Subscript indices must either be real positive

Hi it is a common error, yes I saw the other Q/A here, but still could not figure it out. the matrix indices cannot be zero, but I have values in the matrix are fractions 0.1,0.2,0.3etc, is that the issue?

I want to extract the (x,y) coordinates of the points on the boundary of the rectangular geometry as illustrated in the attached and save them in a matrix to use them later in an equation for each point.

first I tried to extract them for each side. I used the following procedure but stopped after a few points

   for r=0:0.1:2
  mat(r*10+1,:)=[1 r]
  end
the error was 
  Subscript indices must either be real positive integers or logicals

Thanks

Answer 1

[KALYAN ACHARJYA]

 for r=0:0.1:2
 mat(fix(r*10+1),:)=[1 r] % Considering only integer using *fix*
 end

Answer 2

[Dennis]

The error is actually not that straight forward. Many floating point numbers can described 100% accurate in binary form. So your r might be not exactly 0.6. You could use round to prevent this (although this might not be the cleanest approach).

for r=0:0.1:2
  mat(round(r,2)*10+1,:)=[1 r]
  end

You could aswell just assign the r vector to a matrix:

r=0:0.1:2;
A=ones(length(r),2);
A(:,2)=r';

Answer 3

[KSSV]

r = 0:0.1:2 ;
mat = zeros([],2) ;
for i = 1:length(r)
    mat(i*10+1,:)=[1 r(i)] ;
end

Answer 4

[Walter Roberson]

r=0:0.1:2;
r(find(r*10 ~= fix(r*10)))

You will find that the entry that looks like 0.3 multiplied by 10 does not give exactly 3

You will also find that the entry that looks like 0.3 is not exactly equal to what you would get if you entered 0.3 in text form.

In short, do not run calculations on values with fractions and expect the results to be exactly integers unless you use round or floor or ceil or fix. Sometimes you can get away with it, but if the fraction involved is not exactly 0.5 or 0.25 then you should assume there might be problems.

Better would be to loop over integer indices and convert to r values if needed

for k=1:21
  r=(k-1)/10