Re: Re: ParrallelDo and set::noval

*To*: mathgroup at smc.vnet.net*Subject*: [mg102744] Re: [mg102697] Re: ParrallelDo and set::noval*From*: Fred Simons <f.h.simons at tue.nl>*Date*: Tue, 25 Aug 2009 01:45:29 -0400 (EDT)*References*: <h63cgg$1b6$1@smc.vnet.net> <h6duf8$h1c$1@smc.vnet.net> <200908191102.HAA26430@smc.vnet.net>

Many have already reported that the correct way for solving the problem is using SetSharedVariable instead of DistributeDefinitions. But part of the original posting was also the observation that DistributeDefinitions was not enough to make the definition available to all subkernels. Though in this form that remark is not correct, there is something strange here that I cannot explain. As an introduction, start with a fresh kernel and execute the following. zzz=12; ParallelEvaluate[zzz=RandomInteger[{0,5}]]; Information[zzz]; ParallelEvaluate[Information[zzz]]; The kernel works in the default context Global` and each subkernel has its own context Global`, independent of the normal Global context. Since I use a dual core computer, my output shows three variables zzz, one in the standard context Global` and two in the two subkernel contexts Global`. The three values are different, so the three variables are different, despite that they have the same name and seem to be in the same context. Now another variant of the original problem, that produces the remarkable Set::noval message. m=Range[5]; DistributeDefinitions[m]; ParallelDo[Print[{i, m, m[[i]]}];m[[i]]=0;Print[m], {i,1,5}] In the output we see which subkernel tried to do what. The message that the symbol m in part assignment does not have an immediate value is very strange, since each subkernel was able to compute both m and m[[i]]. It seems to be the assignment to m[[i]] that generates the message, for here are two pieces of code that do work (and of course produce different values for m in each of the three contexts Global`); they assign to m instead of to m[[i]]. m=Range[5]; DistributeDefinitions[m]; ParallelDo[Print[{i, m, m[[i]]}];m=ReplacePart[m, i->0];Print[m], {i,1,5}] m ParallelEvaluate[m] m=Range[5]; DistributeDefinitions[m]; ParallelDo[Print[{m,i}];m=Flatten[{Take[m, i-1],0, Drop[m, i]}];Print[m], {i,1,5}] m ParallelEvaluate[m] Does someone have an explanation why, in this very theoretical situation, assignment to m is possible and assignment to a part of m not? Fred Simons Eindhoven University of Technology

**References**:**Re: ParrallelDo and set::noval***From:*guerom00 <guerom00@gmail.com>