如何简化这个Fortran函数? [英] How to simplify this Fortran function?
问题描述
我有下面的代码来计算第一个逻辑的连续数b $ b逻辑。我想简化这个功能。有
一直在想,也许我可以使用递归,但现在完全相当
。这是否是possilble?
I have the following code that counts the number of consecuitive logicals from the first one. I want to simplify this function. Have been thinking that perhaps I can use recursion somehow but now quite sure. Is that possilble?
Function count_present &
( &
p1, p2, p3, p4, p5, p6, p7, p8 &
) &
Result (n)
Logical, Intent (in) :: p1
Logical, Intent (in), Optional :: p2, p3, p4, p5, p6, p7, p8
Integer :: n
n = 0
If (Present (p8)) Then
If (p8) Then; n = 8
Else If (p7) Then; n = 7
Else If (p6) Then; n = 6
Else If (p5) Then; n = 5
Else If (p4) Then; n = 4
Else If (p3) Then; n = 3
Else If (p2) Then; n = 2
Else If (p1) Then; n = 1
End If
Else If (Present (p7)) Then
If (p7) Then; n = 7
Else If (p6) Then; n = 6
Else If (p5) Then; n = 5
Else If (p4) Then; n = 4
Else If (p3) Then; n = 3
Else If (p2) Then; n = 2
Else If (p1) Then; n = 1
End If
Else If (Present (p6)) Then
If (p6) Then; n = 6
Else If (p5) Then; n = 5
Else If (p4) Then; n = 4
Else If (p3) Then; n = 3
Else If (p2) Then; n = 2
Else If (p1) Then; n = 1
End If
Else If (Present (p5)) Then
If (p5) Then; n = 5
Else If (p4) Then; n = 4
Else If (p3) Then; n = 3
Else If (p2) Then; n = 2
Else If (p1) Then; n = 1
End If
Else If (Present (p4)) Then
If (p4) Then; n = 4
Else If (p3) Then; n = 3
Else If (p2) Then; n = 2
Else If (p1) Then; n = 1
End If
Else If (Present (p3)) Then
If (p3) Then; n = 3
Else If (p2) Then; n = 2
Else If (p1) Then; n = 1
End If
Else If (Present (p2)) Then
If (p2) Then; n = 2
Else If (p1) Then; n = 1
End If
Else
If (p1) n = 1
End If
End Function count_present
推荐答案
递归。请注意,
It is possible to write this using recursion. Note that
count_present(p_1, p_2, ..., p_n, p_{n+1})
返回值 count_present(p_1,p_2,...,p_n)$ c $除非存在所有
, p_1
,..., p_ {n + 1}
并且 .TRUE。
。在后一种情况下,结果是 n + 1
。 count_present(p_1)
如果 p_1
为<$,则返回 1
c $ c> .TRUE。 0
否则。
returns the value count_present(p_1, p_2, ..., p_n)
unless all p_1
, ..., p_{n+1}
are present and .TRUE.
. In this latter case the result is n+1
. count_present(p_1)
returns 1
if p_1
is .TRUE.
, 0
otherwise.
recursive function count_present(p1, p2, p3, p4, p5, p6, p7, p8) result (res)
logical, intent(in) :: p1, p2, p3, p4, p5, p6, p7, p8
optional p2, p3, p4, p5, p6, p7, p8
integer res
if (PRESENT(p8)) then
res = count_present(p1, p2, p3, p4, p5, p6, p7)
if (res.eq.7.and.p8) res = res+1
else if (PRESENT(p7)) then
res = count_present(p1, p2, p3, p4, p5, p6)
if (res.eq.6.and.p7) res = res+1
else if (PRESENT(p6)) then
res = count_present(p1, p2, p3, p4, p5)
if (res.eq.5.and.p6) res = res+1
else if (PRESENT(p5)) then
res = count_present(p1, p2, p3, p4)
if (res.eq.4.and.p5) res = res+1
else if (PRESENT(p4)) then
res = count_present(p1, p2, p3)
if (res.eq.3.and.p4) res = res+1
else if (PRESENT(p3)) then
res = count_present(p1, p2)
if (res.eq.2.and.p3) res = res+1
else if (PRESENT(p2)) then
res = count_present(p1)
if (res.eq.1.and.p2) res = res+1
else
res = COUNT([p1])
end if
end function count_present
这是个好主意吗?那么,这是另一个问题。
Is this a good idea? Well, that's another question.
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