creat stateflow use script

2 views (last 30 days)
wenchao zhang
wenchao zhang on 14 Aug 2023
Commented: wenchao zhang on 14 Aug 2023
hello,when i run the following script,error happens, any one knows what the root reason, following are the script.
sfnew
rt = sfroot;
ch = find(rt,'-isa','Stateflow.Chart');
st = Stateflow.State(ch);
view(st)
j1 = Stateflow.Junction(ch);
j1.Position.Radius = 16.18;
j1.Position.Center = [31.41 27.18];
j2 = Stateflow.Junction(ch);
j2.Position.Radius = 16.18;
j2.Position.Center = [62.41 27.18];
transition = Stateflow.Transition(ch);
transition.Source = j1;
transition.Destination = j2;
transition.LabelString = '{xx;yy;zz;}';
  2 Comments
Fangjun Jiang
Fangjun Jiang on 14 Aug 2023
No error in R2022b. What is the error message?
wenchao zhang
wenchao zhang on 14 Aug 2023
ok,thanks,may be it is caused by lost the state flow license, tomorrow i will check again. may be some command not supported in 2019a, not sure

Sign in to comment.

Answers (0)

Categories

Find more on Simulink Functions in Help Center and File Exchange

Products


Release

R2019a

Community Treasure Hunt

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

Start Hunting!