Re: FORTRAN style, not OK?

*To*: mathgroup at smc.vnet.net*Subject*: [mg26848] Re: FORTRAN style, not OK?*From*: "Allan Hayes" <hay at haystack.demon.co.uk>*Date*: Thu, 25 Jan 2001 01:13:25 -0500 (EST)*References*: <94m96g$3nd@smc.vnet.net>*Sender*: owner-wri-mathgroup at wolfram.com

Toshi, I'm not famiar with Fortran, but here is what is happening with Mathematica: 1) When v has no value the evaluation steps are v = v - c the rule v = v - c is stored v - c is output the rule is then used on this output to give output (v-c)-c the rule is then used on *this* output to give output ((v-c)-c)-c and so on, until the recursion limit is reached. 2) When v has a value, say p (it in initialized) the evaluation steps are v = v - c the right side is evaluated to give v = p - c the rule v = p - c is stored p -c is output. So in this case we can use v = v -c repeatedly: Clear[v] v = p; Do[v = v - c, {7}] v -7 c + p Allan --------------------- Allan Hayes Mathematica Training and Consulting Leicester UK www.haystack.demon.co.uk hay at haystack.demon.co.uk Voice: +44 (0)116 271 4198 Fax: +44 (0)870 164 0565 "Toshiyuki (Toshi) Meshii" <meshii at mech.fukui-u.ac.jp> wrote in message news:94m96g$3nd at smc.vnet.net... > Hello, > > I found out a case in which I cannot directly use FORTRAN statement in > Mathematica programming. > Here is the case good in FORTRAN. > > v=v-c > > However in Mathematica, it seems that I have to use a trick like this. > > temp=v; > v=temp-c > > Is there more smart way for doing the above? > Please let me know. > ___________________________________ > The final goal for me is to do the following. > > Do[ > Do[ > v[i][j][k] = v[i][j][k] - va[j] > ,{j, 3}] > ,{k, 100}]; > > -Toshi > > >