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

[Tomte]

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.

[John M. Dlugosz]

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.

[Tomte]

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.