Simulink - Embedded Matlab Function - Size of function not bounded error

When compiling the following embedded matlab function, i get this error:

Computed maximum size of the output of function 'colon' is not bounded. Static memory allocation requires all sizes to be bounded. The computed size is [1 x :?].

at this line: "x(end):epsi(1)/5:x(1)"

The embedded matlab function (partially):

function [z, stat] = simps5(x,y)

%-- Make sure x and y are column vectors, or y is a matrix. z=0; stat=0;

if sum(isnan(x)) > 0 sum(isnan(y)) > 0

    disp('x or y has zero size!!!');
    return;
end

xorig = x; yorig = y; stat = 1;

if isempty(xorig) == 1 isempty(yorig) == 1

    disp('x or y is empty!!!');
    z = 0;
    stat = 2;
    return;
end

stat = 0; epsi = mean(abs(diff(x))); epsiy = mean(abs(diff(y))); if epsiy(1) == 0

    z = y(1)*abs(x(end)-x(1));
    return;
end
if epsi(1) == 0

    z =0;
    return;
end

stat2 = 0;

coder.varsize('new_res',1000000); if x(1) > x(end)

    assert(x(1)<2);
    assert(x(end)>-2);
    assert(epsi(1)>1e-2);
    assert(epsi(1)<1e2);
    new_res = x(end):epsi(1)/5:x(1);
    new_res = sort(new_res,'descend');
else

    new_res = x(1):epsi(1)/5:x(end); 
end

[Ketan]

Hi Benoit,

I tried the following code:

function y = fcn(x,step)
  if(x(1) > x(end))
     assert(x(1)<2);
     assert(x(end)>-2);
     assert(epsi(1)>1e-2);
     assert(epsi(1)<1e2);
     new_res = x(end):epsi(1)/5:x(1);
  else
     new_res = x(1):epsi(1)/5:x(end);
  end
  y = numel(new_res);
end

Where x is [10,1] input and step is a scalar input. It errored for the colon expression in the else condition. That makes sense to me because asserts are required within the else block similar to the first if-block. After adding asserts to the else block things work.

I am not sure why you are seeing this error for the colon expression in the first if block. Any chance you can provide a full-set of code that reproduces this error?

Ketan

[Benoit]

Hi Ketan,

Thank you for your answer. It is good to know that it works in that way. One thing to add in the code is that:

epsi =  mean(abs(diff(x)))

as I wrote in my last message. Sorry, it was not clear with the code "styles". I think i wrote everything that was necessary to understand the code of the function. The rest is just using this new_res as new resolution for x, so it does not affect the error I keep receiving.

Answer 1

[Benoit]

nobody?