Subscripted assignment dimension mismatch
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Dear All,
I have recently upgraded to R2016b (from R2013b). When I run my code, I get the following error message:
Subscripted assignment dimension mismatch
Since I have not modified my code, I do not see any logic in it. Do you have any idea about what happened?
Thank you in advance, Antonio Sechi
P.S.: I use a MacBook Pro Retina running 10.12.6.
8 件のコメント
Guillaume
2017 年 11 月 13 日
Well, clearly the sizes don't match. objects is (nearly) twice as big as m in the first two dimensions. How is the size of objects determined?
回答 (2 件)
Walter Roberson
2017 年 11 月 13 日
編集済み: Walter Roberson
2017 年 11 月 13 日
What is size(m) and size(Objects) at the time of the failure?
R2016b introduced implicit expansion, similar to bsxfun. For example (1:2).' + (3:5) would previously have generated an error because of trying to add a 2 x 1 to a 1 x 3, but it is now equivalent to bsxfun(@plus, (1:2).', (3:5)) . This change tended to uncover some subtle bugs in code.
But other than that, you might possibly need
objects(:, :, i, :) = permute(m, [1 2 4 3]);
because your 3rd dimension of m is not a scalar to match against the scalar i.
0 件のコメント
Guillaume
2017 年 11 月 13 日
In R2014 (a or b, can't remember) mathworks completely rewrote the graphics engine of matlab. With that rewrite came a lot of changes of behaviour (e.g. graphics handle are no longer numeric). It looks like your code is grabbing some figure frame (with your mygetframe function). Possibly the frame is now a different size than you expect (since we don't see how you declared object we don't know if you hardcoded the size) or possibly your function does not even return a frame anymore.
In any case, since your code involves graphics, you will have to use the debugger to check that all these functions that deal with graphics still work as expected.
0 件のコメント
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