Asked by Mehedi
on 4 Jun 2014

clc clear x1=5; x2=3; x3=2; p=cell(3,1); for i=1:3 if i==1 x1=6; elseif i==2 x1=7; elseif i==3 x1=8; end p{i}=[x1-x2; x1-x3]; if p{i}==[4;5] break end end for n=1:3 z(n)=(p{n} (1)) end

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Answer by Roger Wohlwend
on 4 Jun 2014

In the second iteration the if-condition p{i} == [4;5] becomes true and terminates the loop. p{3} stays empty. In the second for-loop you want to acces that empty field --> p{3}(1). Matlab does not allow it and throws an error.

Hint: instead of

p{i} == [4; 5]

write

isequal(p{i}, [4;5])

or

all (p{i} == [4;5])

And if you post code the next time, please format it as code.

Opportunities for recent engineering grads.

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