Re: Bug in Octave 8.2.0

Lacking a true `null` element in Octave, empty arrays have a special meaning. Essentially, they trigger a "deletion" of the specified LHS elements.

E.g.:

>> v = [1, 5, 8, 0];

>> v([1,3]) = [];

>> v

v =

5 0

Using `[]` or `''` is equivalent in that context.

In your second example, `v` has 0 elements. Trying to delete the first element results in an error. (The error message could be clearer though admittedly.)

For me that the error is slightly different:

>> v = {};

>> v(1)=''

error: A(I) = []: index out of bounds: value 1 out of bound 0

If you'd like to add a cell with the value `''` at the first position in the cell array `v`, you could use

v{1} = '';

or:

v(1) = {''};

Both variants should equivalently add a new cell element to the empty cell array v at the position 1 with the value `''`.

In your first example, the RHS is *not* an empty array. Thus, the deletion rules don't apply. Instead, the right hand side is converted to a 1x1 cell with the character '1' as the value of that single cell element.

I hope that helps.

Markus

Gesendet: Montag, 14. August 2023 um 15:25 Uhr

Von: "Ian McCallion" <[email protected]>

An: [email protected]

Betreff: Bug in Octave 8.2.0

v = {};

v(1)='1'

produces:

v =

{

[1,1] = 1

}

Which is right. However

v = {};

v(1)=''

produces:

error: A(I) = []: index out of bounds: value 2 out of bound 1

This is surely wrong. I should be able to initialise the cell array with empty strings???

Cheers... Ian

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