Fortran：根据参数

Question

更改函数是否可以根据参数值更改调用程序的函数？我正在考虑类似于函数重载的东西，下面的例子显示了我在想什么。我只是想知道是否有更清洁/更好的方法来做到这一点。

function squareArea(s) result(A) 
    real :: s, A 
    A = s*s 
end function squareArea 

function circleArea(r) result(A) 
    real :: r, A 
    A = 3.14159 * r * r 
end function circleArea 

function Area(shape, dim) result(A) 
    character(len = *) shape 
    real dim, A 

    if (shape == 'circle') then 
     A = circleArea(dim) 
    elseif (shape == 'square') then 
     A = squareArea(dim) 
    end if 

end function Area 

program main 
    character(len = 6) :: sh = 'circle' 
    real :: r = 1.4 
    real :: A 

    A = Area(sh, r) 

    write(*,*) sh, r, A 

end program main 

Answer 1

是的 - 你会看到一种可能的方式。

Fortran 2003允许基于用于引用该过程的对象的动态类型来覆盖过程。这是否更好/更清洁取决于你的情况。

MODULE Shapes 
    IMPLICIT NONE 

    TYPE, ABSTRACT :: Shape 
    CONTAINS 
    PROCEDURE(shape_Area), DEFERRED :: Area 
    END TYPE Shape 

    INTERFACE 
    FUNCTION shape_Area(sh) RESULT(area) 
     IMPORT :: Shape 
     IMPLICIT NONE 
     CLASS(Shape), INTENT(IN) :: sh 
     REAL :: area 
    END FUNCTION shape_Area 
    END INTERFACE 

    TYPE, EXTENDS(Shape) :: Circle 
    REAL :: radius 
    CONTAINS 
    PROCEDURE :: Area => circle_Area 
    END TYPE Circle 

    TYPE, EXTENDS(Shape) :: Square 
    REAL :: side 
    CONTAINS 
    PROCEDURE :: Area => square_Area 
    END TYPE Square 
CONTAINS 
    FUNCTION circle_Area(sh) RESULT(area) 
    CLASS(Circle), INTENT(IN) :: sh 
    REAL :: area 
    area = 3.14159 * sh%radius**2 
    END FUNCTION circle_Area 
    FUNCTION square_Area(sh) RESULT(area) 
    CLASS(Square), INTENT(IN) :: sh 
    REAL :: area 
    area = sh%side**2 
    END FUNCTION square_Area 
END MODULE Shapes 

PROGRAM Areas 
    USE Shapes 
    IMPLICIT NONE 
    TYPE(Circle) :: c = Circle(1.4) 
    TYPE(Square) :: s = Square(1.4) 
    CHARACTER(*), PARAMETER :: fmt = "(A,G0,' has area ',G0)" 
    PRINT fmt, 'Circle with radius ', c%radius, c%Area() 
    PRINT fmt, 'Square with side ', s%side, s%Area() 
END PROGRAM Areas