使用Fortran在循环内计数

Question

I have some Fortran code that performs a simulation. The elapsed time is stored in et , and the timestep is stored in dt . Both are defined as type real. There is another real variable tot which holds the maximum time the simulation should run. i is a counting variable of type integer. My first attempt was like this:

real, intent(in) :: dt
real, intent(in) :: tot

real :: et
integer :: i

et = 0.0
i = 0
do
   i = i+1
   et = real(i)*dt
   if (et > tot) exit

   ! main code here
end do

I wanted to get rid of i since it was only used in the one place, however, when I tried this, the program hangs when the total time is large:

real, intent(in) :: dt
real, intent(in) :: tot

real :: et

et = 0.0
do
   et = et + dt
   if (et > tot) exit

   ! main code here
end do

What is the difference between the two code samples that causes the program to respond so differently? My compiler is g77.

EDIT: I have added the declarations and initializations to the code samples above.

EDIT 2: The initial values passed to the subroutine are dt = 1e-6 and tot = 100.

Answer 1

I don't know if this is your error since you don't give the whole program, but in the first code, the first thing you do is set et equal to dt , since at that point i=1 . In the second code however, you are using et without having set it (as far as we can guess). Also, dt seems to be used uninitialized. If the bytes at the memory address of et give rise to a large negative float, it may take much longer to reach tot . That's as far as I can think of anything without have more code.

EDIT thanks for the update.

Well in that case I think just read the answer of haraldkl, I think that's your solution. If you need to reach 100 by adding up 1.0e-6 , this isn't going to work for a 4-bytes real, as that only has about 6-7 meaningful digits in base 10. Your first solution is slighly better, since you can reach about 2e9 with a 4-byte int. One solution is to use 8-byte variables. However, you should always build in an extra check (eg if (et > tot .OR. i > max_iter) ) to allow for a maximum of iterations, so you can safe-guard against this, because even if you use the integer solution, if you would make tot larger, your integer might overflow and you will be stuck in an infinite loop too.

Answer 2

如果dt相对于tot非常小，则也可能是因为dt在某一点上是如此之小，以至于将其加到et上就没有影响（数值精度损失），因此et不会增长超出那一点...

Answer 3

It is hard to conclude anything when you give partial code, skip the declarations and instead of showing the error messages you just give your interpretation of them, while it is clear that if you knew how to interpret them correctly you would not be getting them in the first place.

Your second loop differs from the first in several things worth noting: a) what are the values of the variables at the start of the loop, b) what is the loop counter, c) is et real or an integer? ... et cetera

Here are two ways in which these loops can be written

    program various_do_loops
    integer :: i
    real :: et, tot, dt


    ! DO WHILE LOOP (whoops, I just now see you're using g77
    ! so this may or may not work)
    i = 0
    et = 0.
    tot = 10.
    dt = 1.
    do while (et<tot)
        i = i+1
        et = real(i)*dt
    ! main code goes here
    ! ....
    ! ....
    write(*,'("et is currently ", f5.2)')et
    end do

    ! Old kind of WHILE LOOP
    i = 0
    et = 0.
    tot = 10.
    dt = 1.
    10 if(et<tot) then
        i = i + 1
        et = real(i)*dt
    ! main code goes here
    ! ....
    ! ....
    write(*,'("et is currently ", f5.2)')et
    goto 10
    end if

    end program various_do_loops