How do I get more info on class method?

I'm new to OOP. I can see how to get the list of methods from a class. But don't know how to get info on how to use the method. So two questions: 1) if a method has a helpfile how do I view it? 2) if no helpfile, how do I get info on how to use the method (what it does, what arguments it needs, etc)?

[Image Analyst]

What method? One you're going to write? Or one that some else wrote that you're planning on using?

[Doug]

One that others wrote (even more specific, one that Matlab developed)

[Doug]

But the question isn't for a particular class/method. It is a general question. What are the ways to get info about methods for any given class. Functions that MathWorks developed have help files. And one can see it by typing 'help function' or maybe even 'type function' to see what the function does and its arguments. How does one do that for class methods?

Answer 1

[Daniel Shub]

As an example to get help with the isplaying method of the MATLAB audioplayer class you would do

help audioplayer.isplaying

which can be generalized as

help class.method

EDIT

If you want the documentation you can do

 doc audioplayer/isplaying

which can be generalized as

doc class/method

If you want to see the code, it is a little harder since the code can be located in two places. You should try

 type @audioplayer/isplaying

and

type audioplayer

[Doug]

Thanks, that is helpful in those cases where a helpfile was provided. What about those that don't have a helpfile. How can I see the code to see what is needed?

[Image Analyst]

For files without help provided, you can look at the source code if it is available, just like for any file. Put cursor in the function and hit control-D.

[Doug]

For example, here is some sample code figure(1);ax=gca;plot(1:100,1:100);x=[50 1];y=[50 1]; scribepin=scribe.scribepin('parent',ax,'DataAxes',ax,'DataPosition',[x;y;[0 0]]');figPixelPos=scribepin.topixels; When I type 'scribepin.methods' I see the 'topixels' method, but help is of no use. How do I find out more about it? I can guess what it does, but that is insufficient.

[Doug]

I wrote my last comment without seeing Image Analyst comment. Thanks you both Daniel and Image Analyst. I think my question has now been answered