creat stateflow use script

3 ビュー (過去 30 日間)
wenchao zhang
wenchao zhang 2023 年 8 月 14 日
コメント済み: wenchao zhang 2023 年 8 月 14 日
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 件のコメント
Fangjun Jiang
Fangjun Jiang 2023 年 8 月 14 日
No error in R2022b. What is the error message?
wenchao zhang
wenchao zhang 2023 年 8 月 14 日
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

サインインしてコメントする。

回答 (0 件)

カテゴリ

Help Center および File ExchangeComplex Logic についてさらに検索

製品


リリース

R2019a

Community Treasure Hunt

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

Start Hunting!

Translated by