Re: unbelievable ...
- To: mathgroup at smc.vnet.net
- Subject: [mg16653] Re: [mg16567] unbelievable ...
- From: Clemens Frey <clemens.frey at bitoek.uni-bayreuth.de>
- Date: Fri, 19 Mar 1999 12:54:03 -0500
- References: <Pine.SOL.4.05.9903180923250.9816-100000@btr0x1.rz.uni-bayreuth.de>
- Sender: owner-wri-mathgroup at wolfram.com
Hi,
this indeed i s believable:
If you have a=A[1,2] set and define like
>
> xASet[A[x_,y_],z_]:=(A[x,y]=A[z,y])
>
> The first use seems ok:
>
> {a,xASet[a,3],a}
> => {A[1,2],Null,A[3,2]}
Then you still have a=A[1,2], but now A[1,2]=A[3,2]. Check this by
typing OwnValues[a] and DownValues[A]! So we have a (invisible)
replacement chain of a=A[1,2]=A[3,2].
>
> But now it is impossible to reset a to A[1,2] (and only to A[1,2] !!!)
>
> {a=A[1,2],a=A[1,5]}
> => {A[3,1],A[1,5]}
Now Trace[a=A[1,2]] and see what happens!
A[1,2] on the left hand side gets replaced by its downvalue A[3,2], so
we have a=A[3,2], and nothing seems to change. But at variance to the
upper no we have an immediate replacement rule a=A[3,2].
If you write a=A[1,5] it works like you expect since A[1,5] stays what
it is.
So to prevent all this stuff, xASet has to be changed, so that the first
parameter is changed and not A[..]. The HoldFirst-attribute has to be
used in order not to evaluate the first parameter:
ClearAll[xASet,a,A];
SetAttributes[xASet,HoldFirst];
xASet[AA_Symbol ,z_] /; MatchQ[AA,A[x_,y_]] := ( AA = A[z,AA[[2]]] )
Now, having
a=A[1,2]
xASet[a,3]
results in no annoying definition for A. DownValues[A] is empty.
hope this helped
Clemens
---------------------------------------------------------
Clemens Frey
Post-Graduate Student/Doctoral Candidate
Department of Mathematics/BITOEK
University of Bayreuth(Germany)
clemens.frey at uni-bayreuth.de
http://www.bitoek.uni-bayreuth.de/~Clemens.Frey
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