unexpected junk after else statement at

18778386191

[Fortran] 纯文本查看 复制代码

MODULE MOD_FUNC

! module contains functions for P and S-wave velocities, density and
! quality factors in sediments (1st layer)

IMPLICIT NONE

CONTAINS
!==================== 1D INHOMOGENEOUS MEDIUM =========================
!------------------------------------------------- P or S-wave velocity
  FUNCTION FUNCV (Z)

    USE NRTYPE, ONLY: WP

    REAL(WP), INTENT(IN) :: Z
    REAL(WP)             :: FUNCV

    if (z<5) then
      !layer
      funcv = 180
    else if(5<=z<18) then
      !layer
      funcv = 200
    else if(18<=z<54) then
      !layer
      funcv = 360
    else if(54<=z<92) then
      !layer
      funcv = 400
    else if(92<=z<130) then
      !layer
      funcv = 440
    else if(130<=z<165) then
      !layer
      funcv = 460

    else
      !halfspace
      funcv = 970
    end if

  END FUNCTION FUNCV

!--------------------------------------------------------------- Density
  FUNCTION FUNCRHO (Z)

    USE NRTYPE, ONLY: WP

    REAL(WP), INTENT(IN) :: Z
    REAL(WP)             :: FUNCRHO


    if(z<5) then
      !layer
      funcrho = 1220
    else if(5<=z<18) then
      !layer
      funcrho = 1487
    else if(18<=z<54) then
      !layer
      funcrho = 1945
    else if(54<=z<92) then
      !layer
      funcrho = 1966
    else if(92<=z<130) then
      !layer
      funcrho = 1966
    else if(130<=z<165) then
      !layer
      funcrho = 1966
    else
      !halfspace
      funcrho = 2123
    end if

  END FUNCTION FUNCRHO

!------------------------------------------- P or S-wave quality factor
  FUNCTION FUNCQ (Z)

    USE NRTYPE, ONLY: WP

    REAL(WP), INTENT(IN) :: Z
    REAL(WP)             :: FUNCQ

    if(z<165) then
      !layer
      funcq = 50
    else
      !halfspace
      funcq = 10000
    end if
  END FUNCTION FUNCQ

END MODULE

这是代码，运行后出现图片的错误，unexpected junk after else statement

fcode

130<=z .and. z<165

fcode

其实你的函数可以简化一下。

[Fortran] 纯文本查看 复制代码

  FUNCTION FUNCRHO (Z)
    USE NRTYPE, ONLY: WP
    REAL(WP), INTENT(IN) :: Z
    REAL(WP)             :: FUNCRHO
    Real(WP) , parameter :: zList(*) = [-huge(z),5._WP,18._WP,54._WP,92._WP,130._WP,165._WP,huge(z)]
    Real(WP) , parameter :: fList(*) = [1220._WP,1487._WP,1945._WP,1966._WP,1966._WP,1966._WP,2123._WP]
    integer :: i
    Do i = 1 , size(zList)-1
      if( zList(i) <= z .and. z < zList(i+1) ) then
        funcRho = fList(i)
        return
      end if
    End Do    
  END FUNCTION FUNCRHO

这样更易读、易修改、易扩展

necrohan

5<=z<18
这是新的写法吗？可以编译成功但是结果不对。

18778386191

fcode 发表于 2019-11-14 09:01
其实你的函数可以简化一下。
[mw_shl_code=fortran,true]  FUNCTION FUNCRHO (Z)
    USE NRTYPE, ONLY: WP ...

谢谢了。改了就对了

vvt

necrohan 发表于 2019-11-14 09:02
5

5<=z<18
等效于
(5<=z)<18
那么

情况1：
如果 5<=z 是 .true.
而.true. 一般在内存里是 0x01 或 0x0001 之类，或者 0xFF 之类的
.true. < 18 等效于 1<18 或 -1<18，结果是 .true.

情况2:
如果 5<=z 是 .false
而.false. 一般在内存里是 0x00 或 0x0000 之类的
.false. < 18 等效于 0<18，结果也是 .true.

所以，不管怎样，这个表达式的结果都是 .true.

necrohan

vvt 发表于 2019-11-14 19:38
5

明白了，谢谢