Re: String#upto edge case - empty string causes infinite loop

> -----Original Message-----
> From: Arnaud Bergeron [mailto:[e-mail]] 
> Sent: Tuesday, January 09, 2007 1:36 PM
> To: [e-mail]
> Subject: Re: String#upto edge case - empty string causes infinite loop
> 
> 
> On 1/8/07, Berger, Daniel <[e-mail]> wrote:
> > > -----Original Message-----
> > > From: Arnaud Bergeron [mailto:[e-mail]]
> > > Sent: Monday, January 08, 2007 12:49 PM
> > > To: [e-mail]
> > > Subject: Re: String#upto edge case - empty string causes infinite 
> > > loop
> >
> > <snip>
> >
> > > Since "\377".succ yields "\001\000" I belive it would make more 
> > > sense having "".succ return "\000" and continue on from there.
> >
> > Hm, it does seem to be a tradition among other languages:
> 
> These examples are not exactly the same situation, as none of 
> these languages have the equivalent of succ for strings.
> 
> > # Perl
> > my $str = "";
> > print ++$str; # 1
> 
> Perl is the only language I'm not familiar with here, so I am 
> not really sure what happens here but I heard about automagic 
> conversion of string to numbers.
> 
> > /* C */
> > #include <stdio.h>
> >
> > int main(){
> >    char* empty = "";
> >    printf("Value: %i\n", atoi(empty) + 1); /* 1 */
> >    return 0;
> > }
> 
> Here you are explicitly converting the string to a number 
> (and due to stupidity, atoi interprets "" as being the number 
> 0).  After adding 1 it is no suprise that it prints 1.

Well, I guess my point was that atoi("") returns 0, stupid or not.
 
> > /* Java */
> > public class EmptyTest{
> >    public static void main(String[] args){
> >       String empty = "";
> >       System.out.println(empty + 1); /* 1 */
> >    }
> > }
> 
> Here you get string concatenation.  You could have written "" 
> + "1" and it would have only been more explicit.  The number 
> 1 gets converted to a string and if you concatenate that 
> string with the empty string you get "1".

Oops, right.

> The .succ function is a different beast than these examples 
> because it returns the next string in order.
> 
> > > However there might be a problem because letters are treated 
> > > differently.  "z".succ yield "aa" and not "}" and the 
> same is true 
> > > for "Z" so this might need more thought, but in my 
> opinion yielding 
> > > the empty string once is the last expected behavior.
> >
> > As you suggest, I'm not sure what repurcussions changing the return 
> > value of "".succ might have.  My approach at least has the 
> advantage 
> > of being safe in that regard.  However, I can live with either 
> > approach. It is just an edge case after all. :)
> 
> I would have it return "\000" because "" < "\000" and is the 
> lowest value for a string for which this inegality is true.  
> Having "".succ return "" breaks the implicit assuption that 
> foo < foo.succ.
> 
> irb(main):001:0> a = 1
> => 1
> irb(main):002:0> a < a.succ
> => true
> irb(main):003:0> a = ""
> => ""
> irb(main):004:0> a < a.succ
> => false

I understand your point, but I could just as well argue that there is no
logical successor to an empty string, and therefore it's a special case.
Another point to consider is that, if we have "".succ return "0", we get
this result from String#upto:

"".upto(2) => ["", "0", "1", "2"]

Either approach feels more or less arbitrary. I vote to err on the side
of caution and stick with my original patch. :)

Regards,

Dan


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