On May 5, Collin Rogowski said:
>rand(123 - 48) + 48:
I am glad you used this construct (and even moreso that you explained the
mathematics and logic used therein). I have dealt with programmers who
used to write simple mIRC scripts, and expect
rand(10, 30);
to return a number from 10 to 30. Much to their chagrin, this code is
seen by Perl as
rand(30);
because of the way the , operator works in scalar context (for more on
context, please read http://www.pobox.com/~japhy/articles/pm/2000-02.html
"List is a Four-Letter Word").
Even "worse", rand(EXPR) returns a real number greater than or equal to 0
and less than EXPR
0 <= x < EXPR
So technically, rand(10,30), if it "worked", would return something
10 <= x < 30
This is usually not the desired case. So, using rather simple math, we
can construct the desired result.
First, though, you have to see how rand(x+y) and rand(x)+y produce VERY
different results:
0 <= rand(x + y) < x + y
y <= rand(x) + y < x + y
Ok, now to the gooey stuff:
to get a random number from A to B (which can be A or B)
i.e. randBound(5,15) = 5, 6, 7, ..., 13, 14, or 15
0 <= rand(B) + 0 < B + 0
A <= rand(B) + A < B + A
5 <= rand(B) + 5 < B + A
5 <= rand(10) + 5 < 15
5 <= rand(11) + 5 < 16
So we get to:
5 <= rand(15 + 1 - 5) + 5 < 16
which, if $a = 5 and $b = 15, is:
rand($b + 1 - $a) + $a;
to return a number $x such that
$a <= $x < $b+1
And when we take the integer value of this value, it will be
$a <= $x <= $b
So there we are. Now for our function.
sub randBound {
my ($lower, $upper) = @_;
return int( rand($upper + 1 - $lower) + $lower );
}
--
Jeff "japhy" Pinyan [EMAIL PROTECTED] http://www.pobox.com/~japhy/
Are you a Monk? http://www.perlmonks.com/ http://forums.perlguru.com/
Perl Programmer at RiskMetrics Group, Inc. http://www.riskmetrics.com/
Acacia Fraternity, Rensselaer Chapter. Brother #734
