Re: Fw: FORTRAN style, not OK?

*To*: mathgroup at smc.vnet.net*Subject*: [mg26890] Re: [mg26806] Fw: FORTRAN style, not OK?*From*: BobHanlon at aol.com*Date*: Fri, 26 Jan 2001 01:27:33 -0500 (EST)*Sender*: owner-wri-mathgroup at wolfram.com

There is only a problem if v is not assigned a value prior to using it in a recursive looking assignment. It is the lack of an initial assignment that causes it to go looking forever. v=5; v = v - c 5 - c v = a; v = v - c a - c SubtractFrom (v = 0; Table[v -= c, {5}]) {-c, -2*c, -3*c, -4*c, -5*c} AddTo (v = 0; Table[v += c, {5}]) {c, 2*c, 3*c, 4*c, 5*c} TimesBy (v = 1; Table[v *= c, {5}]) {c, c^2, c^3, c^4, c^5} DivideBy (v = 1; Table[v /= c, {5}]) {c^(-1), c^(-2), c^(-3), c^(-4), c^(-5)} Clear[v]; Table[v[i,j,k] = 0, {i, 2}, {j,2},{k,2}]; Table[v[i,j,k] = v[i,j,k] - va[j], {i, 2}, {j,2},{k,2}] {{{-va[1], -va[1]}, {-va[2], -va[2]}}, {{-va[1], -va[1]}, {-va[2], -va[2]}}} Bob Hanlon In a message dated 2001/1/24 5:24:18 AM, meshii at mech.fukui-u.ac.jp writes: >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}]; >