Works as designed.
Is that what you were after? Or at least something like it?$ perl -e 'use 5.10.0; say int(6*rand())+1 foreach(1..10);' 6 5 4 6 2 2 4 6 3 5
As MidLifeXis pointed out:
rand EXPR rand Returns a random fractional number greater than or equa +l to 0 and less than the value of EXPR. (EXPR should be posit +ive.) If EXPR is omitted, the value 1 is used. Currently EXP +R with the value 0 is also special-cased as 1 (this was undocu +mented before Perl 5.8.0 and is subject to change in future ve +rsions of Perl). Automatically calls "srand" unless "srand" h +as already been called. See also "srand". Apply "int()" to the value returned by "rand()" if you +want random integers instead of random fractional numbers. +For example, int(rand(10)) returns a random integer between 0 and 9, inclusive. (Note: If your rand function consistently returns numbe +rs that are too large or too small, then your version of Perl w +as probably compiled with the wrong number of RANDBITS.) "rand()" is not cryptographically secure. You should n +ot rely on it in security-sensitive situations. As of this wri +ting, a number of third-party CPAN modules offer random number generators intended by their authors to be cryptographi +cally secure, including: Math::Random::Secure, Math::Random::MT::Perl, and Math::TrulyRandom.
In reply to Re: Random generator fail
by blue_cowdawg
in thread Random generator fail
by Anonymous Monk
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