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I am working on a program to calculate minimum nozzle length in supersonic nozzles (Method of Characteristics). I can't seem to figure out why my code won't write to my output file (the lines "write (6,1000)....). My code is below:
program tester
c----------------------------------------------------------------------c
implicit real (a-h,o-z)
integer count
real kminus(0:100),kplus(0:100),theta(0:100),nu(0:100),mach(0:100)
+ ,mu(0:100)
open (5,file='tester.in')
open (6,file='tester.out')
read (5,*) me
read (5,*) maxturn
read (5,*) nchar
read (5,*) theta0
close(5)
c.....set count to 0 and calculate dtheta
count=0
dtheta=(maxturn-theta0)/nchar
c.....first characteristic
do 10 an=1,nchar+1
count=count+1
c........these are already known
theta(count)=theta0+dtheta*(an-1)
nu(count)=theta(count)
c........trivial, but we will "calculate" them anyways
kminus(count)=2*theta(count)
kplus(count)=0
c........we feed it nu(count) and get m out
call pm_hall_approx(nu(count),m)
mach(count)=m
c........does not work with sqrt(...) for some reason
mu(count)=atan((1/(mach(count)**2)-1)**0.5)
write (6,1000) count,kminus(count),kplus(count),theta(count)
+ ,nu(count),mach(count),mu(count)
10 continue
c.....the other characteristics
do 30 bn=1,nchar
do 20 cn=1,(nchar+1-bn)
count=count+1
c...........these are given
kminus(count)=kminus(cn+bn)
kplus(count)=-1*kplus(bn+1)
c...........if this is the last point, copy the previous values
if (cn.eq.(nchar+1-bn)) then
kminus(count)=kminus(count-1)
kplus(count)=kplus(count-1)
endif
c...........calculate theta and nu
theta(count)=0.5*(kminus(count)+kplus(count))
nu(count)=0.5*(kminus(count)-kplus(count))
c...........calculate m
call pm_hall_approx(nu(count),m)
mach(count)=m
mu(count)=atan((1/(mach(count)**2)-1)**0.5)
write (6,1000) count,kminus(count),kplus(count),theta(count)
+ ,nu(count),mach(count),mu(count)
20 continue
30 continue
close(6)
stop
1000 format (11(1pe12.4))
end
c======================================================================c
include 'pm_hall_approx.f'
And my subroutine is given here:
subroutine pm_hall_approx(nu,mach)
c----------------------------------------------------------------------c
c Given a Mach number, use the Hall Approximation to calculate the
c Prandtl-Meyer Function.
c----------------------------------------------------------------------c
implicit real (a-h,o-z)
c.....set constants
parameter (a=1.3604,b=0.0962,c=-0.5127,d=-0.6722,e=-0.3278)
parameter (numax=2.2769)
y=nu/numax
mach=(1+a*y+b*y*y+c*y*y*y)/(1+d*y+e*y*y)
return
end
And here are the contents of tester.in
2.4 = mache
5.0 = maxturn
7 = nchar
0.375 = theta0
In fortran, unit number 6 is reserved for screen. Never use that as the unit number. Try changing it to some other number like write(7,1000) and then it should work.
I want to calculate julian day for just one year (not from 4713 BC!). For example, If it is 3 of March, it will give as a result the number 62 If February has 28 days, and 63 if it has 29. Until now, I have written the following code in Fortran 95, which unfortunately gives wrong result. Could you please help me debug it?
program task
!============================
!filename: task.f95
implicit none
integer day, sum_month, i, month(12), jd, year, mon
month = [31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31]
write(6,*) "Give day"
read(5,*) day
write(6,*) "Give mon"
read(5,*) mon
write(6,*) "Give year"
read(5,*) year
sum_month = 0
if (mod(year,4)==0 .and. mod(year,100)/=0) then
do i = month(1), mon-1
sum_month = sum_month + month(1) + day + 1
end do
else
do i = month(1), mon-1
sum_month = sum_month + month(1) + day
end do
end if
write(6,*) sum_month
end program task
This loop
do i = month(1), mon-1
sum_month = sum_month + month(1) + day + 1
end do
and the other one like it, is seriously broken. Every time you call it month(1)==31 and mon is likely to be between 1 and 12 (inclusive). So the program will try to execute, for (for example) mon==12
do i = 31,12
and under the rules of Fortran this loop will execute 0 times. I'm almost certain you should write:
do i = 1, mon-1
sum_month = sum_month + month(i) ...
Fix that, sort out the rest of your code, and you'll probably fix it without further help.
EDIT:
Your code is getting worse ! Well, it was getting worse until VladimirF rolled back your latest changes. Do as he and I have suggested. Ask a new question concerning your current problem.
You're not being careful enough about the difference between your variable mon (which you use to hold the number of the month input by the user) and month (which is the array you use to hold the number of days in each month of the year).
These statements
write(6,*) "Give mon"
read(5,*) month(i)
will read the value which belongs in mon and put it into month(i). Or rather, that's what it tries to do but at that place in the code i doesn't have any value and the code attempts to write it into some arbitrary location in the computer's memory -- this is what a segmentation fault is and this is a great way to generate one. Don't do that !
You're going to have a problem later too, when your code reaches
do i = 1, month(i)-1
Go back to the top of my answer and ask yourself What value does month(i) have when this line of code is executed ?
I am trying to use the code below to read a formatted file and write it into another. However, on running it shows the following error
$ ./conv.sac.farm < i_conv.farm
# stn comp Delta Tr-time Start in record
At line 54 of file Main/conv.sac.farm.f (unit = 5, file = 'stdin')
Fortran runtime error: Bad real number in item 1 of list input
The source code is as follows
PARAMETER (nd0=100000,pi=3.1415926)
IMPLICIT COMPLEX*8 (Z)
CHARACTER name*6,comp*6,fname*60,event*20
- ,cmp(0:3)*5,fname0*60,charac*15,scode*60
REAL*8 GFACT(500),PP0(500),depth0
integer hr0,mnu0,yr,month,day,hr,mnu
REAL x(nd0),y(nd0)
DIMENSION Z(nd0),zpole(50),zero(50)
data np,cmp/8,'disp.','vel. ','acc. ','orig.'/
common /tbl/ip(110,14),is(110,14),secp(110,14),secs(110,14)
read(5,'(a)') event
read(5,*) alats,alons,depth,hr0,mnu0,sec0,id,delmin,delmax
depth0=depth
write(22,'(a,a5,3f7.2,2i3,f6.2)')
# event,cmp(id),alats,alons,depth,hr0,mnu0,sec0
* << J-B travel time table >>
OPEN(11,FILE='jb.ptime')
OPEN(12,FILE='jb.stime')
1000 read(11,*,end=1001) n,(ip(n,i),secp(n,i),i=1,14)
goto 1000
1001 read(12,*,end=1002) n,(is(n,i),secs(n,i),i=1,14)
goto 1001
1002 continue
close(11)
close(12)
* << Geometrical factor >>
OPEN(15,FILE='jb.table')
CALL GEOM(GFACT,PP0,depth0)
close(15)
nstn=0
print *,' # stn comp Delta Tr-time Start in record'
5 read(5,'(a)') fname
read(5,'(a)') scode
* ta=advance of start-time relative the standard P/S arrival
* du=duration
c
if(fname.eq.'dummy') goto 90
read(5,*) ta,du,dt,f1,f2,iph,nr,iuni
open(1,file=fname)
READ(1,'(g15.7)') dt0
read(1,'(/////5g15.7)') dum, alat, alon, elev
read(1,'(///////5i10)') yr, nday, hr,mnu, nsec
read(1,'(5i10)') nmsec,ndum,ndum,ndum,nd
read(1,'(/////)')
read(1,'(a6,2x,a13)') name,charac
read(1,'(////)')
And so on..
Line 54 is
read(5,*) ta,du,dt,f1,f2,iph,nr,iuni
I found a similar question following this link
Fortran runtime error: Bad real number
However, if I understand correctly, the corrections mentioned were pertaining to reading unformatted data. Despite this, I tried and failed as expected, given that the file I am trying to read is formatted.
here is a little trick if you can't readily find the offending line in the data file:
replace your read that throws the error with this:
read(5,'(a)')line
read(line,*,iostat=ios) ta,du,dt,f1,f2,iph,nr,iuni
if(ios>0)then
write(*,*)'error reading line:',line
stop
endif
with the declarations
integer ios
character*(200) line
Probably just do that for debugging then revert to the original once you fix the problem.
I have a problem while using a do loop in fortran,
REAL W,V,X
DO 50 W = 0.5,5.0,0.5
DO 50 V = 10.0,1000.0,10.0
DO 50 X = 1.0,10,1.0
C=(W*V*X)/1000.0
WRITE(*,*) W,V,X,C
50 CONTINUE
STOP
END
If I gave this it is showing that only integers needs to be used in do loop, is there any way to give integers in do loop or any other way to do it?
Use integers as your looping indices
REAL W,V,X
INTEGER I,J,K
DO 50 I = 1,10
DO 50 J = 1,100
DO 50 K = 1,10
W = 0.5 * I
V = 10.0 * J
X = 1.0 * K
C=(W*V*X)/1000.0
WRITE(*,*) W,V,X,C
50 CONTINUE
STOP
END
You should be able to accomplish the same thing by incrementing a real variable by adding your step value, and using an if then to exit the loop. Clunky, but should work.
The last time I programmed in Fortran, I used punch cards and an IBM-360, so I'm not going to pretend I remember the syntax.
I have this program that is supposed to search for perfect numbers.
(X is a perfect number if the sum of all numbers that divide X, divided by 2 is equal to X)
sum/2 = x
Now It has found the first four, which were known in Ancient Greece, so it's not really a anything awesome.
The next one should be 33550336.
I know it is a big number, but the program has been going for about 50 minutes, and still hasn't found 33550336.
Is it because I opened the .txt file where I store all the perfect numbers while the program was running, or is it because I don't have a PC fast enough to run it*, or because I'm using Python?
*NOTE: This same PC factorized 500 000 in 10 minutes (while also running the perfect number program and Google Chrome with 3 YouTube tabs), also using Python.
Here is the code to the program:
i = 2
a = open("perfect.txt", 'w')
a.close()
while True:
sum = 0
for x in range(1, i+1):
if i%x == 0:
sum += x
if sum / 2 == i:
a = open("perfect.txt", 'a')
a.write(str(i) + "\n")
a.close()
i += 1
The next one should be 33550336.
Your code (I fixed the indentation so that it does in principle what you want):
i = 2
a = open("perfect.txt", 'w')
a.close()
while True:
sum = 0
for x in range(1, i+1):
if i%x == 0:
sum += x
if sum / 2 == i:
a = open("perfect.txt", 'a')
a.write(str(i) + "\n")
a.close()
i += 1
does i divisions to find the divisors of i.
So to find the perfect numbers up to n, it does
2 + 3 + 4 + ... + (n-1) + n = n*(n+1)/2 - 1
divisions in the for loop.
Now, for n = 33550336, that would be
Prelude> 33550336 * (33550336 + 1) `quot` 2 - 1
562812539631615
roughly 5.6 * 1014 divisions.
Assuming your CPU could do 109 divisions per second (it most likely can't, 108 is a better estimate in my experience, but even that is for machine ints in C), that would take about 560,000 seconds. One day has 86400 seconds, so that would be roughly six and a half days (more than two months with the 108 estimate).
Your algorithm is just too slow to reach that in reasonable time.
If you don't want to use number-theory (even perfect numbers have a very simple structure, and if there are any odd perfect numbers, those are necessarily huge), you can still do better by dividing only up to the square root to find the divisors,
i = 2
a = open("perfect.txt", 'w')
a.close()
while True:
sum = 1
root = int(i**0.5)
for x in range(2, root+1):
if i%x == 0:
sum += x + i/x
if i == root*root:
sum -= x # if i is a square, we have counted the square root twice
if sum == i:
a = open("perfect.txt", 'a')
a.write(str(i) + "\n")
a.close()
i += 1
that only needs about 1.3 * 1011 divisions and should find the fifth perfect number in a couple of hours.
Without resorting to the explicit formula for even perfect numbers (2^(p-1) * (2^p - 1) for primes p such that 2^p - 1 is prime), you can speed it up somewhat by finding the prime factorisation of i and computing the divisor sum from that. That will make the test faster for all composite numbers, and much faster for most,
def factorisation(n):
facts = []
multiplicity = 0
while n%2 == 0:
multiplicity += 1
n = n // 2
if multiplicity > 0:
facts.append((2,multiplicity))
d = 3
while d*d <= n:
if n % d == 0:
multiplicity = 0
while n % d == 0:
multiplicity += 1
n = n // d
facts.append((d,multiplicity))
d += 2
if n > 1:
facts.append((n,1))
return facts
def divisorSum(n):
f = factorisation(n)
sum = 1
for (p,e) in f:
sum *= (p**(e+1) - 1)/(p-1)
return sum
def isPerfect(n):
return divisorSum(n) == 2*n
i = 2
count = 0
out = 10000
while count < 5:
if isPerfect(i):
print i
count += 1
if i == out:
print "At",i
out *= 5
i += 1
would take an estimated 40 minutes on my machine.
Not a bad estimate:
$ time python fastperf.py
6
28
496
8128
33550336
real 36m4.595s
user 36m2.001s
sys 0m0.453s
It is very hard to try and deduce why this has happened. I would suggest that you run your program either under a debugger and test several iteration manually to check if the code is really correct (I know you have already calculated 4 numbers but still). Alternatively it would be good to run your program under a python profiler just to see if it hasn't accidentally blocked on a lock or something.
It is possible, but not likely that this is an issue related to you opening the file while it is running. If it was an issue, there would have probably been some error message and/or program close/crash.
I would edit the program to write a log-type output to a file every so often. For example, everytime you have processed a target number that is an even multiple of 1-Million, write (open-append-close) the date-time and current-number and last-success-number to a log file.
You could then Type the file once in a while to measure progress.