Subscripted assignment dimension mismatch. error is coming in array(k,1)=su; pls help me out??thanks 4 any help

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

Answer 1

[Walter Roberson]

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

[Walter Roberson]

At the MATLAB command line, command

dbstop if error

and then run the program. When it stops, example the value of "i" and "su" and size(p)

[Sony]

the error is at array(k,1)=su; when i print su it shows Empty matrix: 0-by-1,the size of p is 10cross2,the size of su is 0cross1,pls help me with error i dnt knw how to proceed.

[Walter Roberson]

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.