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Re: ClearAll error message

  • To: mathgroup at smc.vnet.net
  • Subject: [mg92428] Re: ClearAll error message
  • From: Jens-Peer Kuska <kuska at informatik.uni-leipzig.de>
  • Date: Tue, 30 Sep 2008 21:51:05 -0400 (EDT)
  • Organization: Uni Leipzig
  • References: <gbnpep$pr3$1@smc.vnet.net> <gbqcv3$ebb$1@smc.vnet.net> <gbt2tq$lgc$1@smc.vnet.net>
  • Reply-to: kuska at informatik.uni-leipzig.de

Hi,

can it be the case that the first part of the message below has nothing
to do with the question you have posted here ?

And

ClearAll[Symbol["A"]]

can't work because to get the symbol A the argument must evaluated
but Mathematica has an infinite evaluation loop and it can't stop
the evaluation after the first time so that
Symbol["A"] would evaluate to A and this would evaluated to the value
of A.

That's why ClearAll[] also accept strings and ClearAll["A"] work as 
expected.

Regards
   Jens


carlos at colorado.edu wrote:
> Actually this was not a big deal.  The group working with the program was
> puzzled by the message but soon it was fixed. BTW, one student came
> with an elegant one-liner to create a nxn symmetric array:
> 
>   S=Table[rootname[Min[i,j],Max[i,j]],{i,n},{j,n}]
> 
> The Clear behavior seems largely a question of esthetics.  One of the basic
> rules of functional programming is function composition equivalence:
> 
>         y:=f(x); z:=g(y)    equiv    z:=g(f(x))
> 
> Since   name=Symbol["A"]; ClearAll[name] is not equivalent
> to   ClearAll[Symbol["A"]] the rule is violated here, and likewise
> for any function with HoldAll attribute.  Correct?
> 


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