Fortran 在另一个函数中调用函数（在子例程中定义）

我正在用fortran编写函数func（x，y）的二维积分代码，y的极限从y1（x）到y2（x），x的极限从x1=3到x2=5。 基本假设如下： 积分[func（x，y），{y=y1到y2}，{x=x1到x2}]=积分[funcx（x），{x=x1到x2}]。其中funcx=积分[func，{y=y1（x）到y2（x）}]

请在代码中间找到我的问题。我写代码如下

implicit real*8(a-h,o-z)

external func
external y1
external y2

x1 = 3.d0
x2 = 5.d0

call twodint(func,y1,y2,x1,x2,result)
print*, result

stop
end

!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

subroutine twodint(origfunc,y1,y2,x1,x2,result) 
implicit real*8(a-h,o-z)
external funcx
call simpsonintegral(funcx,x1,x2,ss) ! my integral routine funcx = function, x1 and x2 = limits, ss = output 
result = ss
return
end

function funcx(x)
implicit real*8(a-h,o-z)
external funcy
external y1
external y2
common/xvalues/xx
xx = x
call simpsonintegral(funcy,y1(x),y2(x),ss)
funcx = ss
return
end

function funcy(y)
implicit real*8(a-h,o-z)
common/xvalues/x

function func(x,y)
implicit real*8(a-h,o-z)
func = (x**2)*y*dsin(x*y) ! some function of x and y
return
end

function y1(x)
implicit real*8(a-h,o-z)
y1 = x ! some limit
return
end

function y2(x)
implicit real*8(a-h,o-z)
y2 = x*x ! some limit
return
end

!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

subroutine twodint(origfunc,y1,y2,x1,x2,result) 
implicit real*8(a-h,o-z)
external funcx
call simpsonintegral(funcx,x1,x2,ss) ! my integral routine funcx = function, x1 and x2 = limits, ss = output 
result = ss
return
end

function funcx(x)
implicit real*8(a-h,o-z)
external funcy
external y1
external y2
common/xvalues/xx
xx = x
call simpsonintegral(funcy,y1(x),y2(x),ss)
funcx = ss
return
end

function funcy(y)
implicit real*8(a-h,o-z)
common/xvalues/x

function func(x,y)
implicit real*8(a-h,o-z)
func = (x**2)*y*dsin(x*y) ! some function of x and y
return
end

function y1(x)
implicit real*8(a-h,o-z)
y1 = x ! some limit
return
end

function y2(x)
implicit real*8(a-h,o-z)
y2 = x*x ! some limit
return
end

!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

subroutine twodint(origfunc,y1,y2,x1,x2,result) 
implicit real*8(a-h,o-z)
external funcx
call simpsonintegral(funcx,x1,x2,ss) ! my integral routine funcx = function, x1 and x2 = limits, ss = output 
result = ss
return
end

function funcx(x)
implicit real*8(a-h,o-z)
external funcy
external y1
external y2
common/xvalues/xx
xx = x
call simpsonintegral(funcy,y1(x),y2(x),ss)
funcx = ss
return
end

function funcy(y)
implicit real*8(a-h,o-z)
common/xvalues/x

function func(x,y)
implicit real*8(a-h,o-z)
func = (x**2)*y*dsin(x*y) ! some function of x and y
return
end

function y1(x)
implicit real*8(a-h,o-z)
y1 = x ! some limit
return
end

function y2(x)
implicit real*8(a-h,o-z)
y2 = x*x ! some limit
return
end

!！我的问题是：如果我在这里写funcy=func（x，y），我的代码运行良好。但我想在这里写一些东西，比如funcy=origfunc（x，y），这样它就可以从名为twodint的子例程接收函数。但这是行不通的。请帮忙。。。 回来 结束

!~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

subroutine twodint(origfunc,y1,y2,x1,x2,result) 
implicit real*8(a-h,o-z)
external funcx
call simpsonintegral(funcx,x1,x2,ss) ! my integral routine funcx = function, x1 and x2 = limits, ss = output 
result = ss
return
end

function funcx(x)
implicit real*8(a-h,o-z)
external funcy
external y1
external y2
common/xvalues/xx
xx = x
call simpsonintegral(funcy,y1(x),y2(x),ss)
funcx = ss
return
end

function funcy(y)
implicit real*8(a-h,o-z)
common/xvalues/x

function func(x,y)
implicit real*8(a-h,o-z)
func = (x**2)*y*dsin(x*y) ! some function of x and y
return
end

function y1(x)
implicit real*8(a-h,o-z)
y1 = x ! some limit
return
end

function y2(x)
implicit real*8(a-h,o-z)
y2 = x*x ! some limit
return
end

---

在我看来，

external

主要是因为遗留的原因而包含在现代Fortran中的。这不是将函数传递给子例程的最佳方式。而且，隐式类型声明是危险的。。。最好使用

隐式无

并显式键入所有变量。。。那样你会犯很多错误。所以下面是如何将各种函数传递给子例程（如积分器）的示意图

module MyStuff

implicit none

abstract interface
  function funcX (x)
      real, intent (in) :: x
      real :: funcX
  end function funcX
end interface



contains

subroutine twodint(origfunc,y1,y2,x1,x2,result)

real :: y1, y2, x1, x2, result, ss
procedure (funcX) :: origfunc

call simpsonintegral (origfunc,x1,x2,ss) ! my integral routine funcx = function, x1 and x2 = limits, ss = output
result = ss
return
end subroutine twodint


real function funcA (x)
real, intent (in) :: x
funcA = x ** 2
end function funcA

real function funcB (x)
real, intent (in) :: x
funcB = x ** 2
end function funcB

end module MyStuff


program test1

use MyStuff

implicit none



real :: x1, x2, y1, y2, result


x1 = 3.d0
x2 = 5.d0

call twodint(funcA,y1,y2,x1,x2,result)
print*, result

call twodint(funcB,y1,y2,x1,x2,result)
print*, result

stop
end program test1

---

它怎么不起作用？它应该通过哪个子程序接收原始文件？请不要用大写字母书写@VladimirF我通过名为origfunc的子程序twodint传递了该函数，并希望它在我提出问题的地方被接收。如果需要进一步澄清，请告诉我。谁知道呢，您的程序充满了难看的

隐式
和
外部
语句，很难理解。至少对我来说，我也有同样的感觉，摆脱了
隐式的
和
外部的
。请注意，在代码末尾有一些伪星，如果用双精度常量初始化，reals可能会更高。我编辑以删除伪字符。