Unexpected behavior of anonymous function

The anonymous function k below behaves correcltly except for the last two cases k(1,1,:) and k(1,2,:), where it interprets the semicolon as a charcter (':'=58, 58^2=3364), while it should return the handle in the first case and error in the second. Any explanations?

>> k=@(varargin) cellfun(@(x) x^2,varargin)

k =

    @(varargin)cellfun(@(x)x^2,varargin)

>> k(:)

ans =

    @(varargin)cellfun(@(x)x^2,varargin)

>> k(1)

ans =

     1

>> k(1,:)

ans =

    @(varargin)cellfun(@(x)x^2,varargin)

>> k(1,2)

ans =

     1     4

>> k(1,1,:)

ans =

           1           1        3364

>> k(1,2,:)

ans =

           1           4        3364

[Cedric]

Could be a trick with subsref that doesn't manage subs well? The substruct that you receive in subsref contains the column as a character. It is also interesting to see that you can use "end": k(end) returns 1.

[Sean de Wolski]

@Cedric, apparently. I'm just puzzled by the discrepancy between the second and third dimension.

[Cedric]

@Sean: yes, it is as if when S.subs is larger than 2, subs are not treated the same way.. and it is not the position of ':' in the subs that matters:

 >> k(:,1,2)
 ans = 3364   1  4

Answer 1

[Steve Eddins]

First, I would like to point out that k(1,1,:) is a valid expression for a subscripting operation on a variable called k, but it is not a valid expression for a call to a function called k.

In other words, you can't pass a "naked" colon as a function argument!

>> sin(:)
Undefined variable sin.

So k(:), k(1,:), k(1,1,:), and k(1,2,:) are all invalid ways to call a function (or a function handle).

The fact that MATLAB isn't just giving a quick error on these expressions is an artifact of the way function handle evaluation syntax using parentheses has been implemented. It's been implemented by overloading the parentheses syntax using subsref. By the time the subsref overload sees things, the "naked" colon has been converted to a character. Like everything else inside the parentheses, this character then gets treated as a function argument.

[Khaled Hamed]

I do agree. It just seemed logical that k(:), k(1,:), k(1,1,:), and k(1,2,:) would be interpreted as indexing (the first three would work for a scalar, while the fourth would error).

Answer 2

[Sean de Wolski]

Please contact us and reference this thread.

That certainly looks like obscure behavior.