RE: Strange ReplaceAll behavior
- To: mathgroup at smc.vnet.net
- Subject: [mg36810] RE: [mg36774] Strange ReplaceAll behavior
- From: "Wolf, Hartmut" <Hartmut.Wolf at t-systems.com>
- Date: Thu, 26 Sep 2002 04:57:24 -0400 (EDT)
- Sender: owner-wri-mathgroup at wolfram.com
>-----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: [mg36810] [mg36774] 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