FORTRAN功能 [英] FORTRAN functions
问题描述
我正在研究一个需要在FORTRAN中实现少数数值方法的项目。为此,我需要编写一些递归函数。这是我的代码。
!
!文件:main.F95
!
RECURSIVE FUNCTION integrate(n)RESULT(rv)
IMPLICIT NONE
双精度:: 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
双精度:: rv1
PRINT *,阶乘(5)
PRINT *,集成(2)
!READ *,rv1
END PROGRAM main
对于这个程序,输出是:
NaN
1
如果我改变打印语句的顺序(第30行&输出将是:
1
-19.000000
输出应该是(对于原始打印语句订单):
120
-19
我从维基百科 Fortran 95语言特性页面。
我是FORTRAN的新手,我不知道我的代码有什么问题。请帮助我们。
- 编译器:使用Cygwin的gfortran 4.5.3
- IDE:Netbeans 7.0 .1
- 平台:Windows 7
预先感谢您。
您的函数写入正确。问题出在主程序中,你没有明确地声明集成
和 factorial
函数的类型,所以你有隐式输入,在这种情况下 REAL
和集成
被假定为 INTEGER
。出于某种原因,你的编译器没有警告你类型不匹配。 Mine did:
$ gfortran recurs.f90
recurs.f90:26.22:
(1)(INTEGER(4)/ REAL(8))
recurs.f90:27.22(2)
1
:
PRINT *,阶乘(5)
1
错误:函数'factorial'在(1)(REAL(4)/ INTEGER(4))的返回类型不匹配
您应该将主程序更改为:
PROGRAM main
IMPLICIT NONE
DOUBLE PRECISION,EXTERNAL :: integrate
INTEGER,EXTERNAL :: factorial
PRINT *,factorial(5 )
PRINT *,integrate(2)
END PROGRAM main
注意 IMPLICIT NONE
行。这个声明语句将禁用任何隐式类型,如果不是所有的变量和函数都被显式声明的话,编译器会抛出一个错误。这是每个Fortran程序中非常重要的一行,如果你拥有它,你自己就会发现自己的问题,因为它会迫使你明确地声明程序中的所有东西。
现在的输出是:
120
-19.0000000000000
如预期的那样。
另外, DOUBLE PRECISION
类型声明并不像使用 REAL
并指定 KIND
参数那么灵活,例如一个 REAL(KIND = myRealKind)
。查看有关如何正确使用 KIND
的问题的答案: Fortran 90种参数。
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. I'm new in FORTRAN, I don't know what's wrong in my code. Please help me guys.
- Compiler : gfortran 4.5.3 with Cygwin
- IDE: Netbeans 7.0.1
- Platform: Windows 7
Thank you in advance.
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.
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