El viernes, 19 de septiembre de 2014 09:26:09 UTC-5, David P. Sanders escribió: > > > > 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? >
I guess some kind of recursion. I'll give it a go... > > >> >> On Fri, Sep 19, 2014 at 9:42 AM, David P. Sanders <dpsa...@gmail.com> >> 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. >>>> >>> >>
