分配中不兼容的等级0和2
我试图用fortran编写代码来计算协调号.但是出现了这些错误
I have tried to write a code in fortran to calculate the coordination number. But a got these errors
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)
这是我尝试过的下面的代码.任何人都可以 帮助我解决这些错误?
This is the code below which I have tried for it. Can anyone please help me to solve these errors?
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
关键是要注意第一个错误在第72列，并且错误消息中的行不完整.
The key is to notice that the first error is at column 72 and that the line is incomplete in the error message.
对于Fortran 77，该行很长，仅允许72个字符.将这些行划分为或使用某些编译器选项以允许更长的行(破坏可移植性).
The line is to long for Fortran 77 which allows only 72 characters. Divide the lines into to or use some compiler option to allow longer lines (breaks portability).
还有其他错误，例如
t1(ia,ib)=ans
t1(ia,ib)是标量(一个数组元素)，ans是一个数组.这是不可能的，您必须保持一致.
t1(ia,ib) is a scalar (one array element) and ans is an array. That is not possible, you must be consistent.
由于隐式键入，过程integrate将其伪参数ans定义为标量实数，但是您将数组ans传递给它.这也是不可能的.
The procedure integrate defines its dummy argument ans as a scalar real due to implicit typing, but you pass an array ans to it. That is also not possible.
我建议您在所有程序中使用implicit none，尽管从技术上讲它不是Fortran77.它将帮助您避免很多错误.
I suggest you to use implicit none in all programs although it is technically not Fortran 77. It will help you to avoid many errors.
最后，num被声明为double precision，但被用作
Finally, num was declared as as double precision but used as
num(ia,ib)
由于这个原因，编译器认为它是一个外部函数，并警告您有关无效的过程参数的信息.您必须保持一致并将其声明为数组或将其用作标量.
for this reason the compiler thinks it is an external function and warns you about the invalid procedure argument. You must be consistent and declare it as an array or use it as a scalar.
可能在代码中还有更多错误，这很杂乱且很难阅读.
Probably, there are many more errors in the code, it is quite messy and very hard to read.