Re: Interaction of Remove and Global variables in a Module

*To*: mathgroup at smc.vnet.net*Subject*: [mg120105] Re: Interaction of Remove and Global variables in a Module*From*: "Oleksandr Rasputinov" <oleksandr_rasputinov at hmamail.com>*Date*: Fri, 8 Jul 2011 04:54:39 -0400 (EDT)*References*: <iv45c7$et1$1@smc.vnet.net>

On Thu, 07 Jul 2011 12:30:15 +0100, blamm64 <blamm64 at charter.net> wrote: > This is what I get for querying SetDelayed > > In[1]:= ?SetDelayed > lhs:=rhs assigns rhs to be the delayed value of lhs. rhs is maintained > in an unevaluated form. When lhs appears, it is replaced by rhs, > evaluated afresh each time. >> > > Note particularly the above reads AFRESH EACH time. It appears then > the following is inconsistent behavior based on the above description: > > In[2]:= f[b_]:=Module[{t},a[t_]=b*t^2;] > In[3]:= a[t] > Out[3]= a[t] > In[4]:= f[3] > In[5]:= a[t]//InputForm > Out[5]//InputForm= > 3*t^2 > In[6]:= f[5] > In[7]:= a[t]//InputForm > Out[7]//InputForm= > 5*t^2 > In[8]:= Remove[a] > In[9]:= f[4] > In[10]:= a[t]//InputForm > Out[10]//InputForm= > a[t] > > Apparently AFRESH is not an accurate description of how SetDelayed > operates in this case, or I am missing something about this particular > interaction of Module, Remove, and global variables inside Modules. > > However, if I go back, after executing the last line above (<a> has > been Removed), and place the cursor in the input line where <f> is > defined and hit Enter, which I thought would be identical to just > evaluating <f> AFRESH again, and then execute the <f[4]> line again, > then the global <a> definition is re-constituted. > > The documentation for Remove reads the name is no longer recognized by > Mathematica. My understanding is that if the same name is defined > AFRESH, it will once again be recognized. > > So if anyone would let me know what I am missing, regarding why the > definition of <a> is not created AFRESH each time <f> is evaluated, I > would appreciate it. > > Please don't construe the definition of <f> as my way of > 'parameterizing' a function definition, I just use that definition to > convey the apparent inconsistency. > > -Brian L. This puzzle is explained by noting that after Remove is called, the definition of f is updated to reflect the removal of a: f[b_] := Module[{t}, Removed[a][t_] = b*t^2;] which coincides with the documented behaviour; you are now free to redefine f if you wish the definition to refer to a "new" a rather than the one that was expunged by Remove. The fact that it does not work as might have been anticipated indicates that Remove is not appropriate in this context; Clear or ClearAll would be better choices.