Find indexes of equal consecutive numbers in vector

I have a row vector, for example: [0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1]

And I want to find the indexes of the values equal to 1.

So, in this case the outputs would be [7 8 9 10 11] and [19 20 21] because I would want this to be in separate outputs (instead of [7 8 9 10 11 19 20 21]). How can I do this automatically? Btw, my row vector has 1650 elements and several blocks of repetitions of 1's, but it should look very similar to this.

Edit: I have manage to do this using the following code, where "mask" is the name of my vector. However, I get a matrix with all the indexes, instead of separate matrices/outputs. Anyone can help me with that?

u = unique(mask(~isnan(mask)));
n = histc(mask,u);
d = u(n > 1);
out = find(ismember(mask,1));

Answer 1

[Ameer Hamza]

If you have image processing toolbox

x = [0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1];

comps = bwconncomp(x);
list = comps.PixelIdxList;

Result

>> list{1}
ans =
     7
     8
     9
    10
    11
>> list{2}
ans =
    19
    20
    21

Answer 2

[Ameer Hamza]

Without the image processing toolbox

x = [0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 1 1 1 0];

x_ = [diff(x) -1];
x1 = find(x_ == 1)+1;
x2 = find(x_ == -1, numel(x1));
list = arrayfun(@(x, y) {x:y}, x1, x2);

Result

>> list{1}
ans =
     7     8     9    10    11
>> list{2}
ans =
    19    20    21