Attempted to access indx(1); index out of bounds because numel(indx)=0.
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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 个评论
Matz Johansson Bergström
2014-8-17
Do you have an example of inputs to the function that gives this error?
Hamid
2014-8-17
采纳的回答
Matz Johansson Bergström
2014-8-17
0 个投票
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.
更多回答（1 个）
Ahmet Cecen
2014-8-17
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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