What is this number? (FORTRAN)

[wyosteve]

Hi,
I'm writing a program in FORTRAN (my first) to find E and B fields due to a spinning sphere of charge. Anyways when printing out E and B field magnitudes I am getting numbers like this: 0.1773372103+245
What the heck does this mean!? Is it like 0.1773372103 x 10^245?
Thanks for any enlightenment!

 

[Bill Simpson]

The old-school FORTRAN formatting was 0.1773372103E245

But we don't know which FORTRAN compiler you are using, we cannot see what your format statement was, if you used one, we don't know what your write statement was, if you used one, etc.

Perhaps for your first FORTRAN program before this program, try assigning a real variable a value and printing it out to see the result. Then try making changes to the value and repeat until you can convince yourself you are sure you can depend on output. If you can find some value that results in output like you have above then that might help explain what is going on.

 

[wyosteve]

Fortran 90
Here's a text file of my program. Large portions of it are copied from an example code my prof gave us for a similar one dimensional problem. I'll admit I know nothing about formatting (nor do I have time to learn that in detail).

 

[wyosteve]

here's the code

 

[AlephZero]

What the heck does this mean!? Is it like 0.1773372103 x 10^245?

Short answer: yes.

Longer answer: In the first versions of Fortran, the exponent field was defined to be 4 characters long, and formatted as E+dd or E-dd (where each "d" is a decinal digit).

When computer hardware developed so that floating point numbers might have 3-digit exponents, the format options were changed to E+dd, E-dd or +0dd, -0dd for 2-digit exponents, and +ddd, -ddd for 3-digit exponents.

That still wasn't enough for some computer hardware (e.g. the Cray supercomputers could represent numbers with 4-digit expoinents) so the format specifier was also given a new option Ew.dEe where the w and d are as before, and e specifies the number of digits printed in the exponent.

 

[wyosteve]

ok great that's what I was hoping to hear! of course that means something is defiantly wrong with my program but at least now I know. Thanks

 

[dlgoff]

Short answer: yes.

Longer answer: In the first versions of Fortran, the exponent field was defined to be 4 characters long, and formatted as E+dd or E-dd (where each "d" is a decinal digit).

When computer hardware developed so that floating point numbers might have 3-digit exponents, the format options were changed to E+dd, E-dd or +0dd, -0dd for 2-digit exponents, and +ddd, -ddd for 3-digit exponents.

That still wasn't enough for some computer hardware (e.g. the Cray supercomputers could represent numbers with 4-digit expoinents) so the format specifier was also given a new option Ew.dEe where the w and d are as before, and e specifies the number of digits printed in the exponent.

You know you're showing your age here?

 

[Mark44]

Mod note: Copied code in attachment to here.

here's the code

program ComputeEB
!
!
implicit none
integer,parameter :: iconst = 6         ! integer used to control the resolution
integer,parameter :: Nz = iconst*5		! half the number of zones in each domain
integer,parameter :: ntot = 3*Nz        ! total number of timesteps
integer,parameter :: printnum = iconst  ! number of timesteps between data printout
integer,parameter :: c = 3E8			! speed of prop in a vacuum
REAL, PARAMETER :: pi = 3.1415927
integer :: i,j,k, n, Nzh, numout
double precision :: x(-Nz:Nz), y(-Nz:Nz), z(-Nz:Nz)
double precision :: Ex(-Nz:Nz), Bx(-Nz:Nz), Ey(-Nz:Nz), By(-Nz:Nz), Ez(-Nz:Nz), Bz(-Nz:Nz) 
double precision :: Exh(-Nz:Nz), Bxh(-Nz:Nz), Eyh(-Nz:Nz), Byh(-Nz:Nz), Ezh(-Nz:Nz), Bzh(-Nz:Nz) 
double precision :: Jx(-Nz:Nz), Jy(-Nz:Nz), Jz(-Nz:Nz)
double precision :: v, L, dt, h, twoh, dth, ed, time, Energy
double precision :: R, rho, omega, Toff, epsilon, mu
character(len=20) :: filenum
!
!
!
! Input parameters and constants
L = 0.1d0			! length of half the domain
h = L/Nz			! length of each zone
twoh = 2.0d0*h 		! 2h
dt = 0.25d0*h/c		! timestep
write(*,*) 'stuff = ', v, h, dt
dth = 0.5d0*dt
Nzh = Nz/2
R = .025d0
rho = 1E10
omega = 1E-3
Toff = 8E-11
epsilon = 8.85E-12
mu = pi*4E-7
!
!  Set up the 3D matrix
do i = -Nz, Nz
  x(i) = i*h
  y(i) = i*h
  z(i) = i*h
enddo
!
! Initial conditions
do i = -Nz, Nz
  Ex(i) = 0.0d0
  Ey(i) = 0.0d0
  Ez(i) = 0.0d0
  Bx(i) = 0.0d0
  By(i) = 0.0d0
  Exh(i) = 0.0d0
  Eyh(i) = 0.0d0
  Ezh(i) = 0.0d0
  Bxh(i) = 0.0d0
  Byh(i) = 0.0d0
  Bzh(i) = 0.0d0
  Jx(i) = 0.0d0
  Jy(i) = 0.0d0
  Jz(i) = 0.0d0
enddo
open(unit=11,file='Ex0')
do i = -Nz, Nz
    write(11,111) x(i), Ex(i) ! write data to file
enddo
close(11)
open(unit=11,file='Ey0')
do i = -Nz, Nz
    write(11,111) y(i), Ey(i) ! write data to file
enddo
close(11)
open(unit=11,file='Ez0')
do i = -Nz, Nz
    write(11,111) z(i), Ez(i) ! write data to file
enddo
close(11)
open(unit=11,file='Bx0')
do i = -Nz, Nz
    write(11,111) x(i), Bx(i) ! write data to file
enddo
close(11)
open(unit=11,file='By0')
do i = -Nz, Nz
    write(11,111) y(i), By(i) ! write data to file
enddo
close(11)
open(unit=11,file='Bz0')
do i = -Nz, Nz
    write(11,111) z(i), Bz(i) ! write data to file
enddo
close(11)
!
! Evolve	
time = 0.0d0
numout = 0
do n = 1, ntot ! Compute E and B
  do i = -Nz+1, Nz-1
    do j = -Nz+1, Nz-1
      do k = -Nz+1, Nz-1
        ! determine if we are within the current
        if (time < Toff .AND. sqrt(x(i)**2+y(j)**2+z(k)**2) < R) then
          Jx(k) = (-y(j)/2)*rho*omega*(1+cos(pi*sqrt(x(i)**2+y(j)**2+z(k)**2)/R))
          Jy(k) = (x(i)/2)*rho*omega*(1+cos(pi*sqrt(x(i)**2+y(j)**2+z(k)**2)/R))
        endif
        ! advance a half timestep
        ! E
        Exh(k) = Ex(k) + (dt/(4*epsilon*mu*h))*(Bz(j+1)-Bz(j-1)-By(k+1)+By(k-1))&
			& - (dt/(2*epsilon))*Jx(k)
        Eyh(k) = Ey(k) + (dt/(4*epsilon*mu*h))*(Bx(k+1)-Bx(k-1)-Bz(i+1)+Bz(i-1))&
			& - (dt/(2*epsilon))*Jy(k)
        Ezh(k) = Ex(k) + (dt/(4*epsilon*mu*h))*(By(i+1)-By(i-1)-Bx(j+1)+Bx(j-1))&
			& - (dt/(2*epsilon))*Jz(k)
        ! B
        Bxh(k) = Bx(k) - (dt/(4*h))*(Ez(j+1)-Ez(j-1)-Ey(k+1)+Ey(k-1))
        Byh(k) = By(k) - (dt/(4*h))*(Ex(k+1)-Ex(k-1)-Ez(i+1)+Ez(i-1))
        Bzh(k) = Bz(k) - (dt/(4*h))*(Ey(i+1)-Ey(i-1)-Ex(j+1)+Ex(j-1))
        ! advance a full timestep
        ! E
        Ex(k) = Ex(k) + (dt/(2*epsilon*mu*h))*(Bzh(j+1)-Bzh(j-1)-Byh(k+1)+Byh(k-1))&
			& - (dt/(2*epsilon))*Jx(k)
        Ey(k) = Ey(k) + (dt/(2*epsilon*mu*h))*(Bxh(k+1)-Bxh(k-1)-Bzh(i+1)+Bzh(i-1))&
			& - (dt/(2*epsilon))*Jy(k)
        Ez(k) = Ex(k) + (dt/(2*epsilon*mu*h))*(Byh(i+1)-Byh(i-1)-Bxh(j+1)+Bxh(j-1))&
			& - (dt/(2*epsilon))*Jz(k)
        ! B
        Bx(k) = Bx(k) - (dt/(2*h))*(Ezh(j+1)-Ezh(j-1)-Eyh(k+1)+Eyh(k-1))
        By(k) = By(k) - (dt/(2*h))*(Exh(k+1)-Exh(k-1)-Ezh(i+1)+Ezh(i-1))
        Bz(k) = Bz(k) - (dt/(2*h))*(Eyh(i+1)-Eyh(i-1)-Exh(j+1)+Exh(j-1))
      enddo
    enddo
  enddo
  time = time + dt
  write(*,*) 'step = ', n, ' time = ', time 
  ! Print out data if n equals a multiple of printnum 
  if (n .eq. (n/printnum)*printnum) then  
    numout = numout + 1                     ! increment numout (number of print-outs)
    write(filenum,*) numout                 ! write integer numout to character filenum
    filenum = adjustl(filenum)              ! remove leading spaces
    open(unit=33,file='Ex'//trim(filenum))	! create file with appropriate filename
    do i = -Nz, Nz
      write(33,111) x(i), Ex(i)         		! write data to file
    enddo
    close(33)
    open(unit=33,file='Ey'//trim(filenum))	! create file with appropriate filename
    do i = -Nz, Nz
      write(33,111) y(i), Ey(i)         		! write data to file
    enddo
    close(33)
    open(unit=33,file='Ez'//trim(filenum))	! create file with appropriate filename
    do i = -Nz, Nz
      write(33,111) z(i), Ez(i)         		! write data to file
    enddo
    close(33)
    open(unit=33,file='Bx'//trim(filenum))	! create file with appropriate filename
    do i = -Nz, Nz
      write(33,111) x(i), Bx(i)         		! write data to file
    enddo
    close(33)
    open(unit=33,file='By'//trim(filenum))	! create file with appropriate filename
    do i = -Nz, Nz
      write(33,111) y(i), By(i)         		! write data to file
    enddo
    close(33)
    open(unit=33,file='Bz'//trim(filenum))	! create file with appropriate filename
    do i = -Nz, Nz
      write(33,111) z(i), Bz(i)         		! write data to file
    enddo
    close(33)
  endif
enddo	
!	 
111 format(2E20.10)
555 format(2E20.10)
999 format(2E20.10)		
end program

 

[Mark44]

Hi,
I'm writing a program in FORTRAN (my first) to find E and B fields due to a spinning sphere of charge. Anyways when printing out E and B field magnitudes I am getting numbers like this: 0.1773372103+245
What the heck does this mean!? Is it like 0.1773372103 x 10^245?
Thanks for any enlightenment!

If you're getting numbers like these, you are probably printing uninitialized variables.

 

[wyosteve]

yeah I realized that! (along with lots of other issues, haha) thanks