Subject: The truth about your ED, recent analysis

From: WebM.D. (Suffering from ED) <[email-masked]>
Date: Fri, 05 Dec 2014 06:36:21 -0800

As a man with E.D. is something you pray you never get. And when you do you're in complete shock and humiliation and think that your life is doomed because you can't perform on-demand anymore.

The reason why I'm telling you this is because I was once in your shoes. The embarrassment and humiliation led me to the root cause of ED, which has nothing to do with Low-T levels, according to Columbia University, and Ivy league research.

More importantly, I would like you to understand the cause and how I managed to overcome my ED. 
 http://www.renaesrealm.com/kumiss/fleeted/drenches.php

Thanks.

PS. Remember, it's normal be shocked at first. Believe me we all are.
And it helps to know that is is hope: http://www.renaesrealm.com/kumiss/fleeted/drenches.php

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This message was mailed to [email-masked] as part of community health PSA.
We make it quick to end further messages please write or go to this http://www.renaesrealm.com/barlows/lempira/amnesiacs.htm

MB-Yal Studies 
2529 Forest Lodge Dr
Fayetteville NC__ 28306

Thank you.

I hope you have a good day.

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Any pack immediately followed by an ellipses is just expanded in place. So A<Ts...> is equivalent to A<T1, T2, ..., TN> and hun(vs...) is similarly equivalent to hun(v1, v2, ..., vn). Where it gets complicated is when rather than a pack followed by ellipses you get something like ((expr)...). This will get expanded into (expr1, expr2, ..., exprN) where expri refers to the original expression with any pack replaced with the ith version of it. So if you had hun((vs+1)...), that becomes hun(v1+1, v2+1, ..., vn+1). Where it gets more fun is that expr can contain more than one pack (as long as they all have the same size!). This is how we implement the standard perfect forwarding model;

foo(std::forward<Args>(args)...)
Here expr contains two packs (Args and args are both packs) and the expansion "iterates" over both:

foo(std::forward<Arg1>(arg1), std::forward<Arg2>(arg2), ..., std::forward<ArgN>(argN));
That reasoning should make it possible to quickly walk through your cases for, say, what happens when you call foo(1, 2, '3').

The first one, gun(A<Ts...>::hun(vs)...); expands Ts "in place" and then there's an expression to expand for the last ellipses, so this calls:

gun(A<int, int, char>::hun(1), 
    A<int, int, char>::hun(2),