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Old   September 10, 2008, 06:28
Default fortran derived type assignment
  #1
Paolo Lampitella
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Hi, probably this is not the best place for such a question but i think that most of you have the right knowledge of fortran (better than me of course) to help me.

I'm writing a genetic algorithm code which makes use of external modules. The code is like the following (i put between quotes "" the not important stuff):

PROGRAM example

USE external_module

IMPLICIT NONE

INTEGER(KIND=my_int_kind) :: "integer stuff"

REAL(KIND=my_real_kind) :: "real stuff"

TYPE(my_der_type) :: type_one,type_two

"Allocation of some components of type_one which have the pointer attribute to have variable dimension"

CALL initialize_type(type_one)

CALL show_me(type_one)

type_two=type_one

CALL modify_type(type_two)

CALL show_me(type_one)

END PROGRAM example

What actually happens is that after having modified type_two, type_one also is modified. I know that there could be thousands reasons for this (some of them have already been checked) but, what i would know is that if, from what i wrote of my code, there is something wrong which i'm doing with the derived types and that determines the wrong modification. I'm using Compaq Visual Fortran 6.0.

Thanks
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Old   September 12, 2008, 04:53
Default Re: fortran derived type assignment
  #2
Paolo Lampitella
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Even if anyone has the answer for my question, maybe some of you could be interested in it (which i found in a manual). Also because i have another question.

The assignment:

type_two=type_one

when the derived type has an allocatable array of pointers as a component, is actually equivalent to

type_two=>type_one

so it is equivalent to a target assignment and this is why type_one was changing with type_two.

Now, the question is:

Because i'm going to modify my whole program to handle this, do you know if assigning my derived type component by component (instead of a single =) is going to produce the same result or what i need for it?

Thanks
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