Index in Position 2 is invalid.
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X=[200:2000]
for i = 1:1801
A = X(1,i)/2
B(1,i)= A
end
i=0
while i<1801
if i==750
break
elseif mod(i,2)==1
C(1,i)=B(1,i)*2
elseif mod(i,2)==0
C(1,i)=B(1,i)/2 %Index in Position 2 is invalid.
end
i=i+1
end
The programm works only without the second elseif. If I add the second elseif, I get the error: "Index in Position 2 is invalid." Can someone help?
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채택된 답변
Ameer Hamza
2020년 5월 8일
편집: Ameer Hamza
2020년 5월 8일
In MATLAB, the indexing starts from 1. Therefore you should initialize 'i' with the value of 1
X=[200:2000];
for i = 1:1801
A = X(1,i)/2;
B(1,i)= A;
end
i=1 % <==== initialize with 1
while i<1801
if i==750
break
elseif mod(i,2)==1
C(1,i)=B(1,i)*2;
elseif mod(i,2)==0
C(1,i)=B(1,i)/2 %Index in Position 2 is invalid.
end
i=i+1;
end
Also, note that I have suppressed the output from printing on the command window using semi-colon (;). Printing an array on command window in each iteration is very inefficient. Suppressing the output will result in a speed increase.
댓글 수: 4
Claudius Simon Appel
2020년 5월 9일
Okay, that makes sense so far.
What I don't get is why this is an error that, according to Bhogal, only occurs while the second elseif
if i==750
break
elseif mod(i,2)==1
C(1,i)=B(1,i)*2;
elseif mod(i,2)==0 %% this line here
C(1,i)=B(1,i)/2 %Index in Position 2 is invalid.
end
exists. What is it about this line that prompts matlab to run into an error. Or, to be precise, why does it not run into the same error before?
I know that the original question has been answered, but could someone explain this to me?
Thank you.
추가 답변 (1개)
Francesco Torricelli
2020년 5월 8일
Hi,
you can easily solve your issue by setting i = 1 just before the while loop (i = 0 leads to the error you encountered in the very first iteration).
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