v		        # prints [[9,8,7],2,3] !
> 
> -- 
> sdmitry -=- Dmitry V. Sabanin
> MuraveyLabs.
> 
> 
> 

Thanks for prompt reply. But why is a treated differentlyi from b[0] ? .


 In C, parameters are  passed by value , unless pointers to variables are used .

 In Perl, params are passed in the '@_' array and contains references to the original
 variables and assignments to the elements of @_ are reflected  in the original
 vars(Execept when the whole '@_' is assigned to at once ).

 How does it work in Ruby?. I couldn't find anything on this in docs.

v.nainar