Re: ReplaceAll behavour

*To*: mathgroup at smc.vnet.net*Subject*: [mg82134] Re: ReplaceAll behavour*From*: Szabolcs Horvát <szhorvat at gmail.com>*Date*: Sat, 13 Oct 2007 03:46:15 -0400 (EDT)*References*: <fen6q6$54t$1@smc.vnet.net>

magma wrote: > I do not understand what Mathematica 6 is doing here: > > a=5 > a/.a->3 > > gives 3 > > But... > > a+2/. a->3 > > gives 7 > > Why is that? > > Thank you for your help > Hi, Use Trance[a /. a->3] or On[] and Off[] to find out how Mathematica evaluates an expression. The following output should explain what happens: In[1]:= On[] During evaluation of In[1]:= On::trace: On[] --> Null. >> In[2]:= a=5 During evaluation of In[2]:= Set::trace: a=5 --> 5. >> Out[2]= 5 In[3]:= a/.a->3 During evaluation of In[3]:= a::trace: a --> 5. >> During evaluation of In[3]:= a::trace: a --> 5. >> During evaluation of In[3]:= Rule::trace: a->3 --> 5->3. >> During evaluation of In[3]:= Rule::trace: 5->3 --> 5->3. >> During evaluation of In[3]:= ReplaceAll::trace: a/.a->3 --> 5/.5->3. >> During evaluation of In[3]:= ReplaceAll::trace: 5/.5->3 --> 3. >> Out[3]= 3 In[4]:= (a+2)/.a->3 During evaluation of In[4]:= a::trace: a --> 5. >> During evaluation of In[4]:= Plus::trace: a+2 --> 5+2. >> During evaluation of In[4]:= Plus::trace: 5+2 --> 7. >> During evaluation of In[4]:= a::trace: a --> 5. >> During evaluation of In[4]:= Rule::trace: a->3 --> 5->3. >> During evaluation of In[4]:= Rule::trace: 5->3 --> 5->3. >> During evaluation of In[4]:= ReplaceAll::trace: a+2/.a->3 --> 7/.5->3. >> During evaluation of In[4]:= ReplaceAll::trace: 7/.5->3 --> 7. >> Out[4]= 7 In[5]:= Off[] Szabolcs P.S. This is why it is a bad idea to do things like sqrt[a_?NumericQ] := x /. FindRoot[x^2 == a, {x, 1}] (Even though this function works correctly almost all the time -- one must come up with a really tricky definition for 'x' to make it fail, like x = Sequence[1]) So use Module[] to localise 'x': sqrt[a_?NumericQ] := Module[{x}, x /. FindRoot[x^2 == a, {x, 1}]]