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From: Jeff Flinn (TriumphSprint2000_at_[hidden])
Date: 2008-02-20 14:49:40

Richard wrote:
> [Please do not mail me a copy of your followup]
> boost-users_at_[hidden] spake the secret code
> <019801c8721d$e1612fc0$6407a80a_at_pdimov2> thusly:
>> It is impossible to overload && in such a way so that in e1 && e2, e2 is not
>> evaluated when e1 is false. But it's possible to overload f && g so that in
>> (f && g)(x) :- f(x) && g(x), g(x) is not evaluated when f(x) is false.
> I guess I'm still a little lost... why does it work for f && g and not
> e1 && e2? Is it because f and g are functors and don't evaluate their
> expressions until their operator() is called?

Yes, and because (f && g) returns another function object, whose
operator() can first evaluate f(x) then decide whether to evaluate g(x).

Jeff Flinn

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