Restart a loop in Fortran - algorithm

Question

I have an algorithm that looks like this:
10 WRITE (*,*) "Start"
DO I = 1, 10
WRITE (*,*) "Step"
IF(I .EQ. 5) then
go to 10
END IF
END DO
I want to restart the loop, when that if statement executes. However, I don't want to have to use a go to, I tried this:
10 WRITE (*,*) "Start"
DO I = 1, 10
WRITE (*,*) "Step"
IF(I .EQ. 5) then
I = 0; CYCLE
END IF
END DO
But then I get the error that I can't redefine the I variable, inside a loop. So I'm not sure how to approach this. Any help would be much appreciated

Answer 1

A conceptually simple way to phrase this problem is: "I want to repeat a loop until it completes, where there is some abort condition".
This "repeat until it completes" is idiomatically a do construct with indeterminate iterations:
do
... ! Our actions; we exit this outer loop when we are satisfied
end do
[This could also be phrased as a do-while loop.]
With the inner loop:
do
do i=1,10
... ! A conditional statement which aborts the inner loop
... ! And some actions
end do
! And once we're complete we exit the outer loop
end do
Now it's just a matter of handling "abort inner" and "exit outer". Here cycle and exit:
outer: do
print*, 'Start'
do i=1,10
print*, 'Step'
if (...) cycle outer ! Abort the inner loop
end do
exit outer ! The inner loop completed, so we're done
end do outer
The outer loop is labelled so that the cycle statement in the inner loop can reference it. Without that label the cycle will cycle the innermost loop containing it.

Answer 2

Likely a typo here somewhere, but this has some other ideas in a "do while" sense.
i.e. One could poll something coming in.
... USE ISO... etc stuff.
REAL(KIND=C_FLOAT), DIMENSION(10) :: A
LOGICAL(KIND=C_BOOL) :: Question1 = .TRUE.
LOGICAL(KIND=C_BOOL) :: Question2 = .TRUE.
INTEGER(KIND=C_INT) :: Index = 0
INTEGER(KIND=C_INT) :: I = 5
WRITE(*,*)'Enter first index to skip:"
READ(*,*) I
Outer_Loop: DO WHILE (Question1) ! .eq. .TRUE.
Inner_Loop: DO WHILE (Question2) ! .EQV. .TRUE.
Index = Index + 1
IF(Index > UBOUND(A,1)) THEN
Question1 = .FALSE.
Question2 = .FALSE.
Exit
ENDIF
IF(Index == I) EXIT
!Whatever like A(INdex) = something....
ENDDO Inner_Loop
IF(Question1) THEN !We must have more to possibly read or do...
WRITE(*,*)'Do more? 1=yes, 0=No"
READ(*,*) I
IF(I == 1) THEN
WRITE(*,*)'Enter next index to skip:"
READ(*,*) I
Question2 = .TRUE.
!and one can do a reset of I=0 here, which then becomes I=1 the next time through the inner loop...
ELSE
Question1 = .FALSE.
Question2 = .FALSE.
EXIT
ENDIF
ELSE !Must have gotten here to exit at the end of array?
EXIT
ENDIF
ENDDO Outer_Loop

Answer 3

I can suggest to ways of solving this issue: either use a while loop or a recursive function (highly depends on your actual algorithm). I am attaching a sample code for both cases (keep in mind that having variables with save qualifier is not the best idea, it's much better to do that with by passing variables as arguments to function - here it is used just for simplicity)
module functions
implicit none
integer :: restarted, save = 0
contains
recursive subroutine restart_loop
integer :: i
do i = 1, 10
print*, i
if ( i == 5 .and. restarted < 5 ) then
restarted = restarted + 1
call restart_loop
exit
end if
end do
end subroutine restart_loop
subroutine while_loop
integer :: i = 1
do while (i <= 10)
print*, i
if ( i == 5 .and. restarted < 5 ) then
i = 1
restarted = restarted + 1
end if
i = i + 1
end do
end subroutine while_loop
end module functions
program test_prog
use functions
implicit none
call while_loop
end program test_prog