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Re: Re: Re: Re: Re: Alpha base-26 to base-10... (golf)

by Tomte (Priest)
on Jul 01, 2003 at 16:10 UTC ( #270521=note: print w/ replies, xml ) Need Help??


in reply to Re: Re: Re: Re: Alpha base-26 to base-10... (golf)
in thread Alpha base-26 to base-10...

To get it, try this:

while(<DATA>) { chomp; $erg = b262b10($_); # scalar context is important! print $erg, "\n"; } # equivalent to # sub b262b10{()=a..lc pop} # but with explaining output sub b262b10{@_=a..lc pop; print join " ", @_, "\n";@_} __DATA__ A Z AA AZ BA ZZ AAA ZBA __END__ # without the debug-output: 1 26 27 52 53 702 703 17629

regards,
tomte


Hlade's Law:

If you have a difficult task, give it to a lazy person --
they will find an easier way to do it.


Comment on Re: Re: Re: Re: Re: Alpha base-26 to base-10... (golf)
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Re: Re: Re: Re: Re: Re: Alpha base-26 to base-10... (golf)
by John M. Dlugosz (Monsignor) on Jul 01, 2003 at 18:04 UTC
    Ah, that's lc the lower-case function, not 1c the hex representation of 28.

    But I still don't understand why you can have the following pop without a delimiter after the assignment statement. I think the purpose of the pop is to subtract one from the result.

      I may be mistaken, but I read it the way, that the pop delivers the scalar parameter for lc, (replace it with shift, the result is the same).

      the basic is that you get a list of all values between a and the parameter, evaluated in scalar-context (I had this wrong at first :-), you get the length of the list, which is of course the decimal representation.

      regards,
      tomte


      Hlade's Law:

      If you have a difficult task, give it to a lazy person --
      they will find an easier way to do it.

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