Index in Position 2 is invalid.

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captainjoker
captainjoker 2020-5-8
评论： Ameer Hamza 2020-5-9
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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Ameer Hamza
Ameer Hamza 2020-5-8
编辑：Ameer Hamza 2020-5-8
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.
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Claudius Simon Appel
Claudius Simon Appel 2020-5-9
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.
Ameer Hamza
Ameer Hamza 2020-5-9
Can you show the code before this if-block? How is B defined?

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更多回答（1 个）

Francesco Torricelli
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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