Re: Evaluation with UpValues

*To*: mathgroup at smc.vnet.net*Subject*: [mg19429] Re: [mg19355] Evaluation with UpValues*From*: "Richard Finley" <rfinley at medicine.umsmed.edu>*Date*: Tue, 24 Aug 1999 01:29:27 -0400*Sender*: owner-wri-mathgroup at wolfram.com

Ted, Given your example, it certainly seems the downvalue takes precedence over the upvalue. On the other hand, that would seem to make sense given the usual order of evaluation of transformation rules or if you look at the results with Trace. But if you define: g[f[x_]]:= Exp[-x] which defines a downvalue of g[f[x]] associated with g and f/:g[f[x]]:= E + x which defines an upvalue for g[f[x]] associated with f then the upvalue g[f[x]] associated with f will always take precedence over the downvalue g[f[x]] associated with g. Perhaps that is the sense in which upvalues take precedence over downvalues?? regards, RF >>> "Ersek, Ted R" <ErsekTR at navair.navy.mil> 08/20/99 09:09PM >>> Section 7.1.3 of the excellent book "Power Programming With Mathematica The Kernel" by David B. Wagner explains the main evaluation loop. In that Section it indicates UpValues for a symbol are applied before DownValues for the symbol are applied. OK then consider the case below where (f) has both an UpValue and a DownValue defined by the user. At Out[4] the UpValue for (f) was used. In that case the DownValue couldn't be used to evaluate f[t]. Then at Out[5] the DownValue for (f) is used to evaluate f[E] --> 1+E before the kernel checked to see if the UpValue would apply (and it would have). I get the same result using Versions 3 and 4. When the kernel evaluates g[f[E]] it must figure out that the Head is (g) before it evaluates f[E]. After determining that the Head is (g) it checks to see if (g) has any Hold* attributes, and continues with evaluation as required. So by the time the kernel evaluates f[E] it has all the information it needs to know that the UpValue for (f) can be used. However, the DownValue is used instead. Wait a minute, aren't UpValues applied before DownValues are applied? Can someone convince me that the main evaluation loop performs as explained by David Wagner when evaluating In[5] below? Also can someone give a different example that more clearly shows that UpValues are used first? ------------------------------ In[1]:= ClearAll[f,g,t]; f[x_?NumericQ]:=1+x; f/:g[f[x_]]:=x+4; In[4]:= g[f[t]] Ou[4]= 4+t In[5]:= g[f[E]] Out[5]= g[1+E] -------------------- Regards, Ted Ersek For Mathematica tips, tricks see http://www.dot.net.au/~elisha/ersek/Tricks.html