Question

我曾尝试在fortran中编写代码来计算协调编号。但是得到了这些错误

 coord.f:43.72:

            read(13,*) ri(i),  g(2,2,i),   g(1,2,i),   g(1,2,i),    g(1,
                                                                        1
Error: Expected array subscript at (1)
coord.f:78.38:

          call integrate(npts-1,ri,gt,ans)                              
                                      1
Warning: Rank mismatch in argument 'ans' at (1) (scalar and rank-2)
coord.f:79.8:

        t1(ia,ib)=ans                                                   
        1
Error: Incompatible ranks 0 and 2 in assignment at (1)
coord.f:52.32:

      call coordination(ri,g,ro,num)                                    
                                1
Warning: Invalid procedure argument at (1)

这是我尝试过的代码。谁能请帮我解决这些错误？

ccccc   Solution for Coordination number: 

c   2 :macro-ion
c   1 :counter-ion

        include "prmts" ! parameter for npts, l, pi, bl!
        character init*10

        integer icont(l,l)
        double precision grid, dm22, dr, dt, num 
        double precision g(l,l,npts),
     &  ro(l,l),z(l),r(l),ri(npts),dm(l,l), 
     &  h(l,l,npts),ans(l,l), t1(l,l),gt(npts)

        open(unit=13,file='grm.out.txt',status='old')
        open(unit=14,file='cor.out',status='unknown')

         read(13,*)(z(i),i=1,l)  ! algebric charge  !
         read(13,*)(r(i),i=1,l)  ! radious of ions  !
         read(13,*)ro(2,2)   ! no.density of 2  !               
        do i = 1, l
          r(i) = r(i)
        enddo
        dm22 = dm22
        dr = r(2)/grid
        ro(2,2) = ro(2,2)
        ro(1,1) = -z(2)*ro(2,2)/z(1)
                   ! From condition of electro neutrality !


        open(unit=13,file='grm.out.txt',status='old')
         do i=1, npts-1
            read(13,*) ri(i),  g(2,2,i),   g(1,2,i),   g(1,2,i),    g(1,1,i)
         enddo     
        close(13)

CCCCC   CALCULATE COORDINATION NUMBER

        call coordination(ri,g,ro,num)

        write(14,*)"# Cornum = ", num(ia,ib) 
        write(14,*)"#----------------------------------------------"
        write(14,*)"#   num22   num21   num12  num11    "
        write(14,*)"#----------------------------------------------" 
999 format( 4f18.6)

        stop
        end    ! end of main !


        subroutine coordination(ri,g,ro,num)
        include 'prmts' ! for l, npts, pi and bl !
        double precision ri(npts), g(l,l,npts), ro(l,l)
     &  , num(l,l), ans(l,l), t1(l,l),gt(npts)
        integer i, ia, ib

CCCCC   COORDINATION NUMBER
CCCCC   Cornum : Coordination number.

        do ia=1,l
        do ib=1,l
        do i=1,npts-1
          gt(i)=g(ia,ib,i)*ri(i)**2
        enddo
          call integrate(npts-1,ri,gt,ans)
          t1(ia,ib)=ans
        enddo
        enddo

CCCCC   COORDINATION NUMBER

        do ia=1,l
        do ib=1,l
          num(ia,ib)= 4.0d0*pi*ro(ib,ib)*t1(ia,ib)
        enddo
        enddo
        write(*,*) 'Cornum = ', num(ia,ib)
        end

      subroutine integrate(n,x,y,ans)
      integer          nin, nout
      parameter        (nin=5,nout=6)
      double precision ans, error
      integer          ifail, n
      double precision x(n), y(n)
         ifail = 1

         call pintegr(x,y,n,ans,error,ifail)

         if (ifail.eq.0) then
            write (nout,99999) 'integral = ', ans,
     +        '     estimated error = ', error
         else if (ifail.eq.1) then
            write (nout,*) 'less than 4 points supplied'
         else if (ifail.eq.2) then
            write (nout,*)
     +        'points not in increasing or decreasing order'
         else if (ifail.eq.3) then
            write (nout,*) 'points not all distinct'
         end if
      return

99999 format (1x,a,e12.4,a,e12.4)
      end

      subroutine pintegr(x,y,n,ans,er,ifail)
      double precision  ans, er
      integer n
      double precision  x(n), y(n)
      double precision c, d1, d2, d3, h1, h2, h3, h4, r1, r2, r3, 
     * r4, s
      integer i, nn
      ans = 0.0d0
      er = 0.0d0
      if (n.ge.4) go to 20
      ifail = 1
      return
        h2 = x(2) - x(1)  20
        do 80 i = 3, n
         h3 = x(i) - x(i-1)
         if (h2*h3) 40, 40, 80
        write(*,*)'points not specified correctly'        40
         ifail = 3
         return
     continue         80

      d3 = (y(2)-y(1))/h2
      h3 = x(3) - x(2)
      d1 = (y(3)-y(2))/h3
      h1 = h2 + h3
      d2 = (d1-d3)/h1
      h4 = x(4) - x(3)
      r1 = (y(4)-y(3))/h4
      r2 = (r1-d1)/(h4+h3)
      h1 = h1 + h4
      r3 = (r2-d2)/h1
      ans = h2*(y(1)+h2*(d3/2.0d0-h2*(d2/6.0d0-(h2+2.0d0*h3)*r3/12.0d0))
     *      )
      s = -(h2**3)*(h2*(3.0d0*h2+5.0d0*h4)+10.0d0*h3*h1)/60.0d0
      r4 = 0.0d0

      nn = n - 1
      do 120 i = 3, nn
         ans = ans + h3*((y(i)+y(i-1))/2.0d0-h3*h3*(d2+r2+(h2-h4)*r3)
     *         /12.0d0)
         c = h3**3*(2.0d0*h3*h3+5.0d0*(h3*(h4+h2)+2.0d0*h4*h2))/120.0d0
         er = er + (c+s)*r4
         if (i.ne.3) s = c
         if (i.eq.3) s = s + 2.0d0*c
         if (i-n+1) 100, 140, 100
         h1 = h2                   100
         h2 = h3
         h3 = h4
         d1 = r1
         d2 = r2
         d3 = r3
         h4 = x(i+2) - x(i+1)
         r1 = (y(i+2)-y(i+1))/h4
         r4 = h4 + h3
         r2 = (r1-d1)/r4
         r4 = r4 + h2
         r3 = (r2-d2)/r4
         r4 = r4 + h1
         r4 = (r3-d3)/r4
       continue       120

       continue       140
          ans = ans + h4*(y(n)-h4*(r1/2.0d0+h4*(r2/6.0d0+(2.0d0*h3+h4)
         *      *r3/12.0d0)))
           er = er - h4**3*r4*(h4*(3.0d0*h4+5.0d0*h2)+10.0d0*h3*(h2+h3+h4))
         *     /60.0d0 + s*r4
           ans = ans + er
           ifail = 0
            return
           end

END of Program

Answer 1

关键是要注意第一个错误是在第72列，并且错误信息中的行不完整。

对于只允许72个字符的Fortran 77，该行很长。将行划分为或使用某些编译器选项以允许更长的行（中断可移植性）。

还有其他错误，例如

t1(ia,ib)=ans

t1(ia,ib)是一个标量（一个数组元素），ans是一个数组。这是不可能的，你必须保持一致。

由于隐式类型，过程integrate将其伪参数ans定义为标量实数，但是将数组ans传递给它。这也是不可能的。

我建议您在所有程序中使用implicit none，虽然技术上不是Fortran 77.它可以帮助您避免许多错误。

最后，num被声明为double precision，但用作

num(ia,ib)

因此编译器认为它是一个外部函数，并警告您无效的过程参数。您必须保持一致并将其声明为数组或将其用作标量。

可能代码中存在更多错误，它非常混乱且难以阅读。

分配中的等级0和2不兼容

1 个答案: