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? >