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hi, i have a big problem :-) in my monte carlo simulations i do have a time warp, how can i convert the integer timer (from another subroutine)to a string timer (character timer) to get the following file output with variable extensions: integer timer common timer...

hi it doesnt work the error message sounds like this: lio: unformatted io not allowed !!! galileo

sorry i forgot something ! i open it without any special "effects" :-) only open(10,file) i know that is the problem, but what is the solution !! thanx in advance

Hi I do open it unformatted. read(10,*) .......... Maybe i must write it out formatted, and read it formatted, but HOW ?? do you have an idea ?

HI what can i do if i have a list of DATA´s in one row that ist larger then 80 characters. I have this problem now !! I have an output of data which are larger then the 80 characters in one row!!! if the end of row is reached, the list continues in the second row (BUT THATS the PROBLEM)...

Hi, i have a problem. I have two vectors, e.g, i(p) and k(q). the two vectors are, e.g.: i(p(i)) = (a,b,c,d,e,f,g) k(q(j)) = (a,b,d,e,g) the letters are also vectors. But now the problem. i need all vectors of k(q), and i want to overwrite the vector i(p). But if you can see, i need also the...

IT WORKS !!!!!!!!!!!! thank you for your help. i wish you a nice week, and weekend ! galileo

ok, here is the complete program: program test integer i,j,k,m(3),n,ksv,a(100),z open(1,FILE='test.out') open(3,FILE='test2.out') ksv = -1 read(1,*) iter do 100 z=1,iter read(1,*) k,m(1),m(2),m(3) if (k.ne.ksv) then c...

the reading is no problem, if i put print *,k,m(1),m(2),m(3) after the read statement. i get the wright output. The Problem is that i only get the last row of the input list. For example i read from the following list: 1 31 3 15 1 3 34 15 1 31 3 55 1 3 34 55 23 22 25 5 23 24 25 5 23...

i think im near the solution: if i kill the following part if (ksv.ne.-1) then print *,ksv,(a(j),j=1,n) n = 0 endif i get this outfile from the following list 1 ...numbbers 1 1 1 2 ...numbers 2 2 2 i get this outfile : 2 .....num,bers i get...

i know where is the problem: if (ksv.ne.-1) then print *,ksv,(a(j),j=1,n) n = 0 endif if i kill this part i think it can work !! do i need this part ?

hi i have further problems i cant understand a aprt of your program if (k.ne.ksv) then if (ksv.ne.-1) then print *,ksv,(a(j),j=1,n) n = 0 endif ksv = k endif i get problems with the a(j). if i start the program i get...

Hi cakiwi, thank you for your program. But I have a problem. during compilation i get this message: do while(1) ^ Type disagreement between expressions at (^) and (^) i hope you know this problem. best regards galielo

hello, i need really help !!! i have implemented a filter, this reads a data file from a Mesh generator and finding of every node its neighbours (3 dimensional). This was difficult enough. But now i get a list of nodes with its neighbours, but some neigbours appears twice or more. For example...