Concatenate array using square brackets

x = zeros(1,9);
x = [1, x(1:9)];

when i run the second line multiple times, it starts to fill in the array x. I want to ask what is the functionality of the second line, and is there any documentation that describes this syntax ?

Answer 1

[Voss]

https://www.mathworks.com/help/matlab/ref/double.horzcat.html

[Quang Nguyen]

this seems like concatenating another array at the end of the first array, instead of replacing the zeros with the value 1 like the code i showed. Or is there any thing i misunderstand ?

[Voss]

% x is a 1-by-9 array (i.e., a vector) of 9 zeros:
x = zeros(1,9)

x = 1×9

0 0 0 0 0 0 0 0 0

% now x is a 1 followed by the first 9 elements of what was x (i.e., 9 zeros)
x = [1, x(1:9)]

x = 1×10

1 0 0 0 0 0 0 0 0 0

% now x is a 1 followed by the first 9 elements of what was x (i.e., a 1 and 8 zeros)
x = [1, x(1:9)]

x = 1×10

1 1 0 0 0 0 0 0 0 0

% now x is a 1 followed by the first 9 elements of what was x (i.e., two ones and 7 zeros)
x = [1, x(1:9)]

x = 1×10

1 1 1 0 0 0 0 0 0 0

% etc.

Perhaps it's useful to see the same code operating with a different starting vector x:

x = 1:9

x = 1×9

1 2 3 4 5 6 7 8 9

x = [1, x(1:9)]

x = 1×10

1 1 2 3 4 5 6 7 8 9

x = [1, x(1:9)]

x = 1×10

1 1 1 2 3 4 5 6 7 8

x = [1, x(1:9)]

x = 1×10

1 1 1 1 2 3 4 5 6 7

Considering that example, it may be more clear now, that the line x = [1, x(1:9)] does not really "replace" zeros with ones, but instead prepends a 1 to the beginning of x and removes anything after element #9, each time it is run.

[Quang Nguyen]

Thank you very much, i really get it now

[Voss]

You're welcome!