Stateflow graphical function won't take variables in ml expression

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Wilfred
Wilfred le 14 Sep 2011
I'm trying to construct a graphical function x=choosefrom(y), with x and y integers, which should return a random integer x, 1<x<=y. I thought it was simple: just create one transition with condition {x=ml('randi(y,1,1)')}. But executing the statechart that uses this function results in an error message "eval of matlab expression randi(y,1,1) didn't return anything (#358) While executing: Transition Guarding Condition."
However, it does work as intended if I create multiple transitions [y==2]{x=ml('randi(2,1,1)')}, [y==3]{x=ml('randi(3,1,1)')}, etc., out of the first junction. But this way, I am limited to a predefined range for y with separate transitions for each value of y, which is impractical.
Thus, it seems I cannot use variables as arguments for randi. This limitation seems to be specific for graphical functions - it doesn't apply to using randi in Stateflow actions anywhere else in the statechart. Why is this, and how could I resolve it?

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Wilfred
Wilfred le 15 Sep 2011
I figured out that I should have used {x=ml.randi(y,1)} instead of using the ml('') notation. Now it works.

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