Addressing arrays between functions

I'm running a while loop and i need to use a value generated within the loop to provide a different variable from a different script and i'm having trouble getting the generated variable into the other functions. Here's what i have, if anyone has any ideas it'd be greatly appreciated.

function R_New = R_New (theta, phi, Params, LaserParams, Orientation)

run_number = 1;

Conv{1} = 0;

while Conv{run_number} == 0

    R_initial = Params(1);

    R_1{run_number} = R_dash(theta, phi, Params, LaserParams, run_number,Orientation);

    Radius = R_Sphere(theta, phi, Params, LaserParams, run_number, Orientation);

    R_2{run_number} = R_1{run_number}.*(R_initial/Radius);

    Conv{run_number} = Convergence(theta, phi, Params, LaserParams, run_number, Orientation);

    run_number = run_number +1;

end

Where R_2 is the generated variable, used in the 2nd iteration of the loop in function R_dash.

Here is R_dash

function R_dash = R_dash(theta, phi, Params, LaserParams, run_number, Orientation, R_2)

I just dont know how to get R_2 into the R_dash function so that i can access the precious values of it.

Thanks

if run_number == 1;

    R = R_0(Params);

else

    R = R_2{run_number};

end

    Pert = P(theta, phi, Params, LaserParams, run_number, Orientation); 

    R_dash = R + Pert;

[Jan]

Are you working with Dougie, who has asked a very similar question? Or do you use multiple accounts? See: http://www.mathworks.de/matlabcentral/answers/37341-for-loop

Please format your code and explain, what the code fragment after "Thanks" means.

Answer 1

[Jan]

The question is not clear. If you want to access a variable inside a function, you have to insert it in the input arguments. In your case the function definition is:

function Output = R_dash(theta, phi, Params, LaserParams, run_number, Orientation, R_2)

I'd avoid to use the same name for the function and its output. This is at least confusing. Anyhow, here you have defined R_2 as last parameter. But you call this function by this:

R_1{run_number} = R_dash(theta, phi, Params, LaserParams, run_number,Orientation);

without the last argument. So perhaps you want to insert it here? But of course you need a value for it in the first iteration - this could be the orphaned code fragment after the "Thanks" in your question?

[Andy McCann]

Yeah sorry, it's just different accounts from different computers.

I've changed the function to call R_2 and given it an an initial value, however now i'm getting an error of

??? Index exceeds matrix dimensions.

Error in ==> R_dash at 30

R = R_2{run_number};

Error in ==> R_Check at 20

R_1{run_number} = R_dash(theta, phi, Params, LaserParams, run_number,Orientation,

R_2);

Here's the tidied code for R_New

function R_New = R_New (theta, phi, Params, LaserParams, Orientation)

run_number = 1;

Conv{1} = 0;

while Conv{run_number} == 0

R_2{1} = R_0(Params);

R_initial = Params(1);

R_1{run_number} = R_dash(theta, phi, Params, LaserParams, run_number,Orientation, R_2);

Radius = R_Sphere(theta, phi, Params, LaserParams, run_number, Orientation);

R_2{run_number} = R_1{run_number}.*(R_initial/Radius);

Conv{run_number} = Convergence(theta, phi, Params, LaserParams, run_number, Orientation);

run_number = run_number +1;

end

end

And here os R_dash

function R_dash = R_dash(theta, phi, Params, LaserParams, run_number, Orientation, R_2)

R = R_2{run_number};

Pert = P(theta, phi, Params, LaserParams, run_number, Orientation);

R_dash = R + Pert

[Jan]

The error message is clear: You want to access R_2{run_number}, but R_2 has less then run_number elements.

As far as I understand the guidelines, it is not allowed to have multiple accounts: http://www.mathworks.com/accesslogin/faq.jsp?s_cid=mwa-cmlndefq

The former question has been a duplicate post already, which was merged by Walter. No you are continuing to ask for the same problem without a link to the former thread and under a different user name.

Please do not post multiple questions for one problem and do not use multiple accounts, because it confused and impedes the contributors, who want to assist you.