create a view based on a .mldatx file.

I have 3 files :

1- Architecture view

2-Logical view

3-Allocation file .mldatx.

I would like to display a view based on the .mldatx file.

The component in the architecture view are linked to several element in the logical view.

SO far i'm able to identify which block are connected, but impossible to display those.

Or is there a better way than a script to do it ?

At first I thought using the stereotype but we can't have several value for one enum, and having the string type doesn't help because there are too much possibility.

Am i missing something ?

here is the code I did so far for the first part.

% Fichier FunctionalBlock.m

% Loading allocation file

allocationSetFile = 'FunctionToDetail.mldatx';

allocationSet = systemcomposer.allocation.load(allocationSetFile);

scenarios = allocationSet.Scenarios;

% Scenario selection

scenarioName = 'Scenario 1';

scenario = [];

for i = 1:numel(scenarios)

if strcmp(scenarios(i).Name, scenarioName)

scenario = scenarios(i);

break;

end

end

if isempty(scenario)

error('Scénario "%s" non trouvé.', scenarioName);

end

fprintf('Scénario d’allocation utilisé : %s\n', scenario.Name);

% Allocations

allocations = scenario.Allocations;

% Functional components extraction

composantsFonctionnels = systemcomposer.arch.Component.empty;

for i = 1:numel(allocations)

src = allocations(i).Source;

if isempty(composantsFonctionnels) || ~any(src == composantsFonctionnels)

composantsFonctionnels(end+1) = src; %#ok<AGROW>

end

end

% Find functional target component

nomComposantFonctionnel = 'sfa_dcl_overvoltage_trip';

composantFonctionnel = [];

for i = 1:numel(composantsFonctionnels)

if strcmp(composantsFonctionnels(i).Name, nomComposantFonctionnel)

composantFonctionnel = composantsFonctionnels(i);

break;

end

end

if isempty(composantFonctionnel)

error('Composant fonctionnel "%s" non trouvé.', nomComposantFonctionnel);

end

fprintf('Composant fonctionnel trouvé : %s\n', composantFonctionnel.Name);

% Allocated logical blocks

blocsLogiquesAlloues = systemcomposer.arch.Component.empty;

for i = 1:numel(allocations)

if allocations(i).Source == composantFonctionnel

blocsLogiquesAlloues(end+1) = allocations(i).Target; %#ok<AGROW>

end

end

fprintf('Composants logiques alloués :\n');

for i = 1:numel(blocsLogiquesAlloues)

fprintf(' - %s\n', blocsLogiquesAlloues(i).Name);

end

Answer 1

[Deepak]

Hi @Remi,

I understand that you are trying to display logical components allocated to an architectural component using a ".mldatx" allocation file in System Composer. The most effective way to visualize these allocations is through the Allocation Editor, which you can open via the System Composer toolbar or using the command

systemcomposer.allocation.open('filename.mldatx')

Within the Allocation Editor, selecting a component and clicking Show Allocations will visually highlight the corresponding linked components across the source and target models, enabling interactive navigation and inspection.

While your script correctly identifies allocations programmatically, for visualization purposes, you can also create a custom architecture view using "createView" and "addElement" to display just the allocated components. It is generally better to use the built-in allocation mechanism rather than stereotypes for linking, since allocations natively support many-to-many relationships. If needed, you can attach metadata to the allocations themselves using allocation stereotypes, which is more scalable than managing multi-value enums or strings on components.

Please find attached the documentation of "Allocation Editor" for reference:

www.mathworks.com/help/systemcomposer/ref/allocationeditor.html

I hope this helps.