failed to determine the appropriate size in find

Hi,

I'm having some problem getting matlab coder to identify a variable used as result from find as a scalar. Below is a simple testcase:

function [ y ] = a_test()
x = zeros(5, 1);
k1 = find(x > 1, 1);
if (isempty(k1))
    k1 = 0;
end

y = x(1:k1);
end

The coder fails to generate code due in the last line because it cannot infer that k1 is a scalar. Is there a way of handling this?

Thanks, -- Octav

Answer 1

[Grzegorz Knor]

What should be the y?

Maybe replace:

if (isempty(k1))
    k1 = 0;
end

with:

if (isempty(k1))
    k1 = 1;
end

Answer 2

[Jan]

While in you code k1 is a scalar, 1:k1 is not when k1 equals 0.

function y = a_test()
x  = zeros(5, 1);
k1 = find(x > 1, 1);
if isempty(k1)
  y = ???;
else
  y = x(1:k1);
end

Answer 3

[Fred Smith]

Try this:

function [ y ] = a_test()
x = zeros(5, 1);
k1 = find(x > 1, 1);
if (isempty(k1))
    k1 = 0;
end

y = x(1:k1(1)); % Added k1(1)
end

HTH,

Fred

[Jan]

After the IF-block, k1 is always a scalar. Therefor k1(1) is not useful. In addition y=x(1:k1(1)) still fails, if k1 is 0, because 1:0 is empty.

[Fred Smith]

k1 is always scalar but MATLAB Coder does not know that. It does know that k1(1) is always a scalar. x(1:0) is legal and returns an empty matrix. I don't know what the intended behavior is in this case but the changed code matches the behavior of Octav's original code.

-Fred