Re: ReplaceAll behavour

*To*: mathgroup at smc.vnet.net*Subject*: [mg82161] Re: ReplaceAll behavour*From*: Bill Rowe <readnewsciv at sbcglobal.net>*Date*: Sat, 13 Oct 2007 04:00:09 -0400 (EDT)

On 10/12/07 at 2:59 AM, maderri2 at gmail.com (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? What is happening in each case is evaluation occurs before replacement. =46or a/.a-> evaluation causes this to become 5/.5->3 then the replacement occurs to give the result 3. =46or a+2/.a->3 evaluation gives 5+2/.5->3 Arithmetic operators such as + or - have higher precedence than patter replacement. So, this becomes 7/.5->3. And finally since there is no 5 to replace, the result is 7. You can easily verify this is the way things work by looking at the output from Trace[a/.a->3] and Trace[a+2/.a->3] -- To reply via email subtract one hundred and four