对变量的含糊不清的引用 [英] Ambiguous reference to variable
问题描述
所以我正在做2个链接到主程序的模块.第一个定义了所有变量,第二个定义了函数.
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
在我这样称呼他们的时候:
And while im calling them like that:
use zmienne; use in
我遇到以下错误:
Name 'inte' at (1) is an ambiguous reference to 'inte' from module 'zmienne'
我已在module1中删除了"inte",但现在出现以下错误:
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)
主程序代码为:
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
推荐答案
在两个模块中都声明了inte
.
The inte
is declared in both modules.
更新.inte(y,beta,r2,r1)
功能在模块in
中定义,并在主程序中使用.此函数需要四个参数,但此调用
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)
仅提供一个参数.我不确定在这种情况下是否应该使用此功能.尝试对变量和函数使用有意义的长名称,以避免类似的问题.
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.
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