[no subject]

[...]

    Geoffrey> Doesn't ord("z") convert "z" to numeric and then do the
    Geoffrey> math?

Ah, I see what you're getting at.  Yes the perl example does a
conversion.  The ruby doesn't, however. 

    Geoffrey> Actually, I believe '5' - '4' in C simply does pointer
    Geoffrey> math.

No, C's closer to the Ruby version in this respect in that the
constant 'a' is internally a one byte immediate value (whereas in Perl
the string part of a scalar is always a char* under the hood).

-- 
Fletch                | "If you find my answers frightening,       __`'/|
fletch at phydeaux.org|  Vincent, you should cease askin'          \ o.O'
                      |  scary questions." -- Jules                =(___)=
                      |                                               U


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