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Hi Mihael,

It's always worked this way. I've always worked around it by coding like this:

dsply %char(%elem(top.eans));
dsply %char(%elem(sub1.eans));
dsply %char(%elem(sub2.eans));

Since top.sub.eans, by definition, will always have the same number of elements as sub1.eans, this should work fine. It's silly that we have to do this, but that's the way it's been since this support was released in V5R2, I don't even think about it anymore.


Schmidt, Mihael wrote:
D top ds qualified
D sub likeds(sub1)
D eans 13A dim(10)

D sub1 ds qualified
D sub likeds(sub2)
D eans 13A dim(10)

D sub2 ds qualified
D dan 7P 0
D ean 13A
D eans 13A dim(10)

/free
top.sub.sub.eans(1) = '1234567890123';
dsply %char(%elem(top.eans)); // works
dsply %char(%elem(top.sub.eans)); // does not work RNF0623E
dsply %char(%elem(top.sub.sub.eans)); // does not work RNF0623E

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