unexpected junk after else statement at
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
130<=z .and. z<165 其实你的函数可以简化一下。
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(*) =
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 5<=z<18
这是新的写法吗?可以编译成功但是结果不对。 fcode 发表于 2019-11-14 09:01
其实你的函数可以简化一下。
FUNCTION FUNCRHO (Z)
USE NRTYPE, ONLY: WP ...
谢谢了。改了就对了 本帖最后由 vvt 于 2019-11-14 19:40 编辑
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.
vvt 发表于 2019-11-14 19:38
5
明白了,谢谢
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