我是fortran的新手,我正在尝试编写和编译一个我在this address中找到的简单示例来读取一个简单的2列文本文件:
Date Water-flow
717976 7.140
717977 6.570
717978 6.040
717979 5.780
717980 5.530
程序粘贴在下面,但是当我尝试编译并运行时,我真的不明白产生的错误,请你指教一下吗?
PROGRAM READHPOP
IMPLICIT NONE
INTEGER, PARAMETER :: lun = 10
INTEGER :: res,i
CHARACTER(len=80) :: cbuffer
INTEGER :: flength
INTEGER,ALLOCATABLE,DIMENSION(:) :: dates
REAL,ALLOCATABLE,DIMENSION(:) :: water_flow
INTEGER :: c_position,string_length
OPEN(UNIT=lun,FILE="st.dat",FORM="FORMATTED",IOSTAT=res)
IF(res/=0) THEN
PRINT*,'error in opening file, status: ',res
STOP
END IF
READ(UNIT=lun,FMT='(A)',IOSTAT=res) cbuffer
IF(res /=0) THEN
PRINT *,'ERROR IN READING file, status: ',res
CLOSE(UNIT=lun)
STOP
END IF
string_length=LEN_TRIM(cbuffer)
c_position=INDEX(cbuffer,':')
READ(cbuffer(c_position+1:string_length),FMT='(A10)') flength
ALLOCATE(dates(flength),STAT=res)
IF (res/=0) THEN
PRINT*,'ERROR IN ALLOCATING MEMORY, status:',res
CLOSE(UNIT=lun)
STOP
END IF
READ(UNIT=lun,FMT='(A)',IOSTAT=res) cbuffer
DO i=1,cbuffer
READ(UNIT=lun,FMT='(I6,1X,F6.3)') dates(i),water_flow(i)
PRINT*,'DIAS ',dates(i)
END DO
end PROGRAM readhpop
编辑:感谢所有输入,只是为了解决问题,我粘贴下面的工作代码,问候!
PROGRAM READHPOP
IMPLICIT NONE
INTEGER, PARAMETER :: lun = 10
INTEGER :: res,i
CHARACTER(len=80) :: cbuffer
INTEGER :: flength
INTEGER,ALLOCATABLE,DIMENSION(:) :: dates
REAL,ALLOCATABLE,DIMENSION(:) :: water_flow
INTEGER :: c_position,string_length
OPEN(UNIT=lun,FILE="st.dat",FORM="FORMATTED",IOSTAT=res)
IF(res/=0) THEN
PRINT*,'error in opening file, status: ',res
STOP
END IF
READ(UNIT=lun,FMT='(A)',IOSTAT=res) cbuffer
IF(res /=0) THEN
PRINT *,'ERROR IN READING file, status: ',res
CLOSE(UNIT=lun)
STOP
END IF
string_length=LEN_TRIM(cbuffer)
c_position=INDEX(cbuffer,':')
READ(cbuffer(c_position+1:string_length),FMT='(I10)') flength
ALLOCATE(dates(flength),water_flow(flength),STAT=res)
IF (res/=0) THEN
PRINT*,'ERROR IN ALLOCATING MEMORY, status:',res
CLOSE(UNIT=lun)
STOP
END IF
DO i=1,flength
READ(UNIT=lun,FMT='(I6,F6.3)') dates(i),water_flow(i)
PRINT*,'LINE OF FILE',i,' DAYS ',dates(i),' WATERFLOW ',water_flow(i)
END DO
PAUSE
END PROGRAM READHPOP
答案 0 :(得分:3)
好的,这是一个工作版本:
PROGRAM READHPOP
IMPLICIT NONE
INTEGER, PARAMETER :: lun = 10
INTEGER :: res,i
CHARACTER(len=80) :: cbuffer
INTEGER :: flength
INTEGER,ALLOCATABLE,DIMENSION(:) :: dates
REAL,ALLOCATABLE,DIMENSION(:) :: water_flow
INTEGER :: c_position,string_length
OPEN(UNIT=lun,FILE="st.dat",FORM="FORMATTED",IOSTAT=res)
IF(res/=0) THEN
PRINT*,'error in opening file, status: ',res
STOP
END IF
READ(UNIT=lun,FMT='(A)',IOSTAT=res) cbuffer
IF(res /=0) THEN
PRINT *,'ERROR IN READING file, status: ',res
CLOSE(UNIT=lun)
STOP
END IF
string_length=LEN_TRIM(cbuffer)
c_position=INDEX(cbuffer,':')
READ(cbuffer(c_position+1:string_length),FMT='(I10)') flength
ALLOCATE(dates(flength),water_flow(flength),STAT=res)
IF (res/=0) THEN
PRINT*,'ERROR IN ALLOCATING MEMORY, status:',res
CLOSE(UNIT=lun)
STOP
END IF
DO i=1,flength
READ(UNIT=lun,FMT='(I6,F6.3)') dates(i),water_flow(i)
PRINT*,'DIAS ',dates(i)
END DO
end PROGRAM readhpop
的问题:
water_flow
需要分配flength
时:flength
是一个整数,因此请将其读作整数(此处为:'(I10)'
)。冒号后的日期数量,因此请将st.dat
更改为:Date Water-flow: 5
717976 7.140
717977 6.570
717978 6.040
717979 5.780
717980 5.530
flength
而不是cbuffer
答案 1 :(得分:2)
在这一行
READ(cbuffer(c_position+1:string_length),FMT='(A10)') flength
您在字符编辑描述符下读取整数值。这可能会返回flength
中的值,该值超出可分配数组的任何合理值。例如,在快速测试中,我得到了538981169
的值。将该行更改为
READ(cbuffer(c_position+1:string_length),'(i)') flength