Fortran functions returning unexpected types and v

2020-02-06 08:17发布

问题:

I'm working on a project that needs to implement few numerical methods in Fortran. For this, I need to write some recursive functions. Here is my code.

!     
! File:   main.F95
!

RECURSIVE FUNCTION integrate(n) RESULT(rv)
    IMPLICIT NONE
    DOUBLE PRECISION :: rv
    INTEGER, INTENT(IN) :: n
    DOUBLE PRECISION, PARAMETER :: minusone = -1.0
    IF (n == 1) THEN
        rv = 10 !exp(minusone)
        RETURN
    ELSE
        rv = 1 - (n * integrate(n - 1))
        RETURN
    END IF
END FUNCTION integrate

RECURSIVE FUNCTION factorial(n) RESULT(res)
    INTEGER res, n
    IF (n .EQ. 0) THEN
        res = 1
    ELSE
        res = n * factorial(n - 1)
    END IF
END

PROGRAM main
    DOUBLE PRECISION :: rv1
    PRINT *, factorial(5)
    PRINT *, integrate(2)

    !READ *, rv1

END PROGRAM main

For this program the output is:

         NaN
       1

If I change the order of the print statements (line 30 & 31), the output will be:

         1
-19.000000

Output should be (for the original print statement order):

  120  
  -19 

I took the factorial function from the Wikipedia Fortran 95 language features page.

  • Compiler : gfortran 4.5.3 with Cygwin
  • IDE: Netbeans 7.0.1
  • Platform: Windows 7

回答1:

Your functions are written correctly. The problem is in the main program, where you do not explicitly declare the type of integrate and factorial functions, so you have implicit typing, in which case factorial is assumed REAL and integrate is assumed INTEGER. For some reason, your compiler did not warn you about type mismatch. Mine did:

$ gfortran recurs.f90 
recurs.f90:26.22:

    PRINT *, integrate(2)
                      1
Error: Return type mismatch of function 'integrate' at (1) (INTEGER(4)/REAL(8))
recurs.f90:27.22:

    PRINT *, factorial(5)
                      1
Error: Return type mismatch of function 'factorial' at (1) (REAL(4)/INTEGER(4))

You should change your main program to:

PROGRAM main
    IMPLICIT NONE
    DOUBLE PRECISION, EXTERNAL :: integrate
    INTEGER, EXTERNAL :: factorial
    PRINT *, factorial(5)
    PRINT *, integrate(2)
END PROGRAM main

Notice the IMPLICIT NONE line. This declaration statement will disable any implicit typing, and the compiler would throw an error if not all variables and functions are explicitly declared. This is a very important line in every Fortran program, and if you had it, you would've figured out your problem yourself, because it would force you to explicitly declare everything in your program.

The output now is:

         120
  -19.0000000000000     

as expected.

As a side note, the DOUBLE PRECISION type declaration is not as flexible as using REAL with KIND parameter specified instead, e.g. anREAL(KIND=myRealKind). See answers to this question about how to use KIND properly: Fortran 90 kind parameter.



回答2:

As one of the comments mentions, a better solution is to put your subroutines and functions into a module, then use that module from your main program. This will make the interface of those procedures known to the caller -- "explicit" in Fortran terminology. Not only will the compiler correctly handle the type of the function, it will be able to check type-agreement between the arguments in the call and the arguments in the callee ("dummy arguments") for consistency.

If you use as many debugging options as possible the compiler will help you find mistakes. With gfortran, try: -O2 -fimplicit-none -Wall -Wline-truncation -Wcharacter-truncation -Wsurprising -Waliasing -Wimplicit-interface -Wunused-parameter -fwhole-file -fcheck=all -std=f2008 -pedantic -fbacktrace

module factorial_procs

   IMPLICIT NONE

contains

   RECURSIVE FUNCTION integrate(n) RESULT(rv)
       DOUBLE PRECISION :: rv
       INTEGER, INTENT(IN) :: n

       IF (n == 1) THEN
           rv = 10
           RETURN
       ELSE
           rv = 1 - (n * integrate(n - 1))
           RETURN
       END IF
   END FUNCTION integrate

   RECURSIVE FUNCTION factorial(n) RESULT(res)
       INTEGER res, n
       IF (n .EQ. 0) THEN
           res = 1
       ELSE
           res = n * factorial(n - 1)
       END IF
   END

end module factorial_procs

PROGRAM main

    use factorial_procs

    implicit none

    PRINT *, factorial(5)
    PRINT *, integrate(2)

END PROGRAM main

You'll probably find that you can only calculate factorials of very small integers by straight forward multiplication using regular integers. One fix is to use a larger integer type, e.g.,

integer, parameter :: BigInt_K = selected_int_kind (18)

Just as you could modernize and use selected_real_kind instead of Double Precision.



回答3:

I would like to highlight some points while using RECURSIVE functions or non recursive function. 

1. Make sure the function has an explicit interface with the calling program. This can be achieved by putting function in a module and USE association. This is explained in aforementioned answer. 2. you can use INTERFACE to build an explicit interface with the main calling program, this is useful when you have a very few numbers of function which have implicit interface, like in your case. The example is given below.

PROGRAM main
IMPLICIT NONE
  INTERFACE 
   FUNCTION factorial(n)
     INTEGER:: factorial 
     INTEGER, INTENT(IN):: n
   END FUNCTION factorial

  FUNCTION integrate(n)
    DOUBLE PRECISION:: integrate
    INTEGER, INTENT(IN):: n
  END FUNCTION integrate
END INTERFACE 

PRINT *, factorial(5) 
PRINT *, integrate(2)

END PROGRAM main

note it always better to define kind parameter and then using KIND clause as!explained by @milancurcic

There is another very simple way to solve your problem: Just define factorial and itegrate in the main program as follows and you are ready to go

PROGRAM main
IMPLICIT NONE
    DOUBLE PRECISION :: integrate
    INTEGER:: factorial
    PRINT *, factorial(5)
    PRINT *, integrate(2)

END PROGRAM main