creat stateflow use script

조회 수: 2 (최근 30일)
wenchao zhang
wenchao zhang 2023년 8월 14일
댓글: wenchao zhang 2023년 8월 14일
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;}';
  댓글 수: 2
Fangjun Jiang
Fangjun Jiang 2023년 8월 14일
No error in R2022b. What is the error message?
wenchao zhang
wenchao zhang 2023년 8월 14일
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

댓글을 달려면 로그인하십시오.

답변 (0개)

카테고리

Help CenterFile Exchange에서 Simulink Functions에 대해 자세히 알아보기

제품


릴리스

R2019a

Community Treasure Hunt

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

Start Hunting!

Translated by