Re: Derived types with allocatable arrays; more problems

Thanks for your swift reply.

If you explicitly write
do i = 1, n
a%b(i) = b(i)
end do
it works.

If b(i)%c=c (=integer array) works fine, why shouldn't a%b=b (=derived
type array) work in the same way?

Also it seems that you allocated a%b(n) but you did not allocate
a%b(i)%c(n)

If a derived type (b(i) in this case) contains an allocatable array, it
will automatically become allocated in an assignment, as far as I
understood. So, if I understood correctly, I do not need to explicitly
allocate a%b(i)%c(n).

I think that the problem is in the assignment between two derived
types. You'd better define your own assignment operator (=).

I used to have that when I was still using pointers, but now I switched
to allocatable arrays in derived types and thought/hoped I could live
without my own assignment operators.

I just don't understand why the code doesn't work, and before altering
(large chunks of) code, I would like to know what is going on. I will
keep in mind the extra loop you suggested.

Arno

.

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