octave: using find() on cell array {} subscript and assigning it to another cell array -

this example in section 6.3.1 comma separated lists generated cell arrays of octave documentation (i browsed through doc command on octave prompt) don't quite understand.

in{1} = [10, 20, 30, 40, 50, 60, 70, 80, 90]; in{2} = inf; in{3} = "last"; in{4} = "first"; out = cell(4, 1); [out{1:3}] = find(in{1 : 3}); % line not understand 

so @ end of section, have in looking like:

in = {   [1,1] =      10 20 30 40 50 60 70 80 90   [1,2] = inf   [1,3] = last   [1,4] = first } 

and out looking like:

out = {   [1,1] =     1 1 1 1 1 1 1 1 1   [2,1] =     1 2 3 4 5 6 7 8 9   [3,1] =     10 20 30 40 50 60 70 80 90   [4,1] = [](0x0) } 

here, find called 3 output parameters (forgive me if i'm wrong on calling them output parameters, pretty new octave) [out{1:3}], represents first 3 empty cells of cell array out.

when run find(in{1 : 3}) 3 output parameters, in:

[i,j,k] = find(in{1 : 3}) 

i get:

i = 1  1  1  1  1  1  1  1  1 j = 1  2  3  4  5  6  7  8  9 k = 10 20 30 40 50 60 70 80 90 

which kind of explains why out looks does, when execute in{1:3}, get:

ans = 10 20 30 40 50 60 70 80 90 ans = inf ans = last 

which 1st 3rd elements of in cell array.

my question is: why find(in{1 : 3}) drop off 2nd , 3rd entries in comma separated list in{1 : 3}?

thank you.

the documentation find should answer question:

when called 3 output arguments, find returns row , column indices of non-zero elements (that's i , j) , vector containing non-zero values (that's k). explains 3 output arguments, not why considers in{1}. answer need @ happens when pass 3 input arguments find in find (x, n, direction):

if 3 inputs given, direction should 1 of "first" or "last", requesting first or last n indices, respectively. however, indices returned in ascending order.

so in{1} x (your data if want), in{2} how many indices find should consider (all of them in case since in{2} = inf) , {in3}is whether find should find first or last indices of vector in{1} (last in case).