RE: Releasing several Holds simultaneously
- To: mathgroup at smc.vnet.net
- Subject: [mg82300] RE: Releasing several Holds simultaneously
- From: "Andrew Moylan" <andrew.j.moylan at gmail.com>
- Date: Wed, 17 Oct 2007 04:01:09 -0400 (EDT)
- References: <ff1po5$90n$1@smc.vnet.net> <4714B636.7010506@gmail.com>
Thanks Szabolcs!
Replace[expr, Hold[e___] :> e, {0, Infinity}, Heads -> True]
is is the answer I'm looking for. I had tried
ReplaceAll[expr, Hold[e___] :> e, {0, Infinity}, Heads -> True]
which is insufficient because of this property of ReplaceAll
(ref/ReplaceAll):
"The first rule that applies to a particular part is used; no further rules
are tried on that part, or on any of its subparts."
I failed to think of Replace at all.
-----Original Message-----
From: Szabolcs Horv=E1t [mailto:szhorvat at gmail.com]
Sent: Tuesday, 16 October 2007 11:02 PM
To: Andrew Moylan
Subject: [mg82300] Re: Releasing several Holds simultaneously
Andrew Moylan wrote:
> Hold[a := Hold[1]]
>
> How can I release both of these Holds (and thus execute a:=1)
> simultaneously?
>
> ReleaseHold[%] doesn't work; it evaluates a := Hold[1] before the
> other hold is removed.
>
> % /. Hold[x_]:>x does the same thing, because /. only matches once per
part.
This should work for releasing all Holds in expr:
Replace[expr, Hold[e___] :> e, {0, Infinity}, Heads -> True]
--
Szabolcs