If I have a binary number 101010 - as a decimal number 42 - is there an easy FORTRAN function to convert to a binary number and then determine the order of 1 and 0's --
I'm not sure exactly what you mean. All numbers are stored internally as binary. You can use internal write to put the binary format of a number into a character string which you can print or process somehow.
If you want to test whether a particular bit is set or not you should look up BTEST in the help. The help also points to functions that allow the user to clear, set, or swap a particular bit.
integer(4) :: l1 character(32) :: gchar write(gchar,'(B32.32)') l1
which would set the string to '00000000000000000000000000010001' if L1 was 17
program Console1 implicit none character(6) :: gchar character :: a integer :: i = 2 integer :: pos logical :: bool integer(4) :: l1 l1 = 17 write(gchar,'(B6.6)') l1 do pos=6,0,-1 bool = btest(i, pos) if(bool .eq. .TRUE.) then print *, pos, 1 else print *, pos, 0 endif end do end program Console1
In structures a restraint point can have 6 degrees of - each freedom is either set or released ie 1 or 0 is the usual code. One ends up with a file with every joint having 6 integers usually stored as integer a(6) - one ends up trying to remember binary codes. Now AUTODESK has a neat function in AutoCAD for setting some switches - you pick a decimal - 1, 2, 4, 8 and it sets the switch - I was wondering as I developed Borr - see the long discussion top that now runs to 30 pages if printed - I did - if I could use a integer code 0- 63 to cover all the restraint options. The answer is a resounding yes, but one needs a large case statement to step through the combinations - or look up the binary equivalent of the integer directly. So now I can input a full restraint as a 63 and a free as 0, lot less typing and less likely to miss one. The other common one is 111000 or 56. I was not aware of the BTEST function and my INTEL install on the home computer does not give me a nice help system - a WINDOWS 10 upgrade appears to have played with it - no search no nothing, so I asked.
This code is great - thanks
By the use of a binary bit field, you imply that multiple bits can be set. Therefor you would not have a select case, but rather you would have a filter.
if(bitest(Mask, 0)) call DoWork0()
if(bitest(Mask, 1)) call DoWork1()
However note you can create an array of function pointers, and note, if each DoWork can run independently, you could parallelize the code
!$omp parallel do schedule(dynamic, 1)
if(btest(Mask, i)) FunctionPointerArray(i)%DoWork()