Returning error that there is not

Hi,

I made this function

function [out] = pardin(N,T)

that contains a figure handle as structure with other handles; some of these are generated from for loop 'cause their position properties in figure are loop-depending;

So, if I edit this function and turn it in a simple m-file (providing variables called N and T, of course) it works, but if use it as above (as a function) it returns me this error:

Error using uitable
Width and height must be > 0

Error in uitable (line 52)
        thandle = builtin('uitable', varargin{:});

Error in pardin (line 43)
   S.w(jj,1) = uitable('Parent', S.fig, 'Data', A(:,:,jj), 'Units', 'normalized', 'Position', [0.7
   1-((jj+jj+0.09)*0.1) 0.26 0.15 ]);

I think this is the error that it should return if 'Position' vector has 3rd and 4th element not positive, but my S.w handle, as you can observe, has those positive elements. How is it possible ? I tried to delete this handle and it returns same error for the other handles. As I wrote, if I use it as m-file script it works

[Walter Roberson]

At the command line, please type

dbstop if error

and then run the program. When it stops, please show us what the value of jj is. You might need to use "dbup" to move between the environment of the uitable function and your pardin routine.

[Marco]

It returns:

52          thandle = builtin('uitable', varargin{:});

in the base workspace the only variable is varargin (I never use this commands);

using "dbup' it returns the number of string with S.w(jj,1) with:

In workspace belonging to pardin at 48

Now there are all variables in the base-workspace and "now" jj=1

What do I need jj value (Width and height are 3rd and 4th element of 'Position', are they) ?

[Walter Roberson]

I was working on the hypothesis that jj might have a value that caused the Y to be out of range and the wrong error being returned.

Okay, with the same breakpoint in place, run until error, and then at the uitable level, please show us

varargin{:}

[Marco]

So, I do it:

varargin{:} has these elements:

varargin =

      'Parent'    [2]    'Data'    {4x4 cell}    'Units'    'normalized'    'Position'    [1x4 int8]

where if I type:

K>> varargin(8)

it returns:

ans = 

      [1x4 int8]

so

K>> ans{:,:}

ans =

      1    1    0    0

why are they zeros ?

Answer 1

[Walter Roberson]

The datatype of your variable "jj" is int8, and that is causing the Position calculation to be carried out in int8. That is resulting in the floating point values being rounded to integers, and [1 1 0 0] is the result.

Is there a particular reason you used int8 for that datatype? If not then just let jj be double. But if you need to, use

[0.7 1-((double(jj)+double(jj)+0.09)*0.1) 0.26 0.15 ]

Note that double(jj+jj) would be different than double(jj)+double(jj) if jj was 128 or larger.