If I have a variable x='name', how can I use x to build a structure with name 'name' that looks like:
name.a
name.b
No need to tell me about the hazards of dynamic variable naming.
Thanks.

 Réponse acceptée

DrK
DrK le 17 Mai 2022

0 votes

Actually, found my own answer:
x = 'name';
eval(sprintf('%s = struct;',x));
This creates a structure named 'name' and one can use the same eval format to add fields to it later.

7 commentaires

James Tursa
James Tursa le 17 Mai 2022
How do you plan on using this variable downstream in your code? More "eval" statements everytime you want to use it?
DrK
DrK le 17 Mai 2022
That is a good point :-)
I guess that is what it takes, doesn't it?!
Is there a better way to do this?
DrK
DrK le 17 Mai 2022
Actually, I intend to form these structures into a table and later I can access then using the table. I guess this is analogous to the second level of structure that you proposed earlier, but since table was my plan all along, this eval method works better.
Stephen23
Stephen23 le 17 Mai 2022
"Is there a better way to do this?"
Yes.
DrK
DrK le 17 Mai 2022
do tell
DrK
DrK le 17 Mai 2022
Thanks for the light reading. I think I am going to take this in the direction of indexing instead.
Thanks again for your help.

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Plus de réponses (2)

James Tursa
James Tursa le 16 Mai 2022

1 vote

Would it be acceptable to use x as a dynamic fieldname instead of the top level variable name? E.g.,
v.(x).a
v.(x).b
DrK
DrK le 17 Mai 2022

0 votes

Not really. I know about the paranthesis around field names, but it creates another level of structure. The problem is that if I am creating the structure names inside my program, I don't know how to create the structures using those names.
Thanks for your response.

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R2021a

Question posée :

DrK
le 16 Mai 2022

Commenté :

DrK
le 17 Mai 2022

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