So I am using the taylor series to calculate sin(0.75) in fortran 90 up until a certain point, so I need to run it in a do while loop (until my condition is met). This means I will need to use a factorial, here's my code:
program taylor
implicit none
real :: x = 0.75
real :: y
integer :: i = 3
do while (abs(y - sin(0.75)) > 10.00**(-7))
i = i + 2
y = x - ((x**i)/fact(i))
print *, y
end do
end program taylor
Where i've written fact(i) is where i'll need the factorial. Unfortunately, Fortran doesn't have an intrinsic ! function. How would I implement the function in this program?
Thanks.
The following simple function answers your question. Note how it returns a real, not an integer. If performance is not an issue, then this is fine for the Taylor series.
real function fact(n)
integer, intent(in) :: n
integer :: i
if (n < 0) error stop 'factorial is singular for negative integers'
fact = 1.0
do i = 2, n
fact = fact * i
enddo
end function fact
But the real answer is that Fortran 2008 does have an intrinsic function for the factorial: the Gamma function. For a positive integer n, it is defined such that Gamma(n+1) == fact(n).
(I can imagine the Gamma function is unfamiliar. It's a generalization of the factorial function: Gamma(x) is defined for all complex x, except non-positive integers. The offset in the definition is for historical reasons and unnecessarily confusing it you ask me.)
In some cases you may want to convert the output of the Gamma function to an integer. If so, make sure you use "long integers" via INT(Gamma(n+1), kind=INT64) with the USE, INTRINSIC :: ISO_Fortran_env declaration. This is a precaution against factorials becoming quite large. And, as always, watch out for mixed-mode arithmetic!
Here's another method to compute n! in one line using only inline functions:
product((/(i,i=1,n)/))
Of course i must be declared as an integer beforehand. It creates an array that goes from 1 to n and takes the product of all components. Bonus: It even works gives the correct thing for n = 0.
You do NOT want to use a factorial function for your Taylor series. That would meant computing the same terms over and over. You should just multiply the factorial variable in each loop iteration. Don't forget to use real because the integer will overflow quickly.
See the answer under the question of your schoolmate Program For Calculating Sin Using Taylor Expansion Not Working?
Can you write the equation which gives factorial?
It may look something like this
PURE FUNCTION Bang(N)
IMPLICIT NONE
INTEGER, INTENT(IN) :: N
INTEGER :: I
INTEGER :: Bang
Bang = N
IF(N == 2) THEN
Bang = 2
ELSEIF(N == 1) THEN
Bang = 1
ELSEIF(N < 1) THEN
WRITE(*,*)'Error in Bang function N=',N
STOP
ELSE
DO I = (N-1), 2, -1
Bang = Bang * I
ENDDO
ENDIF
RETURN
END FUNCTION Bang
Related
Is there a simple and quick way to multiply a column of a matrix with element of a vector. We can do this explicitly,
program test
integer :: x(3,3), y(3), z(3,3)
x = reshape([(i,i=1,9)],[3,3])
y = [1,2,3]
do i=1,3
z(:,i) = x(:,i) * y(i)
print *, z(:,i)
enddo
end program test
Is there a way to perform the do loop in one line. For example in Numpy python we can do this to do the job in one shot
z = np.einsum('ij,i->ij',x,y)
#or
z = x*y[:,None]
Try
z = x * spread(y,1,3)
and if that doesn't work (no Fortran on this computer so I haven't checked) fiddle around with spread until it does. In practice you'll probably want to replace the 3 by size(x,1) or suchlike.
I expect that this will cause the compiler to create temporary arrays. And I expect it will be easy to find situations where it underperforms the explicit looping scheme in the question. 'neat' one-liners often have a cost in both time and space. And often tried-and-trusted Fortran approach of explicit looping is the one to go with.
Why replace clear easy to read code with garbage?
program test
implicit none
integer i,j
integer :: x(3,3), y(3), z(3,3)
x = reshape([(i,i=1,9)],[3,3])
y = [1,2,3]
z = reshape ([((x(j,i)*y(i) ,j=1,3),i=1,3)], [3,3])
print *, z(1,:)
print *, z(2,:)
print *, z(3,:)
end program test
I am converting f77 code to f90 code, and part of the code needs to sum over elements of a 3d matrix. In f77 this was accomplished by using 3 loops (over outer,middle,inner indices). I decided to use the f90 intrinsic sum (3 times) to accomplish this, and much to my surprise the answers differ. I am using the ifort compiler, have debugging, check-bounds, no optimization all turned on
Here is the f77-style code
r1 = 0.0
do k=1,nz
do j=1,ny
do i=1,nx
r1 = r1 + foo(i,j,k)
end do
end do
end do
and here is the f90 code
r = SUM(SUM(SUM(foo, DIM=3), DIM=2), DIM=1)
I have tried all sorts of variations, such as swapping the order of the loops for the f77 code, or creating temporary 2D matrices and 1D arrays to "reduce" the dimensions while using SUM, but the explicit f77 style loops always give different answers from the f90+ SUM function.
I'd appreciate any suggestions that help understand the discrepancy.
By the way this is using one serial processor.
Edited 12:13 pm to show complete example
! ifort -check bounds -extend-source 132 -g -traceback -debug inline-debug-info -mkl -o verify verify.f90
! ./verify
program verify
implicit none
integer :: nx,ny,nz
parameter(nx=131,ny=131,nz=131)
integer :: i,j,k
real :: foo(nx,ny,nz)
real :: r0,r1,r2
real :: s0,s1,s2
real :: r2Dfooxy(nx,ny),r1Dfoox(nx)
call random_seed
call random_number(foo)
r0 = 0.0
do k=1,nz
do j=1,ny
do i=1,nx
r0 = r0 + foo(i,j,k)
end do
end do
end do
r1 = 0.0
do i=1,nx
do j=1,ny
do k=1,nz
r1 = r1 + foo(i,j,k)
end do
end do
end do
r2 = 0.0
do j=1,ny
do i=1,nx
do k=1,nz
r2 = r2 + foo(i,j,k)
end do
end do
end do
!*************************
s0 = 0.0
s0 = SUM(SUM(SUM(foo, DIM=3), DIM=2), DIM=1)
s1 = 0.0
r2Dfooxy = SUM(foo, DIM = 3)
r1Dfoox = SUM(r2Dfooxy, DIM = 2)
s1 = SUM(r1Dfoox)
s2 = SUM(foo)
!*************************
print *,'nx,ny,nz = ',nx,ny,nz
print *,'size(foo) = ',size(foo)
write(*,'(A,4(ES15.8))') 'r0,r1,r2 = ',r0,r1,r2
write(*,'(A,3(ES15.8))') 'r0-r1,r0-r2,r1-r2 = ',r0-r1,r0-r2,r1-r2
write(*,'(A,4(ES15.8))') 's0,s1,s2 = ',s0,s1,s2
write(*,'(A,3(ES15.8))') 's0-s1,s0-s2,s1-s2 = ',s0-s1,s0-s2,s1-s2
write(*,'(A,3(ES15.8))') 'r0-s1,r1-s1,r2-s1 = ',r0-s1,r1-s1,r2-s1
stop
end
!**********************************************
sample output
nx,ny,nz = 131 131 131
size(foo) = 2248091
r0,r1,r2 = 1.12398225E+06 1.12399525E+06 1.12397238E+06
r0-r1,r0-r2,r1-r2 = -1.30000000E+01 9.87500000E+00 2.28750000E+01
s0,s1,s2 = 1.12397975E+06 1.12397975E+06 1.12398225E+06
s0-s1,s0-s2,s1-s2 = 0.00000000E+00-2.50000000E+00-2.50000000E+00
r0-s1,r1-s1,r2-s1 = 2.50000000E+00 1.55000000E+01-7.37500000E+00
First, welcome to StackOverflow. Please take the tour! There is a reason we expect a Minimal, Complete, and Verifiable example because we look at your code and can only guess at what might be the case and that is not too helpful for the community.
I hope the following suggestions helps you figure out what is going on.
Use the size() function and print what Fortran thinks are the sizes of the dimensions as well as printing nx, ny, and nz. As far as we know, the array is declared bigger than nx, ny, and nz and these variables are set according to the data set. Fortran does not necessarily initialize arrays to zero depending on whether it is a static or allocatable array.
You can also try specifying array extents in the sum function:
r = Sum(foo(1:nx,1:ny,1:nz))
If done like this, at least we know that the sum function is working on the exact same slice of foo that the loops loop over.
If this is the case, you will get the wrong answer even though there is nothing 'wrong' with the code. This is why it is particularly important to give that Minimal, Complete, and Verifiable example.
I can see the differences now. These are typical rounding errors from adding small numbers to a large sum. The processor is allowed to use any order of the summation it wants. There is no "right" order. You cannot really say that the original loops make the "correct" answer and the others do not.
What you can do is to use double precision. In extreme circumstances there are tricks like the Kahan summation but one rarely needs that.
Addition of a small number to a large sum is imprecise and especially so in single precision. You still have four significant digits in your result.
One typically does not use the DIM= argument, that is used in certain special circumstances.
If you want to sum all elements of foo, use just
s0 = SUM(foo)
That is enough.
What
s0 = SUM(SUM(SUM(foo, DIM=3), DIM=2), DIM=1)
does is that it will make a temporary 2D arrays with each element be the sum of the respective row in the z dimension, then a 1D array with each element the sum over the last dimension of the 2D array and then finally the sum of that 1D array. If it is done well, the final result will be the same, but it well eat a lot of CPU cycles.
The sum intrinsic function returns a processor-dependant approximation to the sum of the elements of the array argument. This is not the same thing as adding sequentially all elements.
It is simple to find an array x where
summation = x(1) + x(2) + x(3)
(performed strictly left to right) is not the best approximation for the sum treating the values as "mathematical reals" rather than floating point numbers.
As a concrete example to look at the nature of the approximation with ifort, we can look at the following program. We need to enable optimizations here to see effects; the importance of order of summation is apparent even with optimizations disabled (with -O0 or -debug).
implicit none
integer i
real x(50)
real total
x = [1.,(EPSILON(0.)/2, i=1, SIZE(x)-1)]
total = 0
do i=1, SIZE(x)
total = total+x(i)
print '(4F17.14)', total, SUM(x(:i)), SUM(DBLE(x(:i))), REAL(SUM(DBLE(x(:i))))
end do
end program
If adding up in strict order we get 1., seeing that anything smaller in magnitude than epsilon(0.) doesn't affect the sum.
You can experiment with the size of the array and order of its elements, the scaling of the small numbers and the ifort floating point compilation options (such as -fp-model strict, -mieee-fp, -pc32). You can also try to find an example like the above using double precision instead of default real.
Here's my factorial function in Fortran.
module facmod
implicit none
contains
function factorial (n) result (fac)
use FMZM
integer, intent(in) :: n
integer :: i
type(IM) :: fac
fac = 1
if(n==0) then
fac = 1
elseif(n==1) then
fac = 1
elseif(n==2) then
fac = 2
elseif(n < 0) then
write(*,*) 'Error in factorial N=', n
stop 1
else
do i = 1, n
fac = fac * i
enddo
endif
end function factorial
end module facmod
program main
use FMZM
use facmod, only: factorial
implicit none
type(IM) :: res
integer :: n, lenr
character (len=:), allocatable :: str
character(len=1024) :: fmat
print*,'enter the value of n'
read*, n
res = factorial(n)
lenr = log10(TO_FM(res))+2
allocate(character(len=lenr) :: str)
write (fmat, "(A5,I0)") "i", lenr
call im_form(fmat, res, str)
print*, trim( adjustl(str))
end program main
I compile using FMZM:
gfortran -std=f2008 fac.F90 fmlib.a -o fac
echo -e "1000" | .fac computes easy. However, if I give this echo -e "3600" | .fac, I already get an error on my machine:
Error in FM. More than 200000 type (FM), (ZM), (IM) numbers
have been defined. Variable SIZE_OF_START in file
FMSAVE.f95 defines this value.
Possible causes of this error and remedies:
(1) Make sure all subroutines (also functions that do not
return type FM, ZM, or IM function values) have
CALL FM_ENTER_USER_ROUTINE
at the start and
CALL FM_EXIT_USER_ROUTINE
at the end and before any other return, and all
functions returning an FM, ZM, or IM function value have
CALL FM_ENTER_USER_FUNCTION(F)
at the start and
CALL FM_EXIT_USER_FUNCTION(F)
at the end and before any other return, where the actual
function name replaces F above.
Otherwise that routine could be leaking memory, and
worse, could get wrong results because of deleting some
FM, ZM, or IM temporary variables too soon.
(2) Make sure all subroutines and functions declare any
local type FM, ZM, or IM variables as saved. Otherwise
some compilers create new instances of those variables
with each call, leaking memory.
For example:
SUBROUTINE SUB(A,B,C,X,Y,RESULT)
TYPE (FM) :: A,B,C,X,Y,RESULT,ERR,TOL,H
Here A,B,C,X,Y,RESULT are the input variables and
ERR,TOL,H are local variables. The fix is:
SUBROUTINE SUB(A,B,C,X,Y,RESULT)
TYPE (FM) :: A,B,C,X,Y,RESULT
TYPE (FM), SAVE :: ERR,TOL,H
(3) Since = assignments for multiple precision variables are
the trigger for cleaning up temporary multiple precision
variables, a loop with subroutine calls that has no =
assignments can run out of space to store temporaries.
For example:
DO J = 1, N
CALL SUB(A,B,C,TO_FM(0),TO_FM(1),RESULT)
ENDDO
Most compilers will create two temporary variables with
each call, to hold the TO_FM values.
One fix is to put an assignment into the loop:
DO J = 1, N
ZERO = TO_FM(0)
CALL SUB(A,B,C,ZERO,TO_FM(1),RESULT)
ENDDO
(4) If a routine uses allocatable type FM, ZM, or IM arrays
and allocates and deallocates with each call, then after
many calls this limit on number of variables could be
exceeded, since new FM variable index numbers are
generated for each call to the routine.
A fix for this is to call FM_DEALLOCATE before actually
deallocating each array, so those index numbers can be
re-used. For example:
DEALLOCATE(T)
becomes:
CALL FM_DEALLOCATE(T)
DEALLOCATE(T)
(5) If none of this helps, try running this program again
after increasing the value of SIZE_OF_START and
re-compiling.
What optimizations or Fortran idioms am I missing that is hurting my performance so much?
For example, in python, I can factorial numbers much larger than 3500:
>>> import math
>>> math.factorial(100000)
Or in Haskell:
Prelude> product [1..100000]
Both these compute, not exactly quickly, but without error.
How can I improve my algorithm or better use existing libraries to improve performance of large integer factorials in Fortran? Is there a more appropriate big integer library than FMZM?
Try this. Apart from minor cosmetic changes, I just followed the recommendations of the error message in your question:
added calls to FM_ENTER_USER_FUNCTION and FM_EXIT_USER_FUNCTION,
added an assignment inside the loop (without this ii = to_im(i), it still fails, but I'm not sure why, as there is already an assignment with fac = fac * i, and accordind to the doc the assignment triggers cleaning up temporaries),
renamed factorial in main program as there is already a function with this name in FMZM.
Tested with ifort and n=100000.
module fac_mod
implicit none
contains
function factorial(n) result(fac)
use FMZM
integer, intent(in) :: n
integer :: i
type(IM) :: fac
type(IM), save :: ii
call FM_ENTER_USER_FUNCTION(fac)
fac = to_im(1)
if (n < 0) then
write (*, *) "Error in factorial N=", n
stop 1
else if (n > 1) then
do i = 1, n
ii = to_im(i)
fac = fac * ii
end do
end if
call FM_EXIT_USER_FUNCTION(fac)
end function factorial
end module fac_mod
program main
use FMZM
use fac_mod, only: f=>factorial
implicit none
type(IM) :: res
integer :: n, lenr
character(:), allocatable :: str
character(1024) :: fmat
print *, "enter the value of n"
read *, n
res = f(n)
lenr = 2 + log10(TO_FM(res))
allocate (character(lenr) :: str)
write (fmat, "(A5,I0)") "i", lenr
call im_form(fmat, res, str)
print *, trim(adjustl(str))
end program main
I am now running a program for a certain iterations. The time step is 0.01. I want to write some information when a specific time is reached. For example:
program abc
implicit none
double precision :: time,step,target
integer :: x
time = 0.d0
step = 0.01
target = 5.d0
do x = 1,6000
time = time + step
"some equations here to calculate the model parameters"
if(time.eq.target)then
write(*,*) "model parameters"
endif
enddo
However, "time" never equals to 1.0 or 2.0 or etc. It shows like "0.999999866" instead of "1.0" and "1.99999845" instead of "2.0".
Although I can use integer "x" to define when to write the information, I prefer to use the time step. Also, I may want to change the time step (0.01/0.02/0.05/etc) or target (5.0/6.0/8.0/etc).
Does anyone knows how to fix this? Thanks ahead.
You have now discovered floating point arithmetic! Just ensure that the time is sufficiently close to the target.
if(abs(time-target) < 0.5d0*step ) then
...
should do the trick.
Floating point arithmetic is not perfect and your variables are always exact up to a certain machine error, depending on your variables' number format (32, 64, 128 bit). The following example illustrates well this characteristic:
PROGRAM main
USE, INTRINSIC :: ISO_FORTRAN_ENV, qp => real128
IMPLICIT NONE
REAL(qp) :: a, b, c
a = 128._qp
b = a/120._qp + 1
c = 120._qp*(b-1)
PRINT*, "a = ", a
PRINT*, "c = ", c
END PROGRAM main
Here is the output to this program with gfortran v.4.6.3:
a = 128.00000000000000000
c = 127.99999999999999999
I have a code in Fortran IV that I need to run. I was told to try to compile it in Fortran 77 and fix the error. So I named the file with a .f extension and tried to compile it with gfortran. I got the next error referring to the Fortran IV function copied below:
abel.f:432.24:
REAL FUNCTION DGDT*8(IX,NV,XNG,FNG,GNG,X)
1
Error: Expected formal argument list in function definition at (1)
Since I'm not too familiar with Fortran I'd appreciate if someone can tell me how to fix this problem .
REAL FUNCTION DGDT*8(IX,NV,XNG,FNG,GNG,X) AAOK0429
C AAOK0430
C THIS SUBROUTINE COMPUTES THE VALUE OF THE DERIVATIVE OF THE AAOK0431
C G-FUNCTION FOR A SLIT TRANSMISSION FUNCTION GIVEN BY A AAOK0432
C PIECE-WISE CUBIC SPLINE , WHOSE PARAMETERS ARE AAOK0433
C CONTAINED IN XNG,FNG AND GNG. AAOK0434
C AAOK0435
IMPLICIT REAL*8(A-H,O-Z) AAOK0436
C AAOK0437
C ALLOWABLE ROUNDING ERROR ON POINTS AT EXTREAMS OF KNOT RANGE AAOK0438
C IS 2**IEPS*MAX(!XNG(1)!,!XNG(NV)!). AAOK0439
INTEGER*4 IFLG/0/,IEPS/-50/ AAOK0440
DIMENSION XNG(1),FNG(1),GNG(1) AAOK0441
C AAOK0442
C TEST WETHER POINT IN RANGE. AAOK0443
IF(X.LT.XNG(1)) GO TO 990 AAOK0444
IF(X.GT.XNG(NV)) GO TO 991 AAOK0445
C AAOK0446
C ESTIMATE KNOT INTERVAL BY ASSUMING EQUALLY SPACED KNOTS. AAOK0447
12 J=DABS(X-XNG(1))/(XNG(NV)-XNG(1))*(NV-1)+1 AAOK0448
C ENSURE CASE X=XNG(NV) GIVES J=NV-1 AAOK0449
J=MIN0(J,NV-1) AAOK0450
C INDICATE THAT KNOT INTERVAL INSIDE RANGE HAS BEEN USED. AAOK0451
IFLG=1 AAOK0452
C SEARCH FOR KNOT INTERVAL CONTAINING X. AAOK0453
IF(X.LT.XNG(J)) GO TO 2 AAOK0454
C LOOP TILL INTERVAL FOUND. AAOK0455
1 J=J+1 AAOK0456
11 IF(X.GT.XNG(J+1)) GO TO 1 AAOK0457
GO TO 7 AAOK0458
2 J=J-1 AAOK0459
IF(X.LT.XNG(J)) GO TO 2 AAOK0460
C AAOK0461
C CALCULATE SPLINE PARAMETERS FOR JTH INTERVAL. AAOK0462
7 H=XNG(J+1)-XNG(J) AAOK0463
Q1=H*GNG(J) AAOK0464
Q2=H*GNG(J+1) AAOK0465
SS=FNG(J+1)-FNG(J) AAOK0466
B=3D0*SS-2D0*Q1-Q2 AAOK0467
A=Q1+Q2-2D0*SS AAOK0468
C AAOK0469
C CALCULATE SPLINE VALUE. AAOK0470
8 Z=(X-XNG(J))/H AAOK0471
C TF=((A*Z+B)*Z+Q1)*Z+FNG(J) AAOK0472
C TG=((3.*A*Z+2.*B)*Z+Q1)/H AAOK0473
C DGDT=(TG-TF/X)/X AAOK0474
DGDT=(3.*A*Z*Z+2.*B*Z+Q1)/H AAOK0475
RETURN AAOK0476
C TEST IF X WITHIN ROUNDING ERROR OF XNG(1). AAOK0477
990 IF(X.LE.XNG(1)-2D0**IEPS*DMAX1(DABS(XNG(1)),DABS(XNG(NV)))) GO AAOK0478
1 TO 99 AAOK0479
J=1 AAOK0480
GO TO 7 AAOK0481
C TEST IF X WITHIN ROUNDING ERROR OF XNG(NV). AAOK0482
991 IF(X.GE.XNG(NV)+2D0**IEPS*DMAX1(DABS(XNG(1)),DABS(XNG(NV)))) GO AAOK0483
1 TO 99 AAOK0484
J=NV-1 AAOK0485
GO TO 7 AAOK0486
99 IFLG=0 AAOK0487
C FUNCTION VALUE SET TO ZERO FOR POINTS OUTSIDE THE RANGE. AAOK0488
DGDT=0D0 AAOK0489
RETURN AAOK0490
END AAOK0491
This doesn't look so bad. Modern compilers still accept the real*8 syntax although it isn't standard. So you should (as mentioned) replace the line
REAL FUNCTION DGDT*8(IX,NV,XNG,FNG,GNG,X) AAOK0429
with
REAL*8 FUNCTION DGDT(IX,NV,XNG,FNG,GNG,X) AAOK0429
which compiled successfully for me using gfortran 4.6.2 using gfortran -c DGDT.f.
Good luck, and be on the lookout for other problems. Just because the code compiles does not mean it is running the same way it was designed!
Not really an answer, see the one from Ross. But I just can't stand the requirement for fixed form. Here is how this code probably would look like in F90 with free form:
function DGDT(IX, NV, XNG, FNG, GNG, X)
! THIS FUNCTION COMPUTES THE VALUE OF THE DERIVATIVE OF THE
! G-FUNCTION FOR A SLIT TRANSMISSION FUNCTION GIVEN BY A
! PIECE-WISE CUBIC SPLINE, WHOSE PARAMETERS ARE
! CONTAINED IN XNG,FNG AND GNG.
implicit none
integer, parameter :: rk = selected_real_kind(15)
integer :: ix, nv
real(kind=rk) :: dgdt
real(kind=rk) :: xng(nv)
real(kind=rk) :: fng(nv)
real(kind=rk) :: gng(nv)
real(kind=rk) :: x
! ALLOWABLE ROUNDING ERROR ON POINTS AT EXTREAMS OF KNOT RANGE
! IS 2**IEPS*MAX(!XNG(1)!,!XNG(NV)!).
integer, parameter :: ieps = -50
integer, save :: iflg = 0
integer :: j
real(kind=rk) :: tolerance
real(kind=rk) :: H
real(kind=rk) :: A, B
real(kind=rk) :: Q1, Q2
real(kind=rk) :: SS
real(kind=rk) :: Z
tolerance = 2.0_rk**IEPS * MAXVAL(ABS(XNG([1,NV])))
! TEST WETHER POINT IN RANGE.
if ((X < XNG(1) - tolerance) .or. (X > XNG(NV) + tolerance)) then
! FUNCTION VALUE SET TO ZERO FOR POINTS OUTSIDE THE RANGE.
iflg = 0
DGDT = 0.0_rk
return
end if
! ESTIMATE KNOT INTERVAL BY ASSUMING EQUALLY SPACED KNOTS.
J = abs(x-xng(1)) / (xng(nv)-xng(1)) * (nv-1) + 1
! ENSURE CASE X=XNG(NV) GIVES J=NV-1
J = MIN(J,NV-1)
! INDICATE THAT KNOT INTERVAL INSIDE RANGE HAS BEEN USED.
IFLG = 1
! SEARCH FOR KNOT INTERVAL CONTAINING X.
do
if ( (x >= xng(j)) .or. (j==1) ) EXIT
j = j-1
! LOOP TILL INTERVAL FOUND.
end do
do
if ( (x <= xng(j+1)) .or. (j==nv-1) ) EXIT
j = j+1
! LOOP TILL INTERVAL FOUND.
end do
! CALCULATE SPLINE PARAMETERS FOR JTH INTERVAL.
H = XNG(J+1) - XNG(J)
Q1 = H*GNG(J)
Q2 = H*GNG(J+1)
SS = FNG(J+1) - FNG(J)
B = 3.0_rk*SS - 2.0_rk*Q1 - Q2
A = Q1 + Q2 - 2.0_rk*SS
! CALCULATE SPLINE VALUE.
Z = (X-XNG(J))/H
DGDT = ( (3.0_rk*A*Z + 2.0_rk*B)*Z + Q1 ) / H
end function DGDT
Note, I did not test this in any way, also there might be some wrong guesses in there, like that ieps should be a constant. Also, I am not so sure about iflg, and the ix argument does not appear to be used at all. So I might got something wrong. For the tolerance it is better to use a factor instead of a difference and a 2.**-50 will not change the value for a the maxval in a double precision number here. Also note, I am using some other F90 features besides the free form now.
DISCLAIMER: Just mentioning a possible solution here, not recommending it...
As much as all other answers are valid and that supporting some Fortran IV code as is is a nightmare, you still might want / need to avoid touching it as much as possible. And since Fortran IV had some strange behaviours when it comes to loops for example (with loops always cycled at least once IINM), using a "proper" Fortran IV compiler might be a "good" idea.
Anyway, all this to say that the Intel compiler for example, supports Fortran IV natively with the -f66 compiler switch, and I'm sure other compilers do as well. This may be worth checking.