displaydata function in ex3 coursera machine learning - octave

Question

I am facing a issue, here is my script. some end or bracket issue but I have checked noting is missing.
function [h, display_array] = displayData(X, example_width)
%DISPLAYDATA Display 2D data in a nice grid
% [h, display_array] = DISPLAYDATA(X, example_width) displays 2D data
% stored in X in a nice grid. It returns the figure handle h and the
% displayed array if requested.
% Set example_width automatically if not passed in
if ~exist('example_width', 'var') || isempty(example_width)
example_width = round(sqrt(size(X, 2)));
end
% Gray Image
colormap(gray);
% Compute rows, cols
[m n] = size(X);
example_height = (n / example_width);
% Compute number of items to display
display_rows = floor(sqrt(m));
display_cols = ceil(m / display_rows);
% Between images padding
pad = 1;
% Setup blank display
display_array = - ones(pad + display_rows * (example_height + pad), ...
pad + display_cols * (example_width + pad));
% Copy each example into a patch on the display array
curr_ex = 1;
for j = 1:display_rows
for i = 1:display_cols
if curr_ex > m,
break;
end
% Copy the patch
% Get the max value of the patch
max_val = max(abs(X(curr_ex, :)));
display_array(pad + (j - 1) * (example_height + pad) +
(1:example_height), ...
pad + (i - 1) * (example_width + pad) +
(1:example_width)) = ...
reshape(X(curr_ex, :),
example_height, example_width) / max_val;
curr_ex = curr_ex + 1;
end
if curr_ex > m,
break;
end
end
% Display Image
h = imagesc(display_array, [-1 1]);
% Do not show axis
axis image off
drawnow;
end
ERROR:
displayData
parse error near line 86 of file C:\Users\ALI\displayData.m
syntax error
Pls guide which is the error in the script, this script is already written in
the coursera so its must be error free.

Answer 1

You seem to have modified the code, and moved the "ellipsis" operator (i.e. ...) or the line that is supposed to follow it, in several places compared to the original code in coursera.
Since the point of the ellipsis operator is to appear at the end of a line, denoting that the line that follows is meant to be a continuation of the line before, then moving either the ellipsis or the line below it will break the code.
E.g.
a = 1 + ... % correct use of ellipsis, code continues below
2 % treated as one line, i.e. a = 1 + 2
vs
a = 1 + % without ellipsis, the line is complete, and has an error
... 2 % bad use of ellipsis; also anything to the right of '...' is ignored
vs
a = 1 + ... % ellipsis used properly so far
% but the empty line here makes the whole 'line' `a = 1 +` which is wrong
2 % This is a new instruction