MATLAB: Help me understand!

errorhelplengthMATLABsolveunderstandvectors

In the 3rd for loop , xloc(k) and yloc(k) are both vectors of length 6. And xp and yp are vectors of length 30 . If done alone the subtraction returns an error but here it doesn't. I wanna know what's so special here that it doesn't return an error?
if true
  % xloc = [1,7,8,17,22,27] ;
yloc = [28,18,16,2,10,8] ;
V = [3,7,4,5,2,6] ;
best_loc_x = 31 ;
best_loc_y = 31 ;
min_cost = 1e+6 ;
anotherx = [] ;
anothery = [] ;
C = [] ;
for xp = 1:30
  for yp = 1:30
      for k = 1:6
          d(k) = sqrt((xloc(k)-xp).^2+(yloc(k)-yp).^2) ;
          cost(k) = d(k)*V(k);
      end
      loc_cost(xp,yp) = sum(cost) ; 
      if loc_cost(xp,yp) < min_cost
          best_loc_x = xp;
          best_loc_y =  yp;
          min_cost = loc_cost(xp,yp);
      elseif (loc_cost(xp,yp)-min_cost) <=1
          anotherx = [anotherx,xp] ;
          anothery = [anothery,yp];
          C = [C,loc_cost(xp,yp)];
      end
  end
end
end

Best Answer

for xp = 1:30          %so xp is a scalar
  for yp = 1:30        %so yp is a scalar
      for k = 1:6      %so k is a scalar
          d(k) = sqrt((xloc(k)-xp).^2+(yloc(k)-yp).^2) ;
vector indexed at a scalar gives a scalar, so xloc(k) is a scalar and yloc(k) is a scalar. You subtract the scalars xp and yp from them, giving scalars for the sub-expressions. You square, add, square-root, giving a scalar output, which you store in a scalar location. No problem.
          for xp = 1:30
does not make xp a vector of 30 elements: it tells MATLAB to iterate making xp hold each of the values 1:30 in turn, so at any one time xp is a scalar.
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