fortran 错误

believefuture

program main
implicit none
integer m,k,s1_,status,i,j  !m代表文件数，k代表每个文件的行数
real sum1,sum2
character,allocatable::filename1(:)*50,filename2(:)*50
integer,allocatable::zhandian(:)
real,allocatable::temp(:),shidu(:)
character,allocatable::riqi(:)*15
character*15 rq
integer zd(56)
data zd /58015,58102,58122,58203,58215,58221,58236,58311,&
58314,58319,58321,58326,58334,58424,58436,58437,&
58531,58027,58040,58138,58144,58150,58238,58241,&
58251,58259,58265,58343,58345,58358,58445,58457,&
58464,58467,58472,58473,58477,58543,58549,58556,&
58562,58569,58633,58646,58647,58653,58659,58660,&
58665,58666,58667,58752,58760,58765,58362,58367/
open(1,file="f:\8089\namelist1.txt")
do while(.true.)
read(1,*,iostat=status)
i=i+1
if(status/=0) exit
enddo
m=i-1
print *,m
allocate(filename2(m))
allocate(filename1(m))
rewind(1)
do i=1,m
read(1,*) filename1(i)
enddo
close(1)
open(1,file="f:\8089\namelist2.txt")
do i=1,m
read(1,*) filename2(i)
enddo
close(1)
do i=1,m
open(2,file=filename1(i))
open(3,file=filename2(i))
k=0
do while(.true.)
read(2,*,iostat=status)
k=k+1
if(status/=0)exit
enddo
k=k-1
print*,k
allocate(riqi(k))
allocate(zhandian(k))
allocate(temp(k))
allocate(shidu(k))
rewind(2)
do j=1,k
read(2,*) zhandian(j),riqi(j),temp(j),shidu(j)
!print*,zhandian(j),riqi(j),shijian(j),q(j),nengjiandu(j)
end do
rewind(2)
s1_=zhandian(1)
do j=1,k
read(1,*) zhandian(j),riqi(j),temp(j),shidu(j)
end do
rewind(1)
do j=1,k
do i=1,56
if (zhandian(j)==s1_.and.zhandian(j)==zd(i)) then
sum1=sum1+temp(j)
sum2=sum2+shidu(j)
m=m+1
else
write(3,'(5i,3x,f6.2,3x,f6.2)') s1_,sum1/m,sum2/m
m=1
s1_=zhandian(j)
end if
end do
end do
write(3,'(5i,3x,f6.2,3x,f6.2)') s1_,sum1/m,sum2/m
close(3)
close(2)
deallocate(zhandian)
deallocate(riqi)
deallocate(temp)
deallocate(shidu)
enddo
end

lqouc

报错说的很清楚了，某个do循环的变量和外层循环的重复了，既然没把定位贴上来就自己改吧，大段程序眼花了也找不到。

believefuture

lqouc 发表于 2014-8-29 16:03
报错说的很清楚了，某个do循环的变量和外层循环的重复了，既然没把定位贴上来就自己改吧，大段程序眼花了也 ...

谢谢

言深深

先精简一下你的程序吧，一次又一次的rewind做什么，为何不能一次性的read到数组里面，然后对数组展开操作呢？