0

在这个fortran程序中，我被告知要调试，我得到了错误：

“在 (1) 处的参数 'p1' 中的类型不匹配；将 REAL(4) 传递给 TYPE(point)”

而且我似乎无法弄清楚错误发生在哪里。我尝试定义不同的变量来传递给每个函数，而不是 p1 和 p2，但错误相同。有任何想法吗？

MODULE PointType

TYPE POINT
REAL:: x
REAL:: y
END TYPE

CONTAINS

FUNCTION arePointsEqual(p1, p2)
REAL:: arePointsEqual
TYPE(POINT), INTENT(IN):: p1
TYPE(POINT), INTENT(IN):: p2
LOGICAL :: isEqual
IF ( p1%x == p2%x .AND. p1%y == p2%y) THEN
isEqual = .TRUE.
ELSE
isEqual = .FALSE.
END IF
END FUNCTION

 FUNCTION arePointsNotEqual(p1,p2)
 REAL:: arePointsNotEqual
 TYPE(POINT), INTENT(IN):: p1
 TYPE(POINT), INTENT(IN):: p2
 LOGICAL :: isNotEqual

 IF ( p1%x == p2%x .AND. p1%y == p2%y) THEN
 isNotEqual = .FALSE.
 ELSE
 isNotEqual = .TRUE.
 END IF
 END FUNCTION

 FUNCTION distance(p1, p2)
 REAL:: distance
 TYPE(POINT), INTENT(IN):: p1
 TYPE(POINT), INTENT(IN):: p2
 distance = SQRT((p2%x - p1%x)**2 + (p2%y - p1%y)**2)
 END FUNCTION

 END MODULE

 !MAIN PROGRAM BELOW THIS LINE

 PROGRAM Project3

 USE PointType

 PRINT *, arePointsEqual(p1, p2)

 PRINT *, arePointsNotEqual(p1, p2)

 PRINT *, distance(p1, p2)

 END PROGRAM Project3
4

1 回答 1

4

提示：尝试“隐式无”。总是一个好主意。

于 2012-11-21T05:34:29.557 回答