Question

I’m trying to use fillOval in my Matlab program

width = 20;
x = round(rand()*size(3)) - width;
y = round(rand()*size(4)) - width;
Screen(wPtr, 'FillOval', [130 130 130], [x y width width]);

and I get the error:

??? Error using ==> Screen
Usage:

Screen('FillOval', windowPtr [,color] [,rect] [,perfectUpToMaxDiameter]);

Error in ==> Test at 18
       Screen(wPtr, 'FillOval', [130 130 130], [x y width width]);

When I replace the x, y, and width variables with values in the Screen() call, I don’t get the error.

What’s wrong?

Answer 1

rect should be a 1×4 vector. When I use your code it is a 1×6 vector. It looks like you’ve accidentally overloaded the size() function with some size variable that you did not include in your example, and this is causing problems. If you name your size variable something else, and ensure that when you test out [x y width width] you get back a 1×4 vector, it should work.

Also, I assume you’re talking about the PsychToolbox, but this isn’t mentioned in your post either.