Subscripted assignment dimension mismatch. error is coming in array(k,1)=su; pls help me out??thanks 4 any help
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su=0;k=1;
array=zeros(size(p,1),2);
j=1;
while(1)
for i=j:size(p,1)
if i>size(p,1)-1
array(k,1)=1;
array(k,2)=p(i,2);
break;
end
if p(i,2)==p(i+1,2)
su=su+p(i,1);
else
break
end
end
su=su+p(i,1);
array(k,1)=su;
array(k,2)=p(i,2);
j=i+1;
k=k+1;
su=0;
end
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Réponse acceptée
Walter Roberson
le 25 Avr 2013
We do not know size(p)
If p starts out empty then size(p,1) would be 0, and
for i = j:size(p,1)
would be for i = 1:0 which would leave i as being []. Then p(i,1) would be p([],1) which would be []; su = su + p(i,1) would be su = su + [] and that makes su = []. array(k,1) = su then becomes a dimension mismatch
3 commentaires
Walter Roberson
le 26 Avr 2013
When i reaches size(p,1), then j gets set to size(p,1)+1, and then the
for i=j:size(p,1)
becomes
for i=size(p,1)+1:size(p,1)
which does not execute the "for i" loop body and leaves i=[]. Then after the body of the for loop, the su=su+p(i,1) becomes su=su+p([],1) which get you su=[], leading to the failure on the next line.
Base problem: you do not have any "break" out of the "while(1)" loop so you eventually get past the end of p.
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