When i execute function my matlab is being struck; is there any wrong with this simple code???

2013 1 9
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0 个投票

In the following arr is a array(1-d) in which each row has three columns of
x-co-ordinate , y-co-ordinate and intensity values ;
m*n is the required size of output image ie fgt here.
In the code initially i have simply assigned all values to 65539(which is not in range of uint16)
function [fgt]= reconstruct ( arr,m,n)
fgt=zeros(m,n);
fgt(:,:)=65539; %
l=size(arr,2)-2;
for i=1:l
fgt(arr(i),arr(i+1))=arr(i+2);
i=i+3;
end
for i=1:m
for j=1:n
if(fgt(i,j)==65539)
fgt(i,j)=0;
end
end
end

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 采纳的回答

Azzi Abdelmalek
Azzi Abdelmalek 2013-1-9

1 个投票

If arr is a real array, then there will be a problem with fgt(arr(i),arr(i+1))

7 个评论
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Azzi Abdelmalek
Azzi Abdelmalek 2013-1-9
编辑：Azzi Abdelmalek 2013-1-9
In Matlab, indices must be a real positive integers or logicals
try
a=1:5
a(1.5)
nayana
nayana 2013-1-9
编辑：nayana 2013-1-9
I mean arr has only positive integers, not complex .arr contains just the co-ordinates and intensity values which are purely integers to my knowledge
nayana
nayana 2013-1-9
there are no error messages, its getting struck and every time this occurs i need to shutdown and restart again
Azzi Abdelmalek
Azzi Abdelmalek 2013-1-9
I tested your code, there is no problem,
arr=randi(10,8,3)
m=8,n=3;
fgt=zeros(m,n);
fgt(:,:)=65539; %
l=size(arr,2)-2;
for i=1:l
fgt(arr(i),arr(i+1))=arr(i+2);
i=i+3;
end
for i=1:m
for j=1:n
if(fgt(i,j)==65539)
fgt(i,j)=0;
end
end
end
nayana
nayana 2013-1-9
Thank you for your immediate response, maybe there's something wrong with my mat lab software.I will re install and see

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更多回答（1 个）

Walter Roberson
Walter Roberson 2013-1-9

1 个投票

Please do not use "l" as a variable name: it is too easy to confuse it with "1".
You have
for i=1:l
i=i+3
end
changing a loop variable inside of the "for" loop has an effect only until the beginning of the next loop iteration. If you want to increment by 3's, then use
for i = 1 : 3 : l

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