Generate code from referenced model without generating from the parent

7 vues (au cours des 30 derniers jours)
Guilherme T Silva
Guilherme T Silva le 8 Avr 2019
When working with large systems made of many referenced models I can either generate code from the top-level model (parent), which creates a separate .c file for each one of the children, or I can generate code only from one of the child models. The step function for the referenced model changes wether the code is generated from the parent or from the child model, so if I update the child I need to generate code from the parent, otherwise the child models will generate code in standalone mode. The problem is that this can be very time consuming, even if the top level model and most of the referenced ones are unchanged.
Is there a way to generate code only for a referenced model, while keeping the interfaces it had when it wasn't generated as a standalone model?
I'm sending a model attached (R2018a) to exemplify the problem I'm trying to solve. Notice how the code in ReferencedModel.c changes when you generate code directly from it or from its parent model (ParentModel.slx).

Réponses (1)

Nick Sarnie
Nick Sarnie le 13 Mai 2019
Hi Guilherme,
Does using "slbuild(modelName, 'ModelReferenceRTWTarget')" solve your issue?
Thanks,
Nick
  1 commentaire
Rodrigo Estrella
Rodrigo Estrella le 17 Avr 2020
Im having more less the same problem. I have a parent model with several childs. I want to generate the code for the parent model without generating the code for the childs models, because the code for the childs has already been generated. As i will be generating the code for a complete collection of parent models, it will be very inneficent to generate the code of the child models each time.

Connectez-vous pour commenter.

Catégories

En savoir plus sur Deployment, Integration, and Supported Hardware dans Help Center et File Exchange

Produits


Version

R2018a

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by