Ambiguous reference from a Fortran module [duplicate] - compilation

Question

So I am doing 2 modules which are linking to the main program. The first one has all the variables defined in it and the second one is with the functions.
Module1:
module zmienne
implicit none
integer, parameter :: ngauss = 8
integer, parameter :: out_unit=1000
integer, parameter :: out_unit1=1001
integer, parameter :: out_unit2=1002, out_unit3=1003
real(10), parameter :: error=0.000001
real(10):: total_calka, division,tot_old,blad
real(10),parameter:: intrange=7.0
real(10),dimension(ngauss),parameter::xx=(/-0.9602898565d0,&
-0.7966664774d0,-0.5255324099d0,-0.1834346425d0,&
0.1834346425d0,0.5255324099d0,0.7966664774d0,0.9602898565d0/)
real(10),Dimension(ngauss),parameter::ww=(/0.1012285363d0,&
0.2223810345d0,0.3137066459d0,0.3626837834d0,&
0.3626837834d0,0.3137066459d0,0.2223810345d0,0.1012285363d0/)
real(10) :: r, u, r6, tempred, f, r2, r1, calka,beta
real(10) :: inte
real :: start, finish
integer:: i,j,irange
real(10),dimension(ngauss)::x,w,integrand
end module zmienne
Module2
module in
implicit none
contains
real(10) function inte(y,beta,r2,r1)
real(kind=10)::r,beta,r6,r2,r1,u,y
r=(r2-r1)*y+r1
r6=(1.0/r)**6
u=beta*r6*(r6-1.0d0)
if (u>100.d0) then
inte=-1.0d0
else
inte=exp(-u)-1.d0
endif
inte=r*r*inte
end function
end module in
And while im calling them like that:
use zmienne; use in
I am getting following error:
Name 'inte' at (1) is an ambiguous reference to 'inte' from module 'zmienne'
I've deleted "inte" in the module1 but now I am getting following error:
irange=inte(intrange/division)
1
Error: Missing actual argument for argument 'beta' at (1)
The main program code is:
program wykres
use zmienne; use in
implicit none
open(unit=out_unit, file='wykresik.dat', action='write', status='replace')
open(unit=out_unit1, file='wykresik1.dat', action='write')
open(unit=out_unit2, file='wykresik2.dat', action='write')
open(out_unit3, file='wykresik3.dat', action='write')
! the gaussian points (xx) and weights (ww) are for the [-1,1] interval
! for [0,1] interval we have (vector instr.)
x=0.5d0*(xx+1.0d0)
w=0.5d0*ww
! plots
tempred = 1.0
call cpu_time(start)
do i=1,1000
r=float(i)*0.01
r6=(1.0/r)**6
u=beta*r6*(r6-1.0)
f=exp(-u/tempred)-1.0
write(out_unit,*) r, u
write(out_unit1,*)r, f
write(out_unit2,*)r, r*r*f
end do
call cpu_time(finish)
print '("Time = ",f6.3," seconds.")',finish-start
! end of plots
! integration 1
calka=0.0
r1=0.0
r2=0.5
do i=1,ngauss
r=(r2-r1)*x(i)+r1
r6=(1.0/r)**6
u=beta*r6*(r6-1.0d0)
! check for underflows
if (u>100.d0) then
f=-1.0d0
else
f=exp(-u)-1.d0
endif
! the array integrand is introduced in order to perform vector calculations below
integrand(i)=r*r*f
calka=calka+integrand(i)*w(i)
enddo
calka=calka*(r2-r1)
write(*,*)calka
! end of integration
! integration 2
calka=0.0
do i=1,ngauss
integrand(i)=inte(x(i),beta,r2,r1)
calka=calka+integrand(i)*w(i)
enddo
calka=calka*(r2-r1)
! end of integration 2
write(*,*)calka
! vector integration and analytical result
write(*,*)sum(integrand*w*(r2-r1)),-(0.5**3)/3.0
!**************************************************************
! tot_calka - the sum of integrals all integration ranges
! dividion the initial length of the integration intervals
! tot_old - we will compare the results fro two consecutive divisions.
! at the beginning we assume any big number
! blad - the difference between two consecutive integrations,
! at the beginning we assume any big number
! error - assumed precission, parameter, it is necassary for
! performing do-while loop
total_calka=0.0
division=0.5
tot_old=10000.0
blad=10000.0
do while (blad>error)
! intrange - the upper integration limit, it should be estimated
! analysing the plot of the Mayer function. Here - 7.
! irange = the number of subintegrals we have to calculate
irange=inte(intrange/division)
total_calka=-(0.5**3)/3.0
! the analytical result for the integration range [0,0.5]
! the loop over all the intervals, for each of them we calculate
! lower and upper limits, r1 and r2
do j=1,irange
r1=0.5+(j-1)*division
r2=r1+division
calka=0.0
! the integral for a given interval
do i=1,ngauss
integrand(i)=inte(x(i),beta,r2,r1)
calka=calka+integrand(i)*w(i)
enddo
total_calka=total_calka+calka*(r2-r1)
enddo
! aux. output: number of subintervals, old and new integrals
write(*,*) irange,division,tot_old,total_calka
division=division/2.0
blad=abs(tot_old-total_calka)
tot_old=total_calka
! and the final error
write(*,*) blad
enddo
open(1,file='calka.dat', access='append')
! the secod viarial coefficient=CONSTANT*total_calka,
! CONSTANT is omitted here
write(1,*)tempred,total_calka
close(1)
end program wykres

Answer 1

The inte is declared in both modules.
Upd. The inte(y,beta,r2,r1) function is defined in the module in, and is used in the main program. This function requires four arguments, but this call
irange=inte(intrange/division)
provides only one argument. I'm not sure if this function should be used in this case. Try to use long meaningful names for variables and functions to avoid similar issues.