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I'm simualting a project where one of the matlab functions states the following error:
"'y' is inferred as a variable-size matrix, but its size is specified as inherited or fixed. Verify 'y' is defined in terms of non-tunable parameters, or select the 'Variable Size' check box and specify the upper bounds in the Size box."
I have seen similar posts to this however couldn't find infomation on why my output is variable size,
The code for this aspect of the project is quite small and the following:
function y = sector(Valpha, Vbeta)
sector = [];
%all potential sectors for Vbeta greater than zero
if Vbeta>=0 & abs(Valpha) < abs(Vbeta/sqrt(3));
sector = 2;
end
if Vbeta>=0 & abs(Valpha) > abs(Vbeta/sqrt(3)) & Valpha >= 0;
sector = 1;
else sector = 3;
end
%all potential states for Vbeta less than or equal to zero
if Vbeta < 0 & abs(Valpha) < abs(Vbeta/sqrt(3));
sector = 5;
end
if Vbeta < 0 & abs(Valpha) > abs(Vbeta/sqrt(3)) & Valpha >= 0;
sector = 6;
else sector = 4;
end
y = sector;
Does anyone happen to know why the data is variable size and how I coulf stop it from being such.
Many thanks

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KSSV
KSSV 2021-6-24
What are Valpha, Vbeta are they vectors?
They should be vectors yes

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KSSV
KSSV 2021-6-24

2 个投票

Note that the function name and the variable are on the same name. This is not allowed.

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Oh sorry the 'sector' being the same right?
I'll try that now thank you!
Still getting the same error sadly
KSSV
KSSV 2021-6-24
If the inputs are vectors, you cannot use the function like this.
Oh okay, do you know what kind of function I could use?
Obsviously you don't need to provide the code but the general name of the function if possible?
Thanks so much!
KSSV
KSSV 2021-6-24
When you are comapring two vectors....you end up with a vector.
a = rand(10,1) ;
b = rand(10,1) ;
val = zeros(size(a)) ;
%
idx = a > b ;
val(idx) = 1 ;
You can proceed like shown above. So your sector values are going to be vectors.
Brilliant thank you so much for your help!

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