On 10/26/07, S. Newhouse <sen1 at math.msu.edu> wrote:
> >> Why does the 'non-evaluation' of f in this application of sublist work?
> >> It seems that both
> >> sublist([1,2,3,4,5],'f);
> >> and
> >> sublist([1,2,3,4,5],f);
> >> work
> >>
> >> Why does this happen?
Because an unbound symbol evaluates to itself,
and Maxima's apply and map functions recognize a symbol as
a function (i.e. it is not necessary to supply a function object).
> I guess one just has to try various examples to see how to apply these things.
Well, Maxima is somewhat irregular, but the above behavior is
the result of straightforward and widely-used evaluation policies.
(Note that the symmetric difference of {x : straightforwardp(x)} and
{x : widelyusedp(x)} is fairly large in Maxima ....)
HTH
Robert Dodier