# Pre allocation do not work ...

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Gauthier Briere 2020 年 2 月 3 日
コメント済み: Raj 2020 年 2 月 3 日
Hi everyone,
I am a bit lost,
I ran a code this morning and the pre allocation do not work, it changes the size of my matrices with any reason,
So I did a small try like this :
n=3
x = zeros(1,n);
for ii=1:10
x( ii ) = ( ii );
end
%
Before I got an error because the size just exceeds... now it works, it changes the size of x, how to avoid that ?
Best regards,
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Adam 2020 年 2 月 3 日
Pre-size it correctly! You pre-size it to 3 then put it in a for loop up to 10 assignging values to it. Where do you expect the next 7 elements to come from? You didn't pre-allocate those so the array grows in the loop.

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### 採用された回答

Raj 2020 年 2 月 3 日
Question is not clear. What exactly are you trying to do here?The pre allocation works fine. You can check it by putting breakpoints in your code. The loop simply overwrites the pre allocated matrix. If you want to stop the loop when the execution reaches the preallocated matrix size then use something like this:
n=3
x = zeros(1,n);
temp=length(x);
for ii=1:10
x( ii ) = ( ii );
if ii==temp
break
end
end
%
##### 4 件のコメント表示非表示 3 件の古いコメント
Raj 2020 年 2 月 3 日
Thanks for the explanation!

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### その他の回答 (1 件)

Gauthier Briere 2020 年 2 月 3 日
That's my point ! before I got normally an error saying that the indices exceed the size of the matrice.
If I do this example
n=3;
m=9;
x = zeros(n,m);
for ii=1:n
for jj=1:m
x(ii,jj) = ii;
end
end
and you change by :
n=3;
m=9;
x = zeros(n,m);
for ii=1:n
for jj=1:240151
x(ii,jj) = ii;
end
end
It will change the size of x which is really weird because before it never changed the size....
##### 3 件のコメント表示非表示 2 件の古いコメント
Gauthier Briere 2020 年 2 月 3 日
Ok I think I got it, thanks for your help, I just misunderstood

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