Re: Conveniently generating random numbers with TR1 random

From:
"Andrei Alexandrescu (See Website For Email)" <SeeWebsiteForEmail@erdani.org>
Newsgroups:
comp.std.c++
Date:
Sat, 3 Nov 2007 17:19:52 CST
Message-ID:
<472CB5F7.4080609@erdani.org>
Walter E Brown wrote:
[snip]

Thanks for the excellent information.

I see that you've separately now posted a question phrased in terms of
the intermediate document N2032, but I don't know why of all the
references I provided you would choose that one, clearly labeled
"version 2" and equally clearly succeeded by a "version 3" and a
"version 4".

More importantly, your question seems to misconstrue the nature of
"param_type" in the distributions. In particular, you wrote: "I see
with chagrin that the operator()(UniformRandomNumberGenerator& urng,
const param_type& parm) is still there. I don't understand why. The nice
thing is that the function is constrained to a specific integral type
param_type."

(1) I don't know why you believe that param_type is an integral type.
There's no such requirement.

(2) As stated in N1933, param_type and related functions "provide a
uniform interface so that distribution parameters can be manipulated
generically: it now becomes possible to write code that is independent
of any specific distribution."


Ok, I understand.

And finally, let me see if I understand your basic concern about the
interface to a distribution. Is it that you believe that the member
templates of the following form are superfluous?
  template<class URNG>
  result_type
    ..._distribution::operator()(URNG&, param_type const &)
If so, please let me know and I will be very happy to address that
specific issue.


Yes, that's the root of my question. More specifically, I'm not sure
what method to use when generating integers in a range [a, b]. My
understanding is that I could go two routes:

a) Call uniform_int<int>(a, b)(gen)

b) Call a + uniform_int<int>(whatever, whatever)(gen, b + 1)

Why are there two ways, and what are the tradeoffs guiding the choice of
using one of them?

Andrei

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