Does Fortran modify input arguments through subroutine multiple calls? - debugging

Question

I have a subroutine called several times in an if-then-goto loop.The subroutine has two input arguments:
1.is a constant
2.is an array of definite size(i.e 200X1) whose elements change in a do loop right before the subroutine is called.
The problem is that subroutine doesn't understand that change and returns the same results every time it is called (i.e the results of the first time that it is called).It seems as if the values of all variables that are calculated inside the sudroutine are somehow saved and do not change although input atgument no.2 changes..
Is something wrong with my code?Is there a Fortran bug i'm not aware of?
My code looks like this:
PROGRAM calcul.f
REAL aa(100000),dd(100000),mm(100000),yy(100000),hh(100000),mn(100000),ss(100000),ml(100000),m0(100000)
INTEGER N,snv
DOUBLE PRECISION m0(100000),excerpt(1000),k1(100000),sqsum,s2,vk1,mk1,sdk1,v
filelength=610
W=200
OPEN (1,file='filename.dat')
DO i=1,filelength
READ (1,*) aa(i),dd(i),mm(i),yy(i),hh(i),mn(i),ss(i),ml(i),m0(i)
END D0
CLOSE (1)
10 FORMAT(g12.6)
11 FORMAT(I5,1x,g12.6)
c1=1
c2=W
snv=0
14 IF ((c2.LT.filelength).AND.(c1.LT.(filelength-(W-1)))) THEN
DO i=c1,c2
excerpt(i)=m0(i)
END DO
CALL calk1(W,excerpt)
OPEN (3,file='meank1.dat')
READ (3,*) N,mk1
OPEN (2,file='resultsk1.dat')
DO i=1,N
READ (2,*) k1(i)
END DO
sqsum=0.0d0
DO i=1,N
sqsum=dble(sqsum+((k1(i)-mk1)**2))
s2=sqsum
END DO
vk1=(s2)/N
sdk1=dsqrt(vk1)
OPEN (4,file='resultsv.dat')
v=dble(sdk1/mk1)
snv=snv+1
WRITE (4,11) snv,v
CLOSE (2)
CLOSE (3)
mk1=0.0d0
vk1=0.0d0
sdk1=0.0d0
v=0.0d0
c1=c1+1
c2=c2+1
GOTO 14
END IF
CLOSE (4)
END
My subroutine is:
SUBROUTINE calk1(winlength,sm)
DOUBLE PRECISION sm(100000),sumk1,sum,s,x,x2,k1,sk1,mk1
INTEGER snk1
OPEN (2,file='resultsk1.dat')
10 FORMAT (g12.6)
start=1
w=winlength
c3=6
snk1=0
sumk1=0.0d0
sk1=0.0d0
13 IF (c3.LE.w) THEN
l1=1
l2=c3
c=c3
12 IF ((l1.LE.(w-5)).AND.(l2.LE.w)) THEN
sum=0.0d0
s=0.0d0
DO k=l1,l2
sum=sum+sm(k)
s=sum
END DO
av=0.0d0
av2=0.0d0
x=0.0d0
x2=0.0d0
DO k=l1,l2
av=av+dble(((k)/(c))*(sm(k)/s))
av2=av2+dble((((k)/(c))**2)*(sm(k)/s))
x=av
x2=av2
END DO
k1=x2-((x)**2)
sumk1=sumk1+k1
snk1=snk1+1
WRITE (2,10) k1
l1=l1+1
l2=l2+1
k1=0.0d0
GOTO 12
ELSE
c3=c3+1
GOTO 13
END IF
END IF
CLOSE (2)
N=snk1
sk1=sumk1
mk1=dble((sk1)/N)
OPEN (3,file='meank1.dat')
WRITE (3,10) N,mk1
CLOSE (3)
RETURN
END

Answer 1

I haven't attempted to compile the code (I see there are still some errors that would upset a compiler), but I can suggest a problem.
However, the first thing to say is: if this is your code you'll make things much easier for yourself if you use much more modern Fortran features.
You say that excerpt changes each time before the subroutine is entered. This is true, but not in a meaningful way. Let's look at what is happening to the array.
This is all looped:
c1=1
c2=W
DO i=c1,c2
excerpt(i)=m0(i)
END DO
CALL calk1(W,excerpt)
c1=c1+1
c2=c2+1
Well, W isn't changed in an interation. In this first iteration you are (using array syntax) setting excerpt(1:W)=m0(1:W); in the second setting excerpt(2:W+1)=m0(2:W+1), and so on. That is: each time you call calk1, excerpt(1:W) is still exactly m0(1:W) which hasn't changed. The only change to excerpt is after the W-th element, which you suggest won't be used in the subroutine anyway.
As to what you should do instead with that excerpt setting loop, I can't say: it depends on what you want to happen. Perhaps
DO i=c1,c2
excerpt(i-c1+1) = m0(i)
END DO
?
But use modern Fortran instead.

Answer 2

When I try to compile this I get numerous compiler warnings and errors. I suggest using maximum compile-time warning options from our compiler and clean up those problems. That will clear out some problems with minimal effort. With gfortran, try -O2 -ffixed-form -ffixed-line-length-none -W -Wall -pedantic -fimplicit-none -Wsurprising -Waliasing -Wimplicit-interface -Wunused-parameter -fcheck=all -pedantic -fbacktrace.
Indenting your code with help you understand it. And why use FORTRAN 77 in 2014? Fortran 95/2003 is much easier to program in.