Re: ReplaceAll (/.)

*To*: mathgroup at smc.vnet.net*Subject*: [mg69203] Re: [mg69197] ReplaceAll (/.)*From*: Andrzej Kozlowski <akoz at mimuw.edu.pl>*Date*: Fri, 1 Sep 2006 18:41:01 -0400 (EDT)*References*: <200609011041.GAA25660@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

On 1 Sep 2006, at 11:41, Bruce Colletti wrote: > Re Mathematica 5.2.0.0. > > Why doesn't: > > (a b) /. x_Times -> List @@ x > > return {a,b}? My (flawed) reasoning is that a b has form Times > [a,b], and so x matches to Times[a,b]. List@@x replaces Times with > List to yield List[a,b], or {a,b}...or so it would seem. > > The explanation is somehow linked to the fact that: > > (a b) /. x_Times :> List @@ x > > returns {a,b}, but I don't see why this works but not the above. > > Although List@@(a,b) returns {a,b}, I want to use the above > ReplaceAll function (this example here is taken from a larger > problem). > > Thankx. > > Bruce > You need to use RuleDelayed (:>) rather than Rule (->) because, if you use just Rule, List@@x is first evaluated and gives: List@@x x So all you are actually doing is (a b) /. x_Times -> x Andrzej Kozlowski

**Follow-Ups**:**Re: Re: ReplaceAll (/.)***From:*"Chris Chiasson" <chris@chiasson.name>

**References**:**ReplaceAll (/.)***From:*Bruce Colletti <vze269bv@verizon.net>