Increasing the double precision values - fortran

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

Related

Why does lapack dtrmm.f seem to not work properly?

I decided to use lapack subroutine dtrmm instead of matmul to multiply lower triangular (d,d) matrix and general (d,n) matrix. However, it doesn't seem to work correctly. The following code compares results of matlum (top) and dtrmm
Program test
implicit none
integer, parameter :: n = 3, d = 3 ! arbitrary numbers, no luck with other values
integer :: i
real(8) :: a(d,d), b(d,n), c(d,n)
call random_number(a)
call random_number(b)
do i=2,d
a(1:i-1,i) = 0
end do
c = matmul(a,b)
call dtrmm('L','L','N','N',d,n,1,a,d,b,d) ! documentation linked in the question
print*, 'correct result : '
do i=1,d
print*, c(i,:)
end do
print*, 'dtrmm yields : '
do i=1,d
print*, b(i,:)
end do
End Program test
returns this
correct result :
0.75678922130735249 0.51830058846965921 0.51177304237548271
1.1974740765385026 0.46115246753697681 0.98237114765741340
0.98027798241945430 0.53718796235743815 1.0328498570683342
dtrmm yields :
6.7262070844500211E+252 4.6065628207770121E+252 4.5485471599475983E+252
1.0642935166471391E+253 4.0986405551607272E+252 8.7311388520015068E+252
8.7125351741793473E+252 4.7744304178222945E+252 9.1797845822711462E+252
Other lapack suboutines I used work fine. What could be causing this to misbehave? Is this a bug, or have I just badly misunderstood something?
It is a simple data type error. The factor alpha must be of type double precision whereas you supplied an integer of default kind.
Thus
...
! call dtrmm('L','L','N','N',d,n,1,a,d,b,d) WRONG
call dtrmm('L','L','N','N',d,n,1d0,a,d,b,d) ! note 1d0 instead of 1
...
gives the correct result.

How to implement factorial function into code?

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

SIGFPE error with gfortran 4.8.5 handling

I am using a computational fluid dynamics software that is compiled with gfortran version 4.8.5 on Ubuntu 16.04 LTS. The software can be compiled with either single precision or double precision and the -O3 optimization option. As I do not have the necessary computational resources to run the CFD software on double precision I am compiling it with single precision and the following options
ffpe-trap=invalid,zero,overflow
I am getting a SIGFPE error on a line of code that contains the asin function-
INTEGER, PARAMETER :: sp = SELECTED_REAL_KIND( 6, 37) !< single precision
INTEGER, PARAMETER :: wp = sp
REAL(KIND=wp) zsm(:,:)
ela(i,j) = ASIN(zsm(ip,jp))
In other words the inverse sin function and this code is part of a doubly nested FOR loop with jp and ip as the indices. Currently the software staff is unable to help me for various other reasons and so I am trying to debug this on my own. The SIGFPE error is only being observed in the single precision compilation not double precision compilation.
I have inserted the following print statements in my code prior to the line of code that is failing i.e. the asin function call. Would this help me with unraveling the problem that I am facing ? This piece of code is executed for every time step and it is occurring after a series of time steps. Alternatively what other steps can I do to help me fix this problem ? Would adding "precision" to the compiler flag help ?
if (zsm(ip,jp) >= 1.0 .or. zsm(ip,jp) <= -1.0) then
print *,zsm(ip,jp),ip,jp
end if
EDIT
I took a look at this answer Unexpected behavior of asin in R and I am wondering whether I could do something similar in fortran i.e. by using the max function. If it goes below -1 or greater than 1 then round it off in the proper manner. How can I do it with gfortran using the max function ?
On my desktop the following program executes with no problems(i.e. it has the ability to handle signed zeros properly) and so I am guessing the SIGFPE error occurs with either the argument greater than 1 or less than -1.
program testa
real a,x
x = -0.0000
a = asin(x)
print *,a
end program testa
We have min and max functions in Fortran, so I think we can use the same method as in the linked page, i.e., asin( max(-1.0,min(1.0,x) ). I have tried the following test with gfortran-4.8 & 7.1:
program main
implicit none
integer, parameter :: sp = selected_real_kind( 6, 37 )
integer, parameter :: wp = sp
! integer, parameter :: wp = kind( 0.0 )
! integer, parameter :: wp = kind( 0.0d0 )
real(wp) :: x, a
print *, "Input x"
read(*,*) x
print *, "x =", x
print *, "equal to 1 ? :", x == 1.0_wp
print *, asin( x )
print *, asin( max( -1.0_wp, min( 1.0_wp, x ) ) )
end
which gives with wp = sp (or wp = kind(0.0) on my computer)
$ ./a.out
Input x
1.00000001
x = 1.00000000
equal to 1 ? : T
1.57079625 (<- 1.5707964 for gfortran-4.8)
1.57079625
$ ./a.out
Input x
1.0000001
x = 1.00000012
equal to 1 ? : F
NaN
1.57079625
and with wp = kind(0.0d0)
$ ./a.out
Input x
1.0000000000000001
x = 1.0000000000000000
equal to 1 ? : T
1.5707963267948966
1.5707963267948966
$ ./a.out
Input x
1.000000000000001
x = 1.0000000000000011
equal to 1 ? : F
NaN
1.5707963267948966
If it is necessary to modify a lot of asin(x) and the program relies on a C or Fortran preprocessor, it may be convenient to define some macro like
#define clamp(x) max(-1.0_wp,min(1.0_wp,x))
and use it as asin( clamp(x) ). If we want to remove such a modification, we can simply change the definition of clamp() as #define clamp(x) (x). Another approach may be to define some asin2(x) function that limits x to [-1,1] and replace the built-in asin by asin2 (either as a macro or a Fortran function).

Compiling Fortran IV code with Fortran 77 compiler

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.

Is there a standard way to check for Infinite and NaN in Fortran 90/95?

I've been trying to find a standards-compliant way to check for Infinite and NaN values in Fortran 90/95 but it proved harder than I thought.
I tried to manually create Inf and NaN variables using the binary representation described in IEEE 754, but I found no such functionality.
I am aware of the intrinsic ieee_arithmetic module in Fortran 2003 with the ieee_is_nan() and ieee_is_finite() intrinsic functions. However it's not supported by all the compilers (notably gfortran as of version 4.9).
Defining infinity and NaN at the beginning like pinf = 1. / 0 and nan = 0. / 0 seems hackish to me and IMHO can raise some building problems - for example if some compilers check this in compile time one would have to provide a special flag.
Is there a way I can implement in standard Fortran 90/95?
function isinf(x)
! Returns .true. if x is infinity, .false. otherwise
...
end function isinf
and isnan()?
The simple way without using the ieee_arithmatic is to do the following.
Infinity: Define your variable infinity = HUGE(dbl_prec_var) (or, if you have it, a quad precision variable). Then you can simply check to see if your variable is infinity by if(my_var > infinity).
NAN: This is even easier. By definition, NAN is not equal to anything, even itself. Simply compare the variable to itself: if(my_var /= my_var).
I don't have enough rep to comment so I'll "answer" regarding Rick Thompson's suggestion for testing infinity.
if (A-1 .eq. A)
This will also be true if A is a very large floating point number, and 1 is below the precision of A.
A simple test:
subroutine test_inf_1(A)
real, intent(in) :: A
print*, "Test (A-1 == A)"
if (A-1 .eq. A) then
print*, " INFINITY!!!"
else
print*, " NOT infinite"
endif
end subroutine
subroutine test_inf_2(A)
real, intent(in) :: A
print*, "Test (A > HUGE(A))"
if (A > HUGE(A)) then
print*, " INFINITY!!!"
else
print*, " NOT infinite"
endif
end subroutine
program test
real :: A,B
A=10
print*, "A = ",A
call test_inf_1(A)
call test_inf_2(A)
print*, ""
A=1e20
print*, "A = ",A
call test_inf_1(A)
call test_inf_2(A)
print*, ""
B=0.0 ! B is necessary to trick gfortran into compiling this
A=1/B
print*, "A = ",A
call test_inf_1(A)
call test_inf_2(A)
print*, ""
end program test
outputs:
A = 10.0000000
Test (A-1 == A)
NOT infinite
Test (A > HUGE(A))
NOT infinite
A = 1.00000002E+20
Test (A-1 == A)
INFINITY!!!
Test (A > HUGE(A))
NOT infinite
A = Infinity
Test (A-1 == A)
INFINITY!!!
Test (A > HUGE(A))
INFINITY!!!
No.
The salient parts of IEEE_ARITHMETIC for generating/checking for NaN's are easy enough to write for gfortran for a particular architecture.
I have used:
PROGRAM MYTEST
USE, INTRINSIC :: IEEE_ARITHMETIC, ONLY: IEEE_IS_FINITE
DOUBLE PRECISION :: number, test
number = 'the expression to test'
test = number/number
IF (IEEE_IS_FINITE(test)) THEN
WRITE(*,*) 'We are OK'
ELSE
WRITE(*,*) 'Got a problem'
END IF
WRITE(*,*) number, test
END PROGRAM MYTEST
This will print 'Got a problem' for number = 0.0D0, 1.0D0/0.0D0, 0.0D0/0.0D0, SQRT(-2.0D0), and also for overflows and underflows such as number = EXP(1.0D800) or number = EXP(-1.0D800). Notice that generally, things like number = EXP(1.0D-800) will just set number = 1.0 and produce a warning at compilation time, but the program will print 'We are OK', which I find acceptable.
OL.
for testing NaN none of the things worked eg.if testing real s2p to see if it is NaN then
if(isnan(s2p))
did not work in gfortran nor did
if(s2p.ne.s2p).
The only thing that worked was
if(.not.s2p<1.and..not.s2p>1)
though to make real sure u may want to add
if(.not.s2p<1.and..not.s2p>1.and..not.s2p==1)
No.
Neither is there a standards-compliant way of checking for infinities or NaNs in Fortran 90/95, nor can there be a standards-compliant way. There is no standards-compliant way of defining either of these quasi-numbers in Fortran 90/95.
For Fortran, 1/infinity=0
thus, divide your variable by zero
i.e
program test
implicit none
real :: res
integer :: i
do i=1,1000000
res=-log((1.+(10**(-real(i))))-1.)
print*,i,res
if ((1./res)==0.) then
exit
end if
end do
end program
there's your infinity check. No complication necessary.
For Inf it seems to work that if (A-1 .eq. A) is true, then A is Inf