RE: Clearing a symbol known only through a definition

• To: mathgroup at smc.vnet.net
• Subject: [mg66683] RE: [mg66658] Clearing a symbol known only through a definition
• From: Andrew Moylan <andrew.j.moylan at gmail.com>
• Date: Sat, 27 May 2006 03:51:11 -0400 (EDT)
• Sender: owner-wri-mathgroup at wolfram.com

```Thanks Carl.

The function "OwnValues" does allow me to easily solve my problem; but I
can't find any documentation for this function. Do you know why it doesn't
appear in the help system together with e.g. UpValues and DownValues?

I realised a little while after posting this question that my "particular
problem" had that simple recursive solution. (I became interested in the
solution to the first problem while trying to write a (more convoluted)
solution to the "particular" problem.)

-----Original Message-----
From: Carl K. Woll [mailto:carlw at wolfram.com]
To: mathgroup at smc.vnet.net
Subject: [mg66683] Re: [mg66658] Clearing a symbol known only through a definition

Andrew Moylan wrote:
> Suppose I have a symbol "a" that is defined to be equal to another
> symbol. (E.g. It might be that a := b.) How can I clear the value of
> the symbol 'pointed to' by "a", without knowing explicity what symbol that
is?
>
> Clear[a] won't do, of course.
> Clear[Evaluate[a]] won't do, because that will evaluate to
> Clear[Evaluate[b]], which will in turn (in general) evalulate to
> Clear["whatever b evalulates to"].
>

In this case, the function OwnValues is useful. For example:

a:=b

In[42]:= OwnValues[a]

Out[42]= {HoldPattern[a] :> b}

So, a function to clear the symbol pointed to by a could be:

SetAttributes[clearownsymbol, HoldFirst];

clearownsymbol[x_] := Module[{own},
own = OwnValues[x];
If[MatchQ[own, {_ :> _Symbol}],
Clear @@ Extract[own, {1, 2}, Hold]
]
]

a:=b
b=3;

In[48]:= b

Out[48]= 3

In[49]:= clearownsymbol[a]

In[50]:= b

Out[50]= b

> ----
> Alternatively, here is the particular problem I want to solve:
>
> For any expression e, either e is a symbol, or Head[e] is, or
> Head[Head[e]] is, etc. Call this the "topmost symbol" in the expression.
> Thus, the topmost symbol in f[x][y][z] is f.
>
> I want to write a function that takes an expression and calls Clear[]
> on its topmost symbol. Can anyone think of a simple way to write such
> a function?

Unless there is more going on here, I don't see how your first issue relates
to this problem. At any rate, here is a function that does what you want:

f=g;

In[57]:= f

Out[57]= g

In[59]:= f

Out[59]= f

Of course, f[x][y][z] without a hold gets evaluated to

g[x][y][z]

and there is no way to clear f based on the input g[x][y][z].

Carl Woll
Wolfram Research

```

• Prev by Date: RE: Graphics3D axes
• Next by Date: Re: How to make results from Integrate and NIntegrate agree
• Previous by thread: Re: Clearing a symbol known only through a definition
• Next by thread: Using Mathematica to locate a PDE singularity