??? Index exceeds matrix dimensions. please help me
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[m n]=size(handles.data1);%menentukan ukuran citra
for o=1:m
for p=1:n
if o<0
o=1;
end
if p<0
p=1;
end
[d c l]=size(handles.data1);
masuk_enkripsi=double(handles.data1(p:p+d-1,o:o+c-1,:));
??? Index exceeds matrix dimensions.
Error in ==> enkripsi>btn_enkripsi_Callback at 148
masuk_enkripsi=double(handles.data1(p:p+d-1,o:o+c-1,:));
댓글 수: 2
ChristianW
2013년 3월 11일
What are you trying to do in this line:
masuk_enkripsi=double(handles.data1(p:p+d-1,o:o+c-1,:));
채택된 답변
ChristianW
2013년 3월 2일
With your image height d and if p gets p=2, then you are trying to address with (p+d-1) the (d+1) pixel. Thats not possible.
댓글 수: 2
ChristianW
2013년 3월 3일
Yes, I've got an infinite amount of solutions. I'll rub my crystal ball for visions regarding the purpose of your code.
추가 답변 (2개)
Jan
2013년 3월 4일
It is confusing to change the loop counter of A FOR loop inside the loop.
Using lowercase and uppercase Oh's as names of variables is confusing also: "if o<0" looks funny, but is to hard to read.
I confirm, that CristianW's answer is exhaustive: Obviously either p+d-1 and/or d+1 exceed the available size. But it is impossible to guess what you want to do instead.
댓글 수: 2
Walter Roberson
2013년 3월 11일
In your line
[m n]=size(handles.data1);%menentukan ukuran citra
you are using size() as if you expect handles.data1 to be two-dimensional. But later you have
[d c l]=size(handles.data1);
which is expecting handles.data1 to be 3 dimensional. If it is 3 dimensional then when you ask for its size() using only two outputs (e.g., [m n]) then size() will set the second output equal to the product of the second and third dimensions -- so although m would be the same as the later d, n would be the product of c and l. This is going to lead to problems when you do your looping over "n" and try to use that value as the index to the second dimension.
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