Fortran program doesn't update a variable's value properly - fortran

Question

I'm trying to implement a Bisection method subroutine in Fortran to solve a computational science program and Fortran is doing some weird things. So the goal of the program is to find the solution of a transcendental equation for some parameter e0 which is updated every step in a for loop and passed to the subroutine.
The problems are:
e0 is not updated as it should. The value of e0 goes from 0.1799... to 0.8999... in one single iteration, while in theory it should take the program 9 interations to get there. Why is this happening?
The print statements aren't printed as they should (see output below). We expect an "out 1" print, some "in" prints (printed when subroutine f1 is called by subroutine Bisection some times), then "out 2" print (with the new e0 value), some "in" prints, etc. But we see the first "out 1" print, some "in" prints and then only "out" prints. Does this mean the subroutine f1 is only called between "out 1" and "out 2"? (it should be called between every "out" print)
I have used Fortran77 for numerical solving for a couple years and never encountered something like this, but it's been nearly 6 months since I don't program anything so maybe I missed an important thing.
Code:
program roots
implicit none
double precision A,B,eps,e0,e1
integer i,niter
external f1
A = 1d-1
B = 9d-1
eps = 10d-6
open(9,file='dades.dat',status='old')
do i=1,9
e0 = i*(B-A)/10 + A
c Here is the first print. 'Out' meaning outside the subroutine
print *, 'out', i, e0
call Bisection(A,B,eps,f1,niter,e1,e0)
write (9,'(2(f10.5))') e0,e1
end do
close(9)
end program roots
subroutine Bisection(A,B,eps,f,niter,xroot,e0)
implicit none
double precision A,B,eps,xroot,fuc,fua,e0
integer niter,i
niter = nint(log((B-A)/eps)/log(2.))+1
do i=1,niter
xroot = (A+B)/2
c Here the subroutine which uses e0 is called twice
call f(xroot,fuc,e0)
call f(A,fua,e0)
if (fuc .eq. 0) return
if (fuc*fua .lt. 0.) then
B = xroot
else
A = xroot
end if
end do
return
end subroutine Bisection
subroutine f1(x,f,e0)
implicit none
double precision x,f,e0
c Here is the second print. 'In' meaning inside the subroutine
print *, 'in', e0
f = e0+1
end subroutine f1
Output:
out 1 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
in 0.17999999999999999
out 2 0.89999511718750003
out 3 0.89999572753906254
out 4 0.89999633789062505
out 5 0.89999694824218746
out 6 0.89999755859374997
out 7 0.89999816894531248
out 8 0.89999877929687500
out 9 0.89999938964843751

Answer 1

I have finally found out the issue after a deep look through the code. The problem is that variables A and B are both input and output because they are modified inside the subroutine. Because of that, when the subroutine is called the first time, it changes A and B's value (which is different from the one set at the beginning of the code). The rest are issues originated by this.
The solution was to reset the values at the beginning each iteration as shown below:
program roots
implicit none
double precision A,B,eps,e0,e1
integer i,niter
external f1
eps = 10d-6
open(9,file='dades.dat',status='old')
do i=1,9
A = 1d-1
B = 9d-1
e0 = i*(B-A)/10 + A
c Here is the first print. 'Out' meaning outside the subroutine
print *, 'out', i, e0
call Bisection(A,B,eps,f1,niter,e1,e0)
write (9,'(2(f10.5))') e0,e1
end do
close(9)
end program roots
subroutine Bisection(A,B,eps,f,niter,xroot,e0)
implicit none
double precision A,B,eps,xroot,fuc,fua,e0
integer niter,i
niter = nint(log((B-A)/eps)/log(2.))+1
do i=1,niter
xroot = (A+B)/2
c Here the subroutine which uses e0 is called twice
call f(xroot,fuc,e0)
call f(A,fua,e0)
if (fuc .eq. 0) return
if (fuc*fua .lt. 0.) then
B = xroot
else
A = xroot
end if
end do
return
end subroutine Bisection
subroutine f1(x,f,e0)
implicit none
double precision x,f,e0
c Here is the second print. 'In' meaning inside the subroutine
print *, 'in', e0
f = e0+1
end subroutine f1