我正在从输入文件创建链接列表。链接列表中的每个节点都在输入的每个块中包含信息。 当我尝试从输入文件中获取值时,我将多行(每次4行)分配到一个1D数组“ tmp”中。我认为我在“ input:DO”循环中使用的DO循环是错误的。但是我不知道该怎么解决。
我使用gfortran进行编译,编译没有错误。 我已经尝试在“ input:DO”循环中进行WRITE测试。结果显示我可以成功打开输入文件。
PROGRAM read
IMPLICIT NONE
INTEGER,PARAMETER :: nat=4
character(len=20) :: filename
!Derived types to store atom data
TYPE :: atom
CHARACTER(LEN=2) :: atom_name
REAL, DIMENSION(3) :: coord
END TYPE atom
!The array info stores info of all atom in one time step
type :: atom_seq
type(atom),dimension(nat):: info
type(atom_seq),pointer :: p
end type atom_seq
TYPE (atom_seq), POINTER :: head
TYPE (atom_seq), POINTER :: tail
type(atom), dimension(nat) :: temp
! Declare variable
INTEGER :: istat
INTEGER :: i=0, n=0
! Open input data file
WRITE(*,*) 'ENTER the file name with the data to be read: '
READ(*,'(A20)') filename
NULLIFY(head)
OPEN( UNIT=9, FILE=TRIM(filename), STATUS="OLD", ACTION="READ", IOSTAT=istat)
! Was the open successful
fileopen: IF (istat == 0) THEN
input: DO
!WRITE(*,*) "OPEN done " ! for testing
READ(9,*) ! <--when run, error is in this line
READ(9,*)
DO i = 1, nat
READ(9,*,IOSTAT=istat) temp(i)%atom_name, temp(i)%coord(1), temp(i)%coord(2), temp(i)%coord(3)
ENDDO
IF (istat /= 0) EXIT
n = n + 1 ! Bump count
IF (.NOT. ASSOCIATED(head) ) THEN ! No values in list
ALLOCATE(head, STAT=istat) ! Allocate new value
tail => head ! Tail points to new value
NULLIFY(tail%p) ! Nullify p in new value
DO i = 1, nat ! Store number
tail%info(i)%atom_name = temp(i)%atom_name
tail%info(i)%coord(1) = temp(i)%coord(1)
tail%info(i)%coord(2) = temp(i)%coord(2)
tail%info(i)%coord(3) = temp(i)%coord(3)
ENDDO
ELSE ! Values already in list
ALLOCATE(tail%p, STAT=istat) ! Allocate new value
tail => tail%p
NULLIFY(tail%p)
DO i = 1, nat ! Store number
tail%info(i)%atom_name = temp(i)%atom_name
tail%info(i)%coord(1) = temp(i)%coord(1)
tail%info(i)%coord(2) = temp(i)%coord(2)
tail%info(i)%coord(3) = temp(i)%coord(3)
ENDDO
END IF
END DO input
ELSE fileopen
WRITE(*,1030) istat
1030 FORMAT ('File open failed --status = ', I6)
END IF fileopen
END PROGRAM read
输入文件:inp
4
Particles:1_0
O 0.8050005000 0.7000000000 3.2350000000
H 1.4750005000 1.2800000000 2.8650000000
H 0.8550005000 -0.0900000000 2.7150000000
O 0.4050005000 0.7500000000 -4.1350000000
4
Particles:1_5
O 0.8799478358 0.6383317306 3.1569568025
H 1.4046592860 1.2232485826 2.4978364665
H 1.1472783731 -0.2687458123 3.0229092093
O 0.5392992531 0.6047144782 -4.0811918365
4
Particles:1_10
O -3.8021765454 3.1600783692 -4.5455655916
H -4.5320715486 3.0937504111 4.9514896261
H -3.5088238380 4.0613340230 -4.5394597924
O -3.3469012765 -0.7064128847 1.2465212113
,错误是
hg@xi /home/hg/pole $ ./read
ENTER the file name with the data to be read:
inp
At line XXX of file read.f95 (unit = 9, file = 'inp')
Fortran runtime error: End of file
Error termination. Backtrace:
#0 0x7f1c1fdbb31a
#1 0x7f1c1fdbbec5
#2 0x7f1c1fdbc68d
#3 0x7f1c1ff32a33
#4 0x7f1c1ff364b7
#5 0x7f1c1ff365b8
#6 0x5566d3dc9daf
#7 0x5566d3dca9ed
#8 0x7f1c1f9d0b96
#9 0x5566d3dc9a79
#10 0xffffffffffffffff
我希望解决该问题。如果我的想法是错误的,请提出一些建议,以设计一个更好的数据结构以将数据保存在输入中(输入文件可能具有数千个块,而不是3个块。它很大并且块数未知,然后运行代码) )
答案 0 :(得分:1)
在循环中第一个执行的IOSTAT
中缺少对READ
的测试。当结果不正确时,例如可以终止循环。更改:
fileopen: IF (istat == 0) THEN
input: DO
!WRITE(*,*) "OPEN done " ! for testing
READ(9,*) ! <--when run, error is in this line
READ(9,*)
在
fileopen: IF (istat == 0) THEN
input: DO
!WRITE(*,*) "OPEN done " ! for testing
READ(9,*,IOSTAT=istat)
IF (istat /=0) EXIT
READ(9,*)