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isCommented

Determine if graphical object is commented out

Description

example

tf = isCommented(graphicalObject) returns a logical value that indicates if a graphical object is commented out. The function returns logical 1 (true) if:

  • The graphical object is explicitly commented out. To explicitly comment out an object, set its IsExplicitlyCommented property to true. Alternatively, you can right-click the graphical object and select Comment Out.

  • The graphical object is implicitly commented out. In this case, its IsImplicitlyCommented property has a value of true. An object is implicitly commented out when you explicitly comment out a state, box, or function that contains the object. Additionally,

    • Transitions are implicitly commented out when you comment out their source or destination.

    • Entry and exit ports are implicitly commented out when you comment out their matching entry or exit junction.

Otherwise, the function returns logical 0 (false).

Examples

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When you explicitly comment out a state, box, or function, you implicitly comment out all the graphical objects that it contains. For example, when you comment out state A in this chart, you also comment out its substates, A1 and A2.

Stateflow chart with a hierarchy of states. The outer state is called A. It contains two inner states called A1 and A2.

Find the Stateflow.State objects named A, A1, and A2.

sA = find(ch,'-isa','Stateflow.State','Name','A');
sA1 = find(ch,'-isa','Stateflow.State','Name','A1');
sA2 = find(ch,'-isa','Stateflow.State','Name','A2');

Check that state A and its substates are not commented out.

[isCommented(sA),isCommented(sA1),isCommented(sA2)]
ans =

  1×3 logical array

   0   0   0

Explicitly comment out state A.

sA.IsExplicitlyCommented = true;

Check that state A and its substates are commented out.

[isCommented(sA),isCommented(sA1),isCommented(sA2)]
ans =

  1×3 logical array

   1   1   1

Input Arguments

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Introduced in R2016a