Additional examples in base/cartesian
On Friday, September 19, 2014 07:26:09 AM David P. Sanders wrote:
> El viernes, 19 de septiembre de 2014 08:58:56 UTC-5, Isaiah escribió:
> > To do what you want, very briefly:
> > > x.args[3] == :pi && x.args[3] = Expr(:call, :BigFloat, :pi)
> >
> > This will make more sense if you look at `xdump` output to see how Exprs
> > are structured
> >
> > (and `xdump` is quite handy to know about in general). Try:
> > > xdump(:(big(3) + big(pi)))
> >
> > ...
>
> Yes, that's what I need, thanks. Now I need to iterate over the (in
> principle arbitrarily complex, i.e. nested) syntax tree and do this
> everywhere.
> Is there a standard method for this kind of iteration?
>
> > On Fri, Sep 19, 2014 at 9:42 AM, David P. Sanders <[email protected]
> >
> > <javascript:>> wrote:
> >> El viernes, 19 de septiembre de 2014 08:34:05 UTC-5, David P. Sanders
> >>
> >> escribió:
> >>> Hi, for a package I'm writing (or, more precisely, trying to write), I
> >>> need to do the following:
> >>>
> >>> Change `:(3 + pi)`
> >>> into `:(3 + big(pi))`
> >>> or `:(BigFloat(3) + BigFloat(pi))`
> >>
> >> I forgot to say that I need to do this so that I can perform the
> >> calculations with different rounding modes.
> >>
> >>> Basically I need to apply `BigFloat` to every symbol in the expression.
> >>> (I believe that it would be sufficient to wrap `MathConst`s in `big`,
> >>> but
> >>> converting everything to BigFloat is probably a good idea.)
> >>>
> >>> This is my first foray into parsing etc. I'm guessing there's a package
> >>> that allows me to do this easily -- could someone please point it out?
> >>> Or how should I go about this?
> >>>
> >>> In principle the expressions could be more complicated, e.g. `(3 + pi)
> >>> * (2.5 - e)`
> >>>
> >>> Thanks,
> >>> David.
