Attempted to access indx(1); index out of bounds because numel(indx)=0.
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Hamid
le 17 Août 2014
Réponse apportée : Matz Johansson Bergström
le 17 Août 2014
Dear all, WHAT IS THE HELL OF THIS ERROR??????
what can I do??
function [Ex,Ey,Ez]=coordxtr(Edof,Coord,Dof,nen)
%[Ex,Ey,Ez]=coordxtr(Edof,Coord,Dof,nen)
[nel,dum]=size(Edof);
ned=dum-1;
[n,nsd]=size(Coord);
[n,nd]=size(Dof);
nend=ned/nen;
%
for i = 1:nel
nodnum=zeros(1,nen);
for j = 1:nen
check=Dof(:,1:nend)-ones(n,1)*Edof(i,(j-1)*nend+2:j*nend+1);
[indx,dum]=find(check==0);
nodnum(j)=indx(1);
end
%
Ex(i,:)=Coord(nodnum,1)';
if nsd>1
Ey(i,:)=Coord(nodnum,2)';
end
if nsd>2
Ez(i,:)=Coord(nodnum,3)';
end
end
%--------------------------end--------------------------------
2 commentaires
Matz Johansson Bergström
le 17 Août 2014
Do you have an example of inputs to the function that gives this error?
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Matz Johansson Bergström
le 17 Août 2014
It would be better if you could give us an example of all the arguments, so we can try it out for ourselves. I know what the immediate problem is but it is more important to understand why it happens.
You are trying to index a vector containing no elements, so the question is: why is this?
To answer the other error you got, the problem seem to be that the type of indices you are passing to the vector is not real positive integers . If you followed Ahmet's solution, you will be passing a 0 as an index, which is not allowed in Matlab. The indexing starts at 1.
So, I want to make sure I understand why the vector is as it is, before I (or anyone else) try to solve the issue. Otherwise you will have error upon error which will take time to answer.
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Plus de réponses (1)
Ahmet Cecen
le 17 Août 2014
That means check sometimes don't have any 0 elements when you do
[indx,dum]=find(check==0);
Try instead:
if length(indx)>0
nodnum(j)=indx(1);
else
nodnum(j)=0;
end
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