OpenEuphoria

Unobvious (?) technique

• Posted by Byron Hinkle <byron.hinkle at TSBBS.COM> Jul 13, 1997
• 1105 views

Jeff,

J>  for i = 1 to length(s)
J>    if s[i] = keyval then
J>      s[i] = newval
J>    end if
J>  end for
J>
J>  There's another way which seems to be somewhat faster:
J>
J>  s = (newval * (s = keyval)) + (s * (s != keyval))
J>
J>  I know _why_ it works, but when I tried to write out an
J>  explanation, it ran to several pages.  If there are any
J>  mathematicians here, they're welcome to try to explain it in
J>  language that we can all understand...

Interesting, yes? Let me give my interpretation.
Only (s = keyval) or (s != keyval) may be true at any given time.
True is a one, not true is a zero. So it will equate to either
   s = (newval * 1) + (s * 0) which is:  s = newval + 0 or
   s = (newval * 0) + (s * 1) which is:  s = 0 + s
since 0 times anything is zero and 1 times anything is itself.
Now what you have here is a Boolean logic equation.
To use it more easily, you can equate '*' to
mean IF and '+' to mean OR. There is more to it, but just reread
and it says: s = newval IF s = keyval, OR s = s IF s != keyval.

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 * SRP 2.00 #2663 *