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MathGroup Archive 2002

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Re: Strange ReplaceAll behavior

  • To: mathgroup at smc.vnet.net
  • Subject: [mg36835] Re: Strange ReplaceAll behavior
  • From: "Allan Hayes" <hay at haystack.demon.co.uk>
  • Date: Sat, 28 Sep 2002 04:34:35 -0400 (EDT)
  • References: <amuj8l$cf3$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

Hartmut,
I add an explicit illustration to your ingenious solution using With.

Hartmut's solution:

    Clear[g];
    ru[a] = a -> x;
    With[{rule = ru[a], arg = x_}, g[arg] := a + b /. rule];
    g[ c]

        b + c

Why is arg = x_  needed?
Without it we get

    Clear[g];
    With[{rule=ru[a]}, g[x_]:=a+b/.rule];
    g [c]

        b+x

The reason for this shows in

   ?g

        Global`g

        g[x$_] := a + b /. a -> x

The x in x_ has been changed to x$ and there is no x$ on the right side.

This is a general feature of scoping.  Taking it further we get

    Clear[g];
    With[{rule=ru[a]},g[x_]:=a+x/.rule];
    g [c]

        c+x


    ?g

        Global`g

        g[x$_] := a + x$ /. a -> x

--
Allan

---------------------
Allan Hayes
Mathematica Training and Consulting
Leicester UK
www.haystack.demon.co.uk
hay at haystack.demon.co.uk
Voice: +44 (0)116 271 4198
Fax: +44 (0)870 164 0565


"Wolf, Hartmut" <Hartmut.Wolf at t-systems.com> wrote in message
news:amuj8l$cf3$1 at smc.vnet.net...
>
> >-----Original Message-----
> >From: Lawrence A. Walker Jr. [mailto:lwalker701 at earthlink.net]
To: mathgroup at smc.vnet.net
> >Sent: Wednesday, September 25, 2002 7:51 AM
> >Subject: [mg36835]  Strange ReplaceAll behavior
> >
> >
> >Hi,
> >
> >For the life of me I am not sure why the following is not
> >working in my
> >v. 4.2:
> >
> >ru[a]=a->x;
> >f[x_]:=(a+b) /. ru[a];
> >
> >Why do I get
> >f[c] = b+x
> >
> >and not
> >f[c] = b+c?
> >
> >What gives?
> >
> >Thanks,
> >Lawrence
> >
> >--
> >Lawrence A. Walker Jr.
> >http://www.kingshonor.com
> >
>
> Lawrence,
>
> in your definition of f, x doesn't show up explicitely. So, in the
> evaluation sequence, when the definition for f[c] is applied, no x appears
> at rhs i.e.
> (a + b) /. ru[a] and such c cannot be inserted. The result is the same as
> directly executing
>
> In[11]:= (a + b) /. ru[a]
> Out[11]= b + x
>
> If you don't like this, you have to make explicit the Value of ru[a] in
the
> definiton of f. One way to do so is to use Set instead of SetDelayed:
>
> In[9]:= f[x_] = (a + b) /. ru[a]
> Out[9]= b + x
>
> In[10]:= f[c]
> Out[10]= b + c
>
> The drawback of this that not only the value of ru[a] is inserted but also
> the whole expression including ReplaceAll is evaluated. If this is not
> wanted, you have to insert the value of ru[a] into the unevaluated rhs at
> the definition. The general means for this are function application, With
or
> Replace:
>
> In[7]:= (g[x_] := (a + b) /. #) &[ru[a]]
> In[8]:= g[c]
> Out[8]= b + c
>
> In[16]:= Clear[g]
> In[20]:=
> Unevaluated[g[x_] := (a + b) /. rule] /. rule -> ru[a]
> In[21]:= g[c]
> Out[21]= b + c
>
> Here we have to prevent evaluation of the defintion before our rule is
> inserted, this is achieved by Unevaluated.
>
> With is a bit more complicated, since the scoping rules for SetDelayed
would
> not allow the substition of an expression at rhs containing a pattern
> variable (the pattern variable is renamed in this case). A simple answer
to
> this is to also substitute the argument variable (the pattern):
>
>
> In[31]:= Clear[g]
> In[32]:=
> With[{rule = ru[a], arg = x_}, g[arg] := (a + b) /. rule]
> In[33]:= g[c]
> Out[33]= b + c
>
> --
> Hartmut Wolf
>
>




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